Answer:
2 hours
Explanation:
[tex]\text{Speed}=\frac{Dis\tan ce}{Time}[/tex]The first bicyclist rides at a speed of 8 mph. Therefore:
[tex]\begin{gathered} 8=\frac{d}{t} \\ \implies d=8t \end{gathered}[/tex]One hour later, the second bicyclist leaves and rides at a speed of 12 mph.
Therefore, the time of the second bicyclist = (t-1) hours.
Therefore:
[tex]\begin{gathered} 12=\frac{d}{t-1} \\ \implies d=12(t-1) \end{gathered}[/tex]Since the second bicyclist will catch up to the first bicyclist, the distance traveled will be the same.
So:
[tex]\begin{gathered} 8t=12(t-1) \\ 8t=12t-12 \\ 8t-12t=-12 \\ -4t=-12 \\ \frac{-4t}{-4}=\frac{-12}{-4} \\ t=3\text{ hours} \end{gathered}[/tex]Therefore, the second bicyclist will have traveled for:
(t-1) = (3-1) =2 hours.
State all integer values of X in the interval that satisfy the following inequality.
Solve the inequality
-5x - 5 < 8
for all integer values of x in the interval [-4,2]
We solve the inequality
Adding 5:
-5x - 5 +5 < 8 +5
Operating:
-5x < 13
We need to divide by -5, but we must be careful to flip the inequality sign. It must be done when multiplying or dividing by negative values
Dividing by -5 and flipping the sign:
x > -13 / 5
Or, equivalently:
x > -2.6
I am here, I'm correcting the answer. the interval was [-4,2] I misread the question. do you read me now?
Any number greater than -2.6 will solve the inequality, but we must use only those integers in the interval [-4,2]
Those possible integers are -4, -3, -2, -1, 0, 1, 2
The integers that are greater than -2.6 are
-2, -1, 0, 1, 2
This is the answer.
Okay so I’m doing this assignment and got stuck ont his question can someone help me out please
ANSWER
[tex]B.\text{ }\frac{256}{3}[/tex]EXPLANATION
We want to find the value of the function for F(4):
[tex]F(x)=\frac{1}{3}*4^x[/tex]To do this, substitute the value of x for 4 in the function and simplify:
[tex]\begin{gathered} F(4)=\frac{1}{3}*4^4 \\ F(4)=\frac{1}{3}*256 \\ F(4)=\frac{256}{3} \end{gathered}[/tex]Therefore, the answer is option B.
(0,1), (2,4), (4,7) (9.1)}Domain:Range:
The domain of an ordered pair are its first elements and its range are all the second elements of the ordered pair.
So, the domain ={0,2,4,9}
Range={1,4,7,1}
Put the following equation of a line into slope-intercept form, simplifying all fractions. 3x+9y=63
Answer: y = 63x - 180
Step-by-step explanation: y = mx + b ------(i)
Step one: y = 9, x = 3
9 = 63 (3) + b
9 = 189 + b
-180 = b
b = -180
y = 63x - 180
Answer is
y = -1/3x-6
Function f is defined by f(x) = 2x – 7 and g is defined by g(x) = 5*
Answer
f(3) = -1, f(2) = -3, f(1) = -5, f(0) = -7, f(-1) = -9
g(3) = 125, g(2) = 25, g(1) = 5, g(0) = 1, g(-1) = 0.2
Step-by-step explanation:
Given the following functions
f(x) =2x - 7
g(x) = 5^x
find f(3), f(2), f(1), f(0), and f(-1)
for the first function
f(x) = 2x - 7
f(3) means substitute x = 3 into the function
f(3) = 2(3) - 7
f(3) = 6 - 7
f(3) =-1
f(2), let x = 2
f(2) = 2(2) - 7
f(2) = 4 - 7
f(2) =-3
f(1) = 2(1) - 7
f(1) = 2 - 7
f(1) =-5
f(0) = 2(0) - 7
f(0) =0 - 7
f(0) = -7
f(-1) = 2(-1) - 7
f(-1) = -2 - 7
f(-1) = -9
g(x) = 5^x
find g(3), g(2), g(1), g(0), and g(-1)
g(3), substitute x = 3
g(3) = 5^3
g(3) = 5 x 5 x 5
g(3) = 125
g(2) = 5^2
g(2) = 5 x 5
g(2) = 25
g(1) = 5^1
g(1) = 5
g(0) = 5^0
any number raised to the power of zero = 1
g(0) = 1
g(-1) = 5^-1
g(-1) = 1/5
g(-1) = 0.2
Instructions: Factor 2x2 + 252 + 50. Rewrite the trinomial with the c-term expanded, using the two factors. Answer: 24 50
Given the polynomial:
[tex]undefined[/tex]What is the value of 12x if x = −5?
−60 −17 −125 −47
Answer:
-60
Step-by-step explanation:
3 2 — · — = _____ 8 5 2 9· — = _____ 3 7 8 — · — = _____ 8 7 x — · y = _____ y a b —— · — = _____ 2b c m n2 —- · —— = _____ 3n mGive the product in simplest form: 1 2 · 2— = _____ 2Give the product in simplest form: 1 2 — · 3 = _____ 4 Give the product in simplest form: 1 1 1— · 1— = _____ 2 2 Give the product in simplest form: 1 2 3— · 2— = _____ 4 3
Given:
[tex]\frac{3}{8}\cdot\frac{2}{5}[/tex]Required:
We need to multiply the given rational numbers.
Explanation:
Cancel out the common terms.
[tex]\frac{3}{8}\cdot\frac{2}{5}=\frac{3}{4}\cdot\frac{1}{5}[/tex][tex]Use\text{ }\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b\cdot d}.[/tex][tex]\frac{3}{8}\cdot\frac{2}{5}=\frac{3}{20}[/tex]Consider the number.
[tex]\frac{7}{8}\cdot\frac{8}{7}=\frac{1}{1}\cdot\frac{1}{1}[/tex]Cancel out the common multiples
[tex]9\cdot\frac{2}{3}[/tex][tex]9\cdot\frac{2}{3}=3\cdot2=6[/tex]Consider the number
[tex]\frac{7}{8}\cdot\frac{8}{7}[/tex]Cancel out the common multiples.
[tex]\frac{7}{8}\cdot\frac{8}{7}=\frac{1}{1}\cdot\frac{1}{1}[/tex][tex]\frac{7}{8}\cdot\frac{8}{7}=1[/tex]Consider the number
[tex]\frac{x}{y}\cdot y=x[/tex][tex]\frac{a}{2b}\cdot\frac{b}{c}=\frac{a}{2}\cdot\frac{1}{c}=\frac{a}{2c}[/tex][tex]\frac{m}{3n}\cdot\frac{n^2}{m}=\frac{1}{3}\cdot\frac{n}{m}=\frac{n}{3m}[/tex]Final answer:
[tex]\frac{3}{8}\cdot\frac{2}{5}=\frac{3}{20}[/tex][tex]9\cdot\frac{2}{3}=6[/tex][tex]\frac{7}{8}\cdot\frac{8}{7}=1[/tex][tex]\frac{x}{y}\cdot y=x[/tex][tex]\frac{a}{2b}\cdot\frac{b}{c}=\frac{a}{2c}[/tex][tex]\frac{m}{3n}\cdot\frac{n^2}{m}=\frac{n}{3m}[/tex]1. How much less is the area of a rectangular field 60 by 20
meters than that of a square field with the same perimeter?
The area of the rectangular field is 400m² less than the area of the square field.
How to find the area of a rectangle and square?A rectangle is a quadrilateral that has opposite sides equal to each other. Opposite side are also parallel to each other.
A square is a quadrilateral that has all sides equal to each other.
Therefore,
area of the rectangular field = lw
where
l = lengthw = widthTherefore,
area of the rectangular field = 60 × 20
area of the rectangular field = 1200 m²
The square field have the same perimeter with the rectangular field.
Hence,
perimeter of the rectangular field = 2(60 + 20)
perimeter of the rectangular field = 2(80)
perimeter of the rectangular field = 160 meters
Therefore,
perimeter of the square field = 4l
160 = 4l
l = 160 / 4
l = 40
Hence,
area of the square field = 40²
area of the square field = 1600 m²
Difference in area = 1600 - 1200
Difference in area = 400 m²
Therefore, the area of the square field is 400 metre square greater than the rectangular field.
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What is the standard form of the complex number that point A represents?
Answer
-3 + 4i
Explanation
The standard form for a complex number is given by:
[tex]\begin{gathered} Z=a+bi \\ \text{Where:} \\ a\text{ is the real part,} \\ b\text{ is the imaginary part} \end{gathered}[/tex]From the graph, the coordinates of A corresponding to the real axis and imaginary axis is traced in blue color in the graph below:
Hence, the standard form of the complex number that a represents is: -3 + 4i
which of the following describes the two spheres A congruentB similarC both congruent and similarD neither congruent nor similar
The two spheres are similar since they have a proportion of their radius. This proportion is 9/6 (3/2) or 6/9 (2/3).
They are not congruent. They do not have the same radius.
Therefore, the spheres are similar.
Janelle is conducting an experiment to determine whether a new medication is effective in reducing sneezing. She finds 1,000 volunteers with sneezing issues and divides them into two groups. The control group does not receive any medication; the treatment group receives the medication. The patients in the treatment group show reduced signs of sneezing. What can Janelle conclude from this experiment?
Answer:
Step-by-step explanation:
Hello! Is it possible to get help on this question?
To determine the graph that corresponds to the given inequality, first, let's write the inequality for y:
[tex]2x\le5y-3[/tex]Add 3 to both sides of the expression
[tex]\begin{gathered} 2x+3\le5y-3+3 \\ 2x+3\le5y \end{gathered}[/tex]Divide both sides by 5
[tex]\begin{gathered} \frac{2}{5}x+\frac{3}{5}\le\frac{5}{5}y \\ \frac{2}{5}x+\frac{3}{5}\le y \end{gathered}[/tex]The inequality is for the values of y greater than or equal to 2/5x+3/5, which means that in the graph the shaded area will be above the line determined by the equation.
Determine two points of the line to graph it:
-The y-intercept is (0,3/5)
- Use x=5 to determine a second point
[tex]\begin{gathered} \frac{2}{5}x+\frac{3}{5}\le y \\ \frac{2}{5}\cdot5+\frac{3}{5}\le y \\ 2+\frac{3}{5}\le y \\ \frac{13}{5}\le y \end{gathered}[/tex]The second point is (5,13/5)
Plot both points to graph the line. Then shade the area above the line.
The graph that corresponds to the given inequality is the second one.
Need help with this.. tutors have been a great help
Given the table in I which represents function I.
x y
0 5
1 10
2 15
3 20
4 25
• Graph II shows Item II which represents the second function.
Let's determine the increasing and decreasing function.
For Item I, we can see that as the values of x increase, the values of y also increase. Since one variable increases as the other increases, the function in item I is increasing.
For the graph which shows item II, as the values of x increase, the values of y decrease, Since one variable decreases as the other variable decreases, the function in item I is decreasing.
Therefore, the function in item I is increasing, and the function in item II is decreasing.
ANSWER:
A. The function in item I is increasing, and the function in item II is decreasing.
A cylinder truck all paint cans to be inches across the top diameter in about 10 inches high. How many cubic inches of pink it all to the nearest hundredth?
Given:
A cylinder truck all paint cans to be inches across the top diameter in about 10 inches high.
[tex]\begin{gathered} r=1.5in \\ h=10in \end{gathered}[/tex]Required:
To find the volume of the cylinder.
Explanation:
The volume of the cylinder is,
[tex]V=\pi r^2h[/tex]Therefore,
[tex]\begin{gathered} V=3.14\times1.5^2\times10 \\ \\ =3.14\times2.25\times10 \\ \\ =70.65in^3 \end{gathered}[/tex]Final Answer:
70.65 cubic inches of paint it hold.
the hypotenuse of a right triangle is 5 ft long. the shorter leg is 1 ft shorter than the longer leg. find the side lengths of the triangle
the hypotenuse of the right angle triangle is h = 5 ft
it is given that
the shorter leg is 1 ft shorter than the longer leg.
let the shorter leg is a and longer leg is b
the
b - a = 1
b = 1 + a
in the traingle using Pythagoras theorem,
[tex]a^2+b^2=h^2[/tex]put he values,
[tex]a^2+(1+a)^2=5^2[/tex][tex]\begin{gathered} a^2+1+a^2+2a=25 \\ 2a^2+2a-24=0 \\ a^2+a-12=0 \end{gathered}[/tex][tex]\begin{gathered} a^2+4a-3a-12=0_{} \\ a(a+4)-3(a+4)=0 \\ (a+4)(a-3)=0 \end{gathered}[/tex]a + 4 = 0
a = - 4
and
a - 3 = 0
a = 3
so, the longer leg is b = a + 1 = 3 + 1 = 4
thus, the answer is
shorter leg = 3 ft
longer length = 4 ft
hypotenuse = 5 ft
Elisa purchased a concert ticket on a website. The original price of the ticket was $95. She used a coupon code to receive a 10% discount. The website applied a 10% service fee to the discounted price. Elisa's ticket was less than the original by what percent?
The price of the ticket after the cupon is:
[tex]95\cdot0.9=85.5[/tex]To this price we have to add 10%, then:
[tex]85.5\cdot1.1=94.05[/tex]Hence the final cost of the ticket is $94.05.
To find out how less is this from the orginal price we use the rule of three:
[tex]\begin{gathered} 95\rightarrow100 \\ 94.05\rightarrow x \end{gathered}[/tex]then this represents:
[tex]x=\frac{94.05\cdot100}{95}=99[/tex]Therefore, Elisas's ticket was 1% less than the orginal price.
An outdoor equipment store surveyed 300 customers about their favorite outdoor activities. The circle graph below shows that 135 customers like fishing best, 75 customers like camping best, and 90 customers like hiking best.
it is given that,
total customer surveyed is 300 customers
also, it is given that,
135 customers like fishing best, 75 customers like camping best, and 90 customers like hiking best.
the total 300 customers representing the whole circle and circle has a complete angle of 360 degrees
so, 300 customers = 360 degrees,
1 customer = 360/300
= 6/5 degrees,
so, for fishing
135 customer = 135 x 6/5 degrees
= 27 x 6
= 162 degrees,
so, for camping
75 x 6/5 = 90 degrees,
for hiking
90 x 6/5 = 108 degrees,
Let f(x) = 8x^3 - 3x^2Then f(x) has a relative minimum atx=
1) To find the relative maxima of a function, we need to perform the first derivative test. It tells us whether the function has a local maximum, minimum r neither.
[tex]\begin{gathered} f^{\prime}(x)=\frac{d}{dx}\mleft(8x^3-3x^2\mright) \\ f^{\prime}(x)=\frac{d}{dx}\mleft(8x^3\mright)-\frac{d}{dx}\mleft(3x^2\mright) \\ f^{\prime}(x)=24x^2-6x \end{gathered}[/tex]2) Let's find the points equating the first derivative to zero and solving it for x:
[tex]\begin{gathered} 24x^2-6x=0 \\ x_{}=\frac{-\left(-6\right)\pm\:6}{2\cdot\:24},\Rightarrow x_1=\frac{1}{4},x_2=0 \\ f^{\prime}(x)>0 \\ 24x^2-6x>0 \\ \frac{24x^2}{6}-\frac{6x}{6}>\frac{0}{6} \\ 4x^2-x>0 \\ x\mleft(4x-1\mright)>0 \\ x<0\quad \mathrm{or}\quad \: x>\frac{1}{4} \\ f^{\prime}(x)<0 \\ 24x^2-6x<0 \\ 4x^2-x<0 \\ x\mleft(4x-1\mright)<0 \\ 0Now, we can write out the intervals, and combine them with the domain of this function since it is a polynomial one that has no discontinuities:[tex]\mathrm{Increasing}\colon-\infty\: 3) Finally, we need to plug the x-values we've just found into the original function to get their corresponding y-values:[tex]\begin{gathered} f(x)=8x^3-3x^2 \\ f(0)=8(0)^3-3(0)^2 \\ f(0)=0 \\ \mathrm{Maximum}\mleft(0,0\mright) \\ x=\frac{1}{4} \\ f(\frac{1}{4})=8\mleft(\frac{1}{4}\mright)^3-3\mleft(\frac{1}{4}\mright)^2 \\ \mathrm{Minimum}\mleft(\frac{1}{4},-\frac{1}{16}\mright) \end{gathered}[/tex]4) Finally, for the inflection points. We need to perform the 2nd derivative test:
[tex]\begin{gathered} f^{\doubleprime}(x)=\frac{d^2}{dx^2}\mleft(8x^3-3x^2\mright) \\ f\: ^{\prime\prime}\mleft(x\mright)=\frac{d}{dx}\mleft(24x^2-6x\mright) \\ f\: ^{\prime\prime}(x)=48x-6 \\ 48x-6=0 \\ 48x=6 \\ x=\frac{6}{48}=\frac{1}{8} \end{gathered}[/tex]Now, let's plug this x value into the original function to get the y-corresponding value:
[tex]\begin{gathered} f(x)=8x^3-3x^2 \\ f(\frac{1}{8})=8(\frac{1}{8})^3-3(\frac{1}{8})^2 \\ f(\frac{1}{8})=-\frac{1}{32} \\ Inflection\: Point\colon(\frac{1}{8},-\frac{1}{32}) \end{gathered}[/tex]Write the sequence {15, 31, 47, 63...} as a function A. A(n) = 16(n-1)B. A(n) = 15 + 16nC. A(n) = 15 + 16(n-1)D. 16n
To find the answer, we need to prove for every sequence as:
Answer A.
If n=1 then:
A(1) = 16(1-1) = 16*0 = 0
Since 0 is not in the sequence so, this is not the answer
Answer B.
If n=1 then:
A(1) = 15 + 16*1 = 31
Since 31 is not the first number of the sequence, this is not the answer
Answer D.
If n=1 then:
16n = 16*1 = 16
Since 16 is not in the sequence so, this is not the answer
Answer C.
If n = 1 then:
A(1) = 15 + 16(1-1) = 15
A(2) = 15 + 16(2-1) = 31
A(3) = 15 + 16(3-1) = 47
A(4) = 15 + 16(4-1) = 63
So, the answer is C
Answer: C. A(n) = 15 + 16(n-1)
Find the measure of the indicated angle to the nearest degree.A. 63B. 25C. 31D. 27
The point of the problem is to remember the cosine relation. It says, in this case, that
[tex]\cos (?)=\frac{\text{adjacent side}}{Hypotenuse}\Rightarrow\begin{cases}\text{adjacent side}=6 \\ \text{Hypotenuse}=13\end{cases}\Rightarrow\cos (?)=\frac{6}{13}[/tex]Converting the last equation by the inverse function, we get
[tex]?=\cos ^{-1}(\frac{6}{13})\approx62.5[/tex]For the first decimal place (5) equals 5, and by the rounding rule to the nearest degree, we get 63. The answer is A.
determine how many vertices and how many edges the graph has
in the given figure,
there are 4 vertices
and there are 3 edges.
thus, the answer is,
vertiev
Khalil has 2 1/2 hours to finish 3 assignments if he divides his time evenly , how many hours can he give to each
In order to determine the time Khalil can give to each assignment, just divide the total time 2 1/2 between 3 as follow:
Write the mixed number as a fraction:
[tex]2\frac{1}{2}=\frac{4+1}{2}=\frac{5}{2}[/tex]Next, divide the previous result by 3:
[tex]\frac{\frac{5}{2}}{\frac{3}{1}}=\frac{5\cdot1}{2\cdot3}=\frac{5}{6}[/tex]Hence, the time Khalil can give to each assignment is 5/6 of an hour.
Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 362 with 54 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.
We have to find the 80% confidence interval for a population proportion.
The sample size is n = 362 and the number of successes is X = 54.
Then, the sample proportion is p = 0.149171.
[tex]p=\frac{X}{n}=\frac{54}{362}\approx0.149171[/tex]The standard error of the proportion is:
[tex]\begin{gathered} \sigma_s=\sqrt{\frac{p(1-p)}{n}} \\ \sigma_s=\sqrt{\frac{0.149171*0.850829}{362}} \\ \sigma_s=\sqrt{0.000351} \\ \sigma_s=0.018724 \end{gathered}[/tex]The critical z-value for a 80% confidence interval is z = 1.281552.
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=p-z\cdot\sigma_s=0.149171-1.281552\cdot0.018724\approx0.1492-0.0240=0.1252[/tex][tex]UL=p+z\cdot\sigma_s=0.1492+0.0240=0.1732[/tex]As the we need to express it as a trilinear inequality, we can write the 80% confidence interval for the population proportion (π) as:
[tex]0.125<\pi<0.173[/tex]Answer: 0.125 < π < 0.173
Macky Pangan invested ₱2,500 at the end of every 3-month period for 5 years, at 8% interest compounded quarterly. How much is Macky’s investment worth after 5 years?
Compound interest with addition formula:
[tex]A=P(1+\frac{r}{n})^{nt}+\frac{PMT(1+\frac{r}{n})^{nt}-1}{\frac{r}{n}}[/tex]where,
A = final amount
P = initial principal balance
r = interest rate
n = number of times interest applied per time period
t = number of time periods elapsed
PMT = Regular contributions (additional money added to investment)
in this example
P = 2500
r = 8% = 0.08
n = 4
t = 5 years
PMT = 2500
[tex]A=2500(1+\frac{0.08}{4})^{4\cdot5}+\frac{2500\cdot(1+\frac{0.08}{4})^{4\cdot5}-1}{\frac{0.08}{4}}[/tex]solving for A:
[tex]A=189408.29[/tex]Therefore, his investment after 5 years will be
$189,408.29
A length of 48 ft. gave Malama an area
of 96 sq. ft. What other length would
give her the same area (96 sq. ft.)?
4
Write a division equation that represents the equation, How many 3/4 are in 10/9?
Given:
The number of 3/4 in 10/9.
To find the division equation that represents the given problem:
That is a number that is multiplied by 3/4 to obtain 10/9.
We need to find the number.
[tex]x\times\frac{3}{4}=\frac{10}{9}[/tex]Thus, the division equation will be,
[tex]x=\frac{10}{9}\div\frac{3}{4}[/tex]3. Identify the solution to the system of equations by graphing:(2x+3y=12y=1/3 x+1)
Given equations are
[tex]2x+3y=12[/tex][tex]y=\frac{1}{3}x+1[/tex]The graph of the equations is
Red line represents the equation 2x=3y=12 and the blue line represents the equation y=1/3 x=1.
f(x) = log 2(x+3) and g(x) = log 2(3x + 1).(a) Solve f(x) = 4. What point is on the graph of f?(b) Solve g(x) = 4. What point is on the graph of g?(c) Solve f(x) = g(x). Do the graphs off and g intersect? If so, where?(d) Solve (f+g)(x) = 7.(e) Solve (f-g)(x) = 3.
Given
[tex]\begin{gathered} f(x)=log_2(x+3) \\ and \\ g(x)=log_2(3x+1) \end{gathered}[/tex]a)
[tex]\begin{gathered} f(x)=4 \\ \Rightarrow log_2(x+3)=4 \\ \Leftrightarrow x+3=2^4 \\ \Rightarrow x+3=16 \\ \Rightarrow x=13 \end{gathered}[/tex]The answer to part a) is x=13. The point on the graph is (13,4)
b)
[tex]\begin{gathered} g(x)=4 \\ \Rightarrow log_2(3x+1)=4 \\ \Leftrightarrow3x+1=2^4 \\ \Rightarrow3x+1=16 \\ \Rightarrow3x=15 \\ \Rightarrow x=5 \end{gathered}[/tex]The answer to part b) is x=5, and the point on the graph is (5,4).
c)
[tex]\begin{gathered} f(x)=g(x) \\ \Rightarrow log_2(x+3)=log_2(3x+1) \\ \Rightarrow\frac{ln(x+3)}{ln(2)}=\frac{ln(3x+1)}{ln(2)}] \\ \Rightarrow ln(x+3)=ln(3x+1) \\ \Rightarrow x+3=3x+1 \\ \Rightarrow2x=2 \\ \Rightarrow x=1 \\ and \\ log_2(1+3)=log_2(4)=2 \end{gathered}[/tex]The answer to part c) is x=1 and graphs intersect at (1,2).
d)
[tex]\begin{gathered} (f+g)(x)=7 \\ \Rightarrow log_2(x+3)+log_2(3x+1)=7 \\ \Rightarrow log_2((x+3)(3x+1))=7 \\ \Leftrightarrow(x+3)(3x+1)=2^7 \\ \Rightarrow3x^2+10x+3=128 \\ \Rightarrow3x^2+10x-125=0 \end{gathered}[/tex]Solving the quadratic equation using the quadratic formula,
[tex]\begin{gathered} \Rightarrow x=\frac{-10\pm\sqrt{10^2-4*3*-125}}{3*2} \\ \Rightarrow x=-\frac{25}{3},5 \end{gathered}[/tex]However, notice that if x=-25/3,
[tex]log_2(x+3)=log_2(-\frac{25}{3}+3)=log_2(-\frac{16}{3})\rightarrow\text{ not a real number}[/tex]Therefore, x=-25/3 is not a valid answer.
The answer to part d) is x=5.
e)
[tex]\begin{gathered} log_2(x+3)-log_2(3x+1)=3 \\ log_2(\frac{x+3}{3x+1})=3 \\ \Leftrightarrow\frac{x+3}{3x+1}=2^3=8 \\ \Rightarrow x+3=24x+8 \\ \Rightarrow23x=-5 \\ \Rightarrow x=-\frac{5}{23} \end{gathered}[/tex]The answer to part e) is x=-5/23
The coordinates of three vertices of a rectangle are (3,7), (-3,5), and (0,-4). What are the coordinates of the fourth vertex?A. (6,-2)B. (-2,6)C. (6,2)D. (-2,-6)
ANSWER
A. (6, -2)
EXPLANATION
Let's graph these three vertices,
The fourth vertex must be at the same distance from (0, -4) as vertex (3, 7) is from (-3, 5),
Note that the horizontal distance between these two points is 6 units and the vertical distance is 2 units. The fourth vertex is,
[tex](0+6,-4+2)=(6,-2)[/tex]Hence, the fourth vertex is (6, -2)