probleme 1-2 show two Parallel lines and a transversal. Find the values of x

Probleme 1-2 Show Two Parallel Lines And A Transversal. Find The Values Of X

Answers

Answer 1

From the blurry picture shown, we can concur that:

x and 123.25 degree angle are interior corresponding angles.

They add up to 180 degrees, thus we can write the equation:

[tex]123.25\degree+x\degree=180\degree[/tex]

We can now easily solve for x:

[tex]\begin{gathered} 123.25\degree+x\degree=180\degree \\ x\degree=180-123.25 \\ x=56.75\degree \end{gathered}[/tex]

The solution:

[tex]x=56.75\degree[/tex]


Related Questions

Question
In a pet store, the small fishbowl holds up to 225 gallons of water. The large fishbowl holds up to 213 times as much water as the small fishbowl.

Eloise draws this model to represent the number of gallons of water the large fishbowl will hold.

How many gallons of water does the large fishbowl hold?

Answers

The number of gallons that the large fishbowl holds would be = 47,925 gallons.

What are fishbowls?

The fishbowls are containers that can be used to transport liquid substance such as water and food products such as fish. This can be measured in Liters, millilitres or in gallons.

The quantity of water the small fishbowl can take = 225 gallons.

The quantity of water the large fish bowl can take = 213(225 gallons)

That is, 213 × 225= 47,925 gallons.

Therefore, the quantity of water that the large fishbowl can hold is 47,925 gallons.

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2. The length of one side of the square is the square root ofits area. Use the table tofind the approximate length of one side of the square. Explain how you used thetable to find this information

Answers

we know that

the area of a square is equal to

A=b^2

where

b is the length side

Apply square root both sides of the formula we have

[tex]\sqrt{A}=b[/tex]

Determine the value for which the function f(u)= -9u+8/ -12u+11 in undefined

Answers

ANSWER

[tex]\frac{11}{12}[/tex]

EXPLANATION

A fraction becomes undefined when its denominator is equal to 0.

Hence, the given function will be undefined when:

[tex]-12u+11=0[/tex]

Solve for u:

[tex]\begin{gathered} -12u=-11 \\ u=\frac{-11}{-12} \\ u=\frac{11}{12} \end{gathered}[/tex]

That is the value of u for which the function is undefined.

(b) The area of a rectangular painting is 5568 cm².If the width of the painting is 58 cm, what is its length?Length of the painting:

Answers

Step 1: Problem

The area of a rectangular painting is 5568 cm².

If the width of the painting is 58 cm, what is its length?

Length of the painting:

Step 2: Concept

Area of a rectangle = Length x Width

Step 3: Method

Given data

Area = 5568 cm square

Width = 58 cm

Length = ?

Area of a rectangle = Length x Width

5568 = 58L

L = 5568/58

L = 96cm

Step 4: Final answer

Length of the painting = 96cm

Given a and b are the first-quadrant angles, sin a=5/13, and cos b=3/5, evaluate sin(a+b)1) -33/652) 33/653) 63/65

Answers

We know that angles a and b are in the first quadrant. We also know this values:

[tex]\begin{gathered} \sin a=\frac{5}{13} \\ \cos b=\frac{3}{5} \end{gathered}[/tex]

We have to find sin(a+b).

We can use the following identity:

[tex]\sin (a+b)=\sin a\cdot\cos b+\cos a\cdot\sin b[/tex]

For the second term, we can replace the factors with another identity:

[tex]\sin (a+b)=\sin a\cdot\cos b+\sqrt[]{1-\sin^2a}\cdot\sqrt[]{1-\cos^2b}[/tex]

Now we know all the terms from the right side of the equation and we can calculate:

[tex]\begin{gathered} \sin (a+b)=\sin a\cdot\cos b+\sqrt[]{1-\sin^2a}\cdot\sqrt[]{1-\cos^2b} \\ \sin (a+b)=\frac{5}{13}\cdot\frac{3}{5}+\sqrt[]{1-(\frac{5}{13})^2}\cdot\sqrt[]{1-(\frac{3}{5})^2} \\ \sin (a+b)=\frac{15}{65}+\sqrt[]{1-\frac{25}{169}}\cdot\sqrt[]{1-\frac{9}{25}} \\ \sin (a+b)=\frac{15}{65}+\sqrt[]{\frac{169-25}{169}}\cdot\sqrt[]{\frac{25-9}{25}} \\ \sin (a+b)=\frac{15}{65}+\sqrt[]{\frac{144}{169}}\cdot\sqrt[]{\frac{16}{25}} \\ \sin (a+b)=\frac{15}{65}+\frac{12}{13}\cdot\frac{4}{5} \\ \sin (a+b)=\frac{15}{65}+\frac{48}{65} \\ \sin (a+b)=\frac{63}{65} \end{gathered}[/tex]

Answer: sin(a+b) = 63/65

Need help with all of them please help me serious

Answers

we have 4,5,6

In a right triangle

c^2=a^2+b^2

where

c is the hypotenuse (greater side)

a and b are the legs

In an acte triangle

c^2 < a^2+b^2

we have

c=6

a=4

b=5

substitute

c^2=6^2=36

a^2=4^2=16

b^2=5^2=25

36 < 16+25

36 < 41

therefore

is an acute triangle

Part 2

10,24,26 and also classify the triangle

we have

c=26

a=10

b=24

so

c^2=676

a^2=100

b^2=576

in this problem

c^2=a^2+b^2

therefore

Is a right triangle

Use properties to rewrite the given equation. Which equations have the same solution as 2.3p – 10.1 = 6.5p – 4 – 0.01p? Select two options. 2.3p – 10.1 = 6.4p – 4 2.3p – 10.1 = 6.49p – 4 230p – 1010 = 650p – 400 – p 23p – 101 = 65p – 40 – p 2.3p – 14.1 = 6.4p – 4

Answers

The required equation has the same solution as 2.3p – 10.1 = 6.5p – 4 – 0.01p is 230p – 1010 = 650p – 400 – p.

What is an equivalent expression?

Equivalent expressions are even though they appear to be distinct, their expressions are the same. when the values are substituted into the expression, both expressions produce the same result and are referred to be equivalent expressions.

We have the given expression below:

⇒ 2.3p – 10.1 = 6.5p – 4 – 0.01p

Convert the decimal into a fraction to get

⇒ (23/10)p – (101/10) = (65/10)p – 4 – (1/100)p

⇒ (23p – 101)/10 = (650p – 400 – p) /100

⇒ 230p – 1010 = 650p – 400 – p

As a result, the equation that has the same answer as 230p – 1010 = 650p – 400 – p.

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Hello! Need some help on part c. The rubric, question, and formulas are linked. Thanks!

Answers

Explanation:

The rate of increase yearly is

[tex]\begin{gathered} r=69\% \\ r=\frac{69}{100}=0.69 \end{gathered}[/tex]

The number of lionfish in the first year is given beow as

[tex]N_0=9000[/tex]

Part A:

To figure out the explicit formula of the number of fish after n years will be represented using the formula below

[tex]P(n)=N_0(1+r)^n[/tex]

By substituting the formula, we will have

[tex]\begin{gathered} P(n)=N_{0}(1+r)^{n} \\ P(n)=9000(1+0.69)^n \\ P(n)=9000(1.69)^n \end{gathered}[/tex]

Hence,

The final answer is

[tex]f(n)=9,000(1.69)^n[/tex]

Part B:

to figure out the amoutn of lionfish after 6 years, we wwill substitute the value of n=6

[tex]\begin{gathered} P(n)=9,000(1.69)^{n} \\ f(6)=9000(1.69)^6 \\ f(6)=209,683 \end{gathered}[/tex]

Hence,

The final answer is

[tex]\Rightarrow209,683[/tex]

Part C:

To figure out the recursive equation of f(n), we will use the formula below

From the question the common difference is

[tex]d=-1400[/tex]

Hence,

The recursive formula will be

[tex]f(n)=f_{n-1}-1400,f_0=9000[/tex]

consider the graph of the function f(x)= 10^x what is the range of function g if g(x)= -f(x) -5 ?

Answers

SOLUTION

So, from the graph, we are looking for the range of

[tex]\begin{gathered} g(x)=-f(x)-5 \\ where\text{ } \\ f(x)=10^x \\ \end{gathered}[/tex]

The graph of g(x) is shown below

[tex]g(x)=-10^x-5[/tex]

The range is determined from the y-axis or the y-values. We can see that the y-values is from negative infinity and ends in -5. So the range is between

negative infinity to -5.

So we have

[tex]\begin{gathered} f(x)<-5\text{ or } \\ (-\infty,-5) \end{gathered}[/tex]

So, comparing this to the options given, we can see that

The answer is option B

in the figure below, RTS is an isosceles triangle with sides SR=RT, TVU is an equilateral triangle, WT is the bisected of angle STV, points S, T, and U are collinear, and c= 40 degrees.I'm completely lost and have to answer for a, b+c, f, b+f, a+d, e+g

Answers

Step 1: Concept

Triangle SRT is an isosceles triangle with equal base angles a = b

Triangle TUV is an equilateral triangle with all angles equal: g = d = h

Step 2: Apply sum of angles in a triangle theorem to find angle a and b.

[tex]\begin{gathered} a+b+c=180^o \\ c=40^o \\ \text{Let a = b = x} \\ \text{Therefore} \\ x\text{ + x + 40 = 180} \\ 2x\text{ = 180 - 40} \\ 2x\text{ = 140} \\ x\text{ = }\frac{140}{2} \\ x\text{ = 70} \\ a\text{ = 70 and b = 70} \end{gathered}[/tex]

Step 3:

2) a = 70

3) b + c = 70 + 40 = 110

Step 4:

Since WT is a bisector of angle STV,

Angle f = e = x

b + f + e = 180 sum of angles on a straight line.

b = 70

70 + x + x = 180

2x = 180 - 70

2x = 110

x = 110/2

x = 55

Hence, f = 55

4) f = 55

5) f + b = 55 + 70 = 125

Step 5:

Since triangle TUV is an equilateral triangle, angle g = h = d = 60

g = 60

h = 60

d = 60

6) Angle a + d = 70 + 60 = 130

7) e + g = 55 + 60 = 115

ok so the question is Write an expression to rubbers in the area of the figure the figure is a right triangle with 2X -2 and 4X plus 2 in the answer to that is 4X to the power of 2 - 2X -2 and that's part a and amp RP is what would the area be if X equals negative 2

Answers

ANSWERS

a) A = 4x² - 2x - 2

b) if x = -2, A = 18 units²

EXPLANATION

The area of a triangle is the length of the base, multiplied by its height and divided by 2:

[tex]A=\frac{b\cdot h}{2}[/tex]

In this triangle, b = 4x + 2 and h = 2x - 2. The area is:

[tex]A=\frac{(4x+2)(2x-2)}{2}[/tex]

We can simplify this expression. First we have to multiply the binomials in the numerator:

[tex]\begin{gathered} A=\frac{4x\cdot2x-4x\cdot2+2\cdot2x-2\cdot2}{2} \\ A=\frac{8x^2-8x+4x-4}{2} \\ A=\frac{8x^2-4x-4}{2} \end{gathered}[/tex]

Now, using the distributive property for the division:

[tex]\begin{gathered} A=\frac{8x^2}{2}-\frac{4x}{2}-\frac{4}{2} \\ A=4x^2-2x-2 \end{gathered}[/tex]

For part b, we just have to replace x with -2 in the expression above and solve:

[tex]\begin{gathered} A=4(-2)^2-2(-2)-2 \\ A=4\cdot4+4-2 \\ A=16+2 \\ A=18 \end{gathered}[/tex]

The one-to-one functions 9 and h are defined as follows.g={(0, 5), (2, 4), (4, 6), (5, 9), (9, 0)}h(x)X +811

Answers

Step 1: Write out the functions

g(x) = { (0.5), (2, 4), (4,6), (5,9), (9,0) }

[tex]h(x)\text{ = }\frac{x\text{ + 8}}{11}[/tex]

Step 2:

For the function g(x),

The inputs variables are: 0 , 2, 4, 5, 9

The outputs variables are: 5, 4, 6, 9, 0

The inverse of an output is its input value.

Therefore,

[tex]g^{-1}(9)\text{ = 5}[/tex]

Step 3: find the inverse of h(x)

To find the inverse of h(x), let y = h(x)

[tex]\begin{gathered} h(x)\text{ = }\frac{x\text{ + 8}}{11} \\ y\text{ = }\frac{x\text{ + 8}}{11} \\ \text{Cross multiply} \\ 11y\text{ = x + 8} \\ \text{Make x subject of formula} \\ 11y\text{ - 8 = x} \\ \text{Therefore, h}^{-1}(x)\text{ = 11x - 8} \\ h^{-1}(x)\text{ = 11x - 8} \end{gathered}[/tex]

Step 4:

[tex]Find(h.h^{-1})(1)[/tex][tex]\begin{gathered} h(x)\text{ = }\frac{x\text{ + 8}}{11} \\ h^{-1}(x)\text{ = 11x - 8} \\ \text{Next, substitute h(x) inverse into h(x).} \\ \text{Therefore} \\ (h.h^{-1})\text{ = }\frac{11x\text{ - 8 + 8}}{11} \\ h.h^{-1}(x)\text{ = x} \\ h.h^{-1}(1)\text{ = 1} \end{gathered}[/tex]

Step 5: Final answer

[tex]\begin{gathered} g^{-1}(9)\text{ = 5} \\ h^{-1}(x)\text{ = 11x - 8} \\ h\lbrack h^{-1}(x)\rbrack\text{ = 1} \end{gathered}[/tex]

please show and explain this please

Answers

only answer ;( c.

Step-by-step explanation:

so hope it help

There are four roots: -2, 1, 1, and 3 [1 is repeated as it is a local maximum]
f(x) = (x + 2)(x - 3)(x - 1)(x - 1)
so the correct answer is C

The table shows the number of hours spent practicingsinging each week in three samples of 10 randomlyselected chorus members.Time spent practicing singing each week (hours)Sample 1 45873 56 579 Mean = 5.9Sample 2 68 74 5 4 8 4 5 7 Mean = 5.8Sample 3 8 4 6 5 6 4 7 5 93 Mean = 5.7Which statement is most accurate based on the data?O A. A prediction based on the data is not completely reliable, becausethe means are not the same.B. A prediction based on the data is reliable, because the means ofthe samples are close together.O C. A prediction based on the data is reliable, because each samplehas 10 data points.D. A prediction based on the data is not completely reliable, becausethe means are too close together.

Answers

The means of three samples are close together. Therefore, option B is the correct answer.

In the given table 3 sample means are given.

What is mean?

In statistics, the mean refers to the average of a set of values. The mean can be computed in a number of ways, including the simple arithmetic mean (add up the numbers and divide the total by the number of observations).

Here, mean of sample 1 is 5.9, mean of sample 2 is 5.8 and mean of sample 3 is 5.7.

Thus, means of these three samples are close together.

The means of three samples are close together. Therefore, option B is the correct answer.

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Translate the sentence into an equation.Eight more than the quotient of a number and 3 is equal to 4.Use the variable w for the unknown number.

Answers

We are to translate into an equation

Eight more than the quotient of a number and 3 is equal to 4.

Let the number be w

Hence, quotient of w and 3 is

[tex]\frac{w}{3}[/tex]

Therefore, eight more than the quotient of a number and 3 is equal to 4

Is given as

[tex]\frac{w}{3}+8=4[/tex]

Solving for w

we have

[tex]\begin{gathered} \frac{w}{3}=4-8 \\ \frac{w}{3}=-4 \\ w=-12 \end{gathered}[/tex]

Therefpore, the equation is

[tex]\frac{w}{3}+8=4[/tex]

what is the equation of the line with x-intercept (6,0) and y-intercept (0, 2)

Answers

Answer:

3y=6-x

Explanation:

The slope-intercept form of a line is y=mx+b.

First, we determine the slope(m) of the line.

[tex]\begin{gathered} m=\frac{2-0}{0-6} \\ =-\frac{2}{6} \\ m=-\frac{1}{3} \end{gathered}[/tex]

Since the y-intercept, b=2

The equation of the line is:

[tex]\begin{gathered} y=-\frac{1}{3}x+2 \\ y=\frac{-x+6}{3} \\ 3y=6-x \end{gathered}[/tex]

Which of the following functions has an amplitude of 3 and a phase shift of pi over 2 question mark

Answers

Remember that f(x) = A f(Bx-C) +D

Where |A| is the Amplitude and C/B is the phase Shift

Options

A, B C all have amplitudes of |3| so we have just eliminated D with the amplitude

We need a phase shift of C/B = pi/2

A has Pi/2

B has -Pi/2

C has pi/2 /2 = pi/4

Choice A -3 cos ( 2x-pi) +4 has a magnitude of 3 and and phase shift of pi/2

Use the graph below to answer the following questionsnegative sine graph with local maxima at about (-3,55) and local minima at (3,55)1. Estimate the intervals where the function is increasing.2. Estimate the intervals where the function is decreasing.3. Estimate the local extrema.4. Estimate the domain and range of this graph.

Answers

Answer:

1. Increasing on ( -inf, -3] and ( 3, inf)

2. decreasing on (-3, 3]

3. Local maximum: 60, Local minimum: -60

4. Domain: (-inf , inf)

Range: [-60, 60]

Explanation:

1.

A function is increasing when its slope is positive. Now, in our case we can see that the slope of f(x) is postive from - infinity to -3 and then it is negatvie from -3 to 3; it again increasing from 3 to infinity.

Therefore, we c

At noon a private plane left Austin for Los Angeles, 2100 km away, flying at 500 km/h. One hour later a jet left Los Angeles for Austin at 700 km/h. At what time did they pass each other?

Answers

So if you take the L and do the w it should work

the points (-4,-2) and (8,r) lie on a line with slope 1/4 . Find the missing coordinate r.

Answers

The points (-4, -2) and (8, r) are located on a line of slope 1/4, We are asked to find the value of "r" that would make suche possible.

So we recall the definition of the slope of the segment that joins two points on the plane as:

slope = (y2 - y1) / (x2 - x1)

in our case:

1/4 = ( r - -2) / (8 - -4)

1/4 = (r + 2) / (8 + 4)

1/4 = (r + 2) / 12

multiply by 12 both sides to cancel all denominators:

12 / 4 = r + 2

operate the division on the left:

3 = r + 2

subtract 2 from both sides to isolate "r":

3 - 2 = r

Then r = 1

A 9-foot roll of waxed paper costs $4.95. What is the price per yard ?

Answers

the answer is 1.65 dollars per yard

Answer:

$0.55 per yard

Step-by-step explanation:

a 9 foot roll is 4.95 so you divide the cost by the amount to get the unit rate which is $0.55 per yard

the sum of the measure of angle m and angle r is 90

Answers

Given:

The sum of measure of angle m and r is 90 degrees.

If (2 +3i)^2 + (2 - 3i)^2 = a + bia =b=

Answers

(2 + 3i)^2 = 4 + 12i + 9(-1)

= 4 + 12i - 9

= -5 + 12i

(2 - 3i)^2 = 4 - 12i - 9(-1)

= 4 - 12i + 9

= 13 - 12i

REsult

= -5 + 12i + 13 - 12i

= 8 - 0i

Then

a = 8 and b = 0

Simplify the following expression 6 + (7² - 1) + 12 ÷ 3

Answers

You have to simplify the following expression

[tex]6+(7^2-1)+12\div3[/tex]

To solve this calculation you have to keep in mind the order of operations, which is:

1st: Parentheses

2nd: Exponents

3rd: Division/Multiplication

4th: Addition/Subtraction

1) The first step is to solve the calculation within the parentheses

[tex](7^2-1)[/tex]

To solve it you have to follows the order of operations first, which means you have to solve the exponent first and then the subtraction:

[tex]7^2-1=49-1=48[/tex]

So the whole expression with the parentheses calculated is:

[tex]6+48+12\div3[/tex]

2) The second step is to solve the division:

[tex]12\div3=4[/tex]

Now the expression is

[tex]6+48+4[/tex]

3) Third step is to add the three values:

[tex]6+48+4=58[/tex]

E is the midpoint of DF, DE = 2x + 4 and EF = 3x - 1 how do I find the value of x, DE, EF and DF

Answers

We know that

E is midpoint of DF, that means DE is equal to EF, so we can form the following equation

[tex]DE=EF[/tex]

Replacing the given equations, we have

[tex]\begin{gathered} 2x+4=3x-1 \\ 4+1=3x-2x \\ x=5 \end{gathered}[/tex]

Now, we replace this value in each equation to find each part of the segment.

[tex]\begin{gathered} DE=2x+4=2(5)+4=10+4=14 \\ EF=3x-1=3(5)-1=15-1=14 \end{gathered}[/tex]Therefore, each part of the segment is 14 units, and DF is 28.

I need this practice problem answered I will provide the answer options in another pic

Answers

The inverse of a matrix can be calculated as:

[tex]\begin{gathered} \text{When} \\ A=\begin{bmatrix}{a} & {b} & {} \\ {c} & {d} & {} \\ {} & {} & \end{bmatrix} \\ \text{Then A\textasciicircum-1 is:} \\ A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}{d} & -{b} & {} \\ {-c} & {a} & {} \\ {} & {} & \end{bmatrix} \end{gathered}[/tex]

Then, let's start by calculating the inverse of the given matrix:

[tex]\begin{gathered} \begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}^{-1}=\frac{1}{4\cdot3-1\cdot(-2)}\begin{bmatrix}{3} & -{1} & {} \\ {-(-2)} & {4} & {} \\ {} & {} & \end{bmatrix} \\ \begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}^{-1}=\frac{1}{14}\begin{bmatrix}{3} & -{1} & {} \\ {2} & {4} & {} \\ {} & {} & \end{bmatrix} \end{gathered}[/tex]

The problem says he multiplies the left side of the coefficient matrix by the inverse matrix, thus:

[tex]\begin{gathered} \begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}^{-1}\begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}\cdot\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}^{-1}\begin{bmatrix}{2} & {} & {} \\ {-22} & {} & {} \\ {} & {} & {}\end{bmatrix} \\ \end{gathered}[/tex]

*These matrices will be the options to put on the first and second boxes.

Then:

[tex]\begin{gathered} \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\frac{1}{14}\begin{bmatrix}{3} & -{1} & {} \\ {2} & {4} & {} \\ {} & {} & \end{bmatrix}\cdot\begin{bmatrix}{2} & {} & {} \\ {-22} & {} & {} \\ {} & {} & {}\end{bmatrix}\text{ This is for the third box} \\ \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\frac{1}{14}\begin{bmatrix}{3\times2+(-1)\times(-22)} & & {} \\ {2\times2+4\times(-22)} & & {} \\ {} & {} & \end{bmatrix}=\frac{1}{14}\begin{bmatrix}{28} & & {} \\ {-84} & & {} \\ {} & {} & \end{bmatrix}\text{ This is the 4th box} \\ \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{28/14} & & {} \\ {-84/14} & & {} \\ {} & {} & \end{bmatrix}=\begin{bmatrix}{2} & & {} \\ {-6} & & {} \\ {} & {} & \end{bmatrix}\text{ And finally this is the last box} \end{gathered}[/tex]

For which pair of triangles would you use ASA to prove the congruence of the two triangles?

Answers

Solution:

Remember that the Angle-Side-Angle Postulate (ASA) states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent. According to this, the correct answer is:

C.

A special deck of cards has 4 blue cards, and 4 red cards. The blue cards are numbered 1, 2, 3, and 4. The red cards are numbered 1, 2, 3, and 4. The cards are well shuffled and you randomly draw one card.A = card drawn is blueB = card drawn is odd-numbereda) How many elements are there in the sample space? b) P(A) = c) P(B) =

Answers

Answer

• a) 8

,

• b) 4/8

,

• c) 4/8

Explanation

Given

• Blue cards: 4, {B1, B2, B3, B4}

,

• Red cards: 4 {R1, R2, R3, R4}

,

• A = card drawn is blue

• B = card drawn is odd-numbered {B1, R1, B3, R3}

Procedure

• a) elements in the sample space

There are: S = {B1, B2, B3, B4, R1, R2, R3, R4}

Thus, the number of elements in the sample space is n(S) = 8.

• b) P(A)

Can be calculated as follows:

[tex]P(A)=\frac{n(A)}{n(S)}=\frac{4}{8}[/tex]

• c) P(B)

Can be calculated as follows:

[tex]P(B)=\frac{n(B)}{n(S)}=\frac{4}{8}[/tex]

In a robotics competition, all robots must be at least 37 inches tall to enter the competition.Read the problem. Which description best represents the heights a robot must be?Any value less than or equal to 37Any value greater than or equal to 37Any value greater than 37Any value less than 37

Answers

Solution

Since the robots must be at least 37 inches tall to enter the competition.

Therefore, the height of any robot must be Any value greater than or equal to 37

Hello, may I have help with finding the maximum or minimum of this quadratic equation? Could I also know the domain and range and the vertex of the equation?

Answers

To solve this problem, we will use the following graph as reference:

From the above graph, we get that the quadratic equation represents a vertical parabola that opens downwards with vertex:

[tex](3,5)\text{.}[/tex]

The domain of the function consists of all real numbers, and the range consists of all numbers smaller or equal to 5.

Answer:

Maximum of 5, at x=3.

Vertex (3,5).

Domain:

[tex](-\infty,\infty).[/tex]

Range:

[tex](-\infty,5\rbrack.[/tex]

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