From a hot-air balloon, Guadalupe measures a 31^{\circ} ∘ angle of depression to a landmark that’s 316 feet away, measuring horizontally. What’s the balloon’s vertical distance above the ground?

Given that f(x) = ax – 5 and f(4) = 15, then the value of a is 5

The given function is:

f(x)  =  ax  -  5

If f(4)  =  15, then x = 4

Substitute x = 4 into f(x) = ax - 5

f(4)  =  a(4)  -  5

15   =  4a   -  5

Solve for a by adding 5 to both sides

15  +  5  =  4a  -  5  +  5

20    =  4a

Divide both sides by 4

4a/4   =  20/4

a    =  5

Given that f(x) = ax – 5 and f(4) = 15, then the value of a is 5

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Based on the information, the best research question that Juan should choose will be "How does playing a team sport affect a child's development?"

A research question simply means a question that a researcher wants to answer. It should be noted that choosing a research question is an important element in research.

From the complete question, it should be noted that since Juan is writing a paper about children who play team sports, and he has different research questions, he should choose the one that conveys the research appropriately.

In this case, the best option will be  "How does playing a team sport affect a child's development?"

Learn more about research questions on:

https://brainly.com/question/25257437

Answer: the answer is 7.5

Step-by-step explanation:

The cost function is an illustration of a linear function

The equation of the cost function is C(x) = 35 + 0.5x

The given parameters are:

[tex]Rentals = 35[/tex] -- the cost of renting a cooler

[tex]Rate = 0.5[/tex] --- the rate per bottle

Assume the number of bottles in a cooler is x.

The cost function for the bottles in a cooler would be:

[tex]C(x) = Rentals + Rate \times x[/tex]

So, we have:

[tex]C(x) = 35 + 0.5\times x[/tex]

Evaluate the products

[tex]C(x) = 35 + 0.5x[/tex]

Hence, the equation is [tex]C(x) = 35 + 0.5x[/tex]

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There are 2 blue marbles out of 6 total marbles.

6-2 = 4 marbles are not blue.

Probability of not being blue = 4/6 which reduces to 2/3

Answer: 2/3

Answer:

-2/15

this should work

Step-by-step explanation:

Answer:

13/15

Step-by-step explanation:

1 1/5 - 1/3 =

= 1 + 1/5 - 1/3

= 5/5 + 1/5 - 1/3

= 6/5 - 1/3

= 18/15 - 5/15

= 13/15

Answer:

C

Step-by-step explanation:

Slope is rise over run. Pick two points and then count the rise and run. The rise in this case is one and the run is 2 and one over 2 is 1/2

[tex]Hiya![/tex]

Sokka is here to help!

[tex]C.[/tex]

Because, All you need to do is.. Find the slope of proportional graph, by y and x, Which the line is [tex]\frac{4}{2}[/tex].

[tex]And, \frac{4}{2} =\frac{1}{2} .[/tex]

Hopefully, this helps you!!

[tex]Sokka[/tex]

Answer:

speed in still water is v=40 speed of current

Answer:

36   and -1/8

Step-by-step explanation:

x^2 = 3^2 = 9 * 4 = 36

y^-3 = (-2)^-3 = -1/8

Answer:

-4.5.

Step-by-step explanation:

4x^2 y^-3

= 4x^2 / y^3

When x = 3 and y = -2

it is 4(3)^2 / (-2)^3

= 36 / -8

= -4.5.

Answer:

200% increase

Step-by-step explanation:

Answer:

this might help

Step-by-step explanation:

this might be kinda similar

Chickens and rabbits

In the yard were chickens and rabbits. Together they had 18 heads and 56 legs. How many chickens and how many rabbits were in the yard?

Correct answer:

chickens:  8

rabbits:  10

Step-by-step explanation:

a+b=18

2a+4b=56

a=8

b=10

a+b = 18

2•a+4•b=56

a+b = 18

2a+4b = 56

a = 8

b = 10

The vertices of the feasible region of the constraints is (0,2.5)

The constraints are given as:

x + 3y ≤ 6

4x + 6y ≥ 9

x ≥ 0,  y ≥ 0

Express the inequalities as equations

x + 3y = 6

4x + 6y = 9

Make x the subject in x + 3y = 6

x = 6 - 3y

Substitute x = 6 - 3y in 4x + 6y = 9

4(6 - 3y) + 6y = 9

Expand

24 - 12y + 6y = 9

Evaluate the like terms

-6y = -15

Divide by 6

y = 2.5

Substitute y = 2.5 in x = 6 - 3y

x = 6 - 3 * 2.5

Evaluate

x = -1.5

So, we have:

(x,y) = (-1.5, 2.5)

Recall that: x ≥ 0,  y ≥ 0

This means that the feasible region must have positive coordinates or zero

So, we set x = 0

(x,y) = (0, 2.5)

Hence, the vertices of the feasible region of the constraints is (0,2.5)

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#SPJ1

Answer:

the answer you're looking for is D. when you move to the left the -4 becomes -6 and when you move up 2 becomes 4.

nCx⋅px⋅(1−p)n−x

llllllllllllllllllllllllllllllllllllll

Answer:

4,-1

Step-by-step explanation:

Please like!♥

Answer:

Step-by-step explanation:

imagine you chop off a right triangle from the left side

now you have a rectangle and a right triangle

the rectangle

height 5

width 12

5 x 12 = 60

triangle

height 5

width 3

3 x 5 = 15

15 / 2 = 7.5

60 + 7.5 = 67.5 m^2

Answer: 67.5 m^2

Step-by-step explanation:

To find out the area of the yard, we split the shape up into a triangle and rectangle. The area of the triangle is,

A = 1 × b × h/2 = 1 × 3 × 5/2 = 7.5 m^2.

And the area of the rectangle is,

A = 12 × 5 = 60 m^2.

We then add the area of the triangle and the area of the rectangle.

A = 7.5 + 60 =67.5 m^2

Rounded to one decimal place, the area of the yard is 67.5 square meters.

Answer:

262

Step-by-step explanation:

4x4x4x4+6

256+6

262

Answer:

15+20=35+15 And then add an additional plus 15 for each hour

Step-by-step explanation:

I'll do the first two parts to get you started.

=====================================================

Part (i)

We can show that the operation [tex]\nabla[/tex] is not commutative by picking two random and different real numbers for a,b. In other words, I'm using a counter-example.

Let's say we go for a = 1 and b = 2

[tex]a \nabla b = \frac{3a+b}{5} - 1\\\\1 \nabla 2 = \frac{3*1+2}{5} - 1\\\\1 \nabla 2 = \frac{5}{5} - 1\\\\1 \nabla 2 = 1 - 1\\\\1 \nabla 2 = 0\\\\[/tex]

Now let's swap the values. We'll try a = 2 and b = 1

[tex]a \nabla b = \frac{3a+b}{5} - 1\\\\2 \nabla 1 = \frac{3*2+1}{5} - 1\\\\2 \nabla 1 = \frac{7}{5} - 1\\\\2 \nabla 1 = 1.4 - 1\\\\2 \nabla 1 = 0.4\\\\[/tex]

We can see that [tex]1 \nabla 2[/tex] and [tex]2 \nabla 1[/tex] are not the same value. In general, [tex]a \nabla b \ne b \nabla a[/tex] Therefore, the operation [tex]\nabla[/tex] is not commutative.

The only time [tex]a \nabla b = b \nabla a\\\\[/tex] is true is when [tex]a = b[/tex], since,

[tex]\frac{3a+b}{5}-1 = \frac{3b+a}{5}-1\\\\\frac{3a+a}{5}-1 = \frac{3a+a}{5}-1\\\\\frac{4a}{5}-1 = \frac{4a}{5}-1\\\\[/tex]

It's when [tex]a \ne b[/tex] is where the operation becomes noncommutative.

Another way to arrive at the [tex]a = b[/tex] condition is to solve the original equation for either 'a' or b.

=====================================================

Part (ii)

Using the previous part as inspiration, we'll do a counter-example to show that the operation is not associative. Pick 3 random values for a,b,c. Here are the values I'll pick.

Then,

[tex]d = b \nabla c = \frac{3b+c}{5}-1 = \frac{3*2+3}{5}-1 = 0.8\\\\a \nabla (b \nabla c) = a \nabla d = \frac{3a+d}{5}-1 = \frac{3*1+0.8}{5}-1 = -0.24\\[/tex]

Next, we can say:

[tex]d = a \nabla b = \frac{3a+b}{5}-1 = \frac{3*1+2}{5}-1 = 0\\\\(a \nabla b) \nabla c = d \ \nabla c = \frac{3d+c}{5}-1 = \frac{3*0+3}{5}-1 = -0.4[/tex]

Let's compare the two outputs:

[tex]a \nabla (b \nabla c) = -0.24\\\\(a \nabla b) \nabla c = -0.4[/tex]

They don't match up, so [tex]a \nabla (b \nabla c) \ne (a \nabla b) \nabla c[/tex] when (a,b,c) = (1,2,3).

In general, the operation is not associative in R.

Answer:

i think BC = 8.5

Step-by-step explanation:

19-(2+6.5+2)

19-10.5

8.5

Answer:

the number 0 Originally called finding the vertical asymptote.

Answer:

616cm³

Step-by-step explanation:

you just need to multiply those numbers

7 x 8 x 11= 616

[tex]\text{Perimeter of the triangle,}\\\\ 2s = 7 + 8 +11 =26}\\\\\\\implies s = \dfrac{26}2 = 13\\\\\\\text{Apply Heron's formula to find the area of the triangle,}\\\\A = \sqrt{s(s-8)(s-7)(s-11)}\\\\\implies A = \sqrt{13(13-8)(13-7)(13-11)}\\\\\implies A = \sqrt{780} \\\\\implies A = 27.93~~ \text{cm}^2[/tex]

Answer:

Yes, it is a function

Step-by-step explanation:

Each x value in the domain has its own output, and it does not have more than 1

Answer:

P = -3, each line in the negative direction is 1 number

so you would have -1, -2, -3 , -4, etc.

P is on the -3 line.

See attached picture.

Step-by-step explanation:

Using the t-distribution, it is found that since the test statistic is more than the critical value for the right-tailed test, it can be conclude that the isolated compound gives more counts than those from background, and hence, we can be 95% confident that the compound is indeed radioactive.

At the null hypothesis, it is tested if isolated compounds do not give more counts than those from background, that is:

[tex]H_0: \mu_1 - \mu_2 \leq 0[/tex]

At the alternative hypothesis, it is tested if the give more counts, that is:

[tex]H_1: \mu_1 - \mu_2 > 0[/tex]

Using a calculator, the mean and the standard deviation for each sample are given by:

[tex]\mu_1 = 33.2, s_1 = 6.058[/tex]

[tex]\mu_2 = 24.25, s_2 = 4.35[/tex]

Considering the sample sizes of 5 and 4, respectively, the standard errors are given by:

[tex]s_1 = \frac{6.058}{\sqrt{5}} = 2.71[/tex]

[tex]s_2 = \frac{4.35}{\sqrt{4}} = 2.175[/tex]

For the distribution of differences, the mean and standard error are given by:

[tex]\overline{x} = \mu_1 - \mu_2 = 33.2 - 24.25 = 8.95[/tex]

[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{2.71^2 + 2.175^2} = 3.4749[/tex]

The standard error was found from the standard deviation for each sample, hence, the t-distribution is used.

The test statistic is given by:

[tex]t = \frac{\overline{x} - \mu}{s}[/tex]

In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis, hence:

[tex]t = \frac{\overline{x} - \mu}{s}[/tex]

[tex]t = \frac{8.95 - 0}{3.4749}[/tex]

[tex]t = 2.58[/tex]

The critical value for a right-tailed test, as we are testing if the mean is greater than a value, with 5 + 4 - 2 = 7 df and a 0.05 significance level is of [tex]t^{\ast} = 1.895[/tex]

Since the test statistic is more than the critical value for the right-tailed test, it can be conclude that the isolated compound gives more counts than those from background, and hence, we can be 95% confident that the compound is indeed radioactive.

A similar problem is given at brainly.com/question/24826023

Answer:

2=-12.2/(-6.1

-2=16/(-8)

-3= -2 2/5/4/5

3= 3 3/7 / 1 1/7

Step-by-step explanation:

Explanation :

________________________________

Given:

(2+√3)+(4-√3)

To Find:

Evaluate the expression

Solution:

⇨Step 1:

Remove the parentheses

(2+√3)+(4-√3)

When there is + or no sign in front of an expression in parentheses, the expression remains the same

= 2 + √3+ 4-√3

⇨Step 2:

Cancel the opposite terms √3 and -√3

"Two numbers are opposites if they have the same absolute value but different signs"

= 2 + 4

⇨Step 3:

Add the numbers

= 6

Hence (2+√3)+(4-√3) = 6.

Answer:

  x = 15

Step-by-step explanation:

Corresponding sides have the same ratio:

  UV/PR = TV/QR

  (x +6)/14 = (x -3)/8

  4(x +6) = 7(x -3) . . . . . . multiply by 56

  4x +24 = 7x -21 . . . . . . eliminate parentheses

  45 = 3x . . . . . . . . . . . add 21-4x

  15 = x . . . . . . . . . . . .divide by 3

_____

Alternate solution

The long-side : short-side ratios for the two triangles are ...

  14 : 8 = (x +6) : (x -3)

If we look at the differences between the ratio numbers we see ...

  14 -8 = 6

  (x +6) -(x -3) = 9

That is, the numbers in the second ratio must be 9/6 = 3/2 times the numbers in the first ratio. In other words, ...

  x -3 = (3/2)(8) = 12

  x = 15

Check: x +6 = 3/2(14) ; 15 +6 = 21

Answer:

A)62x-8

B)11x+2

C)11x-8

D)62x

Step-by-step explanation:

12×3=36

16×2=32

36+32-6=62x-8

Answer:

[tex]x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}[/tex] and [tex]y=-1,\frac{1}{2}[/tex]

The ordered pair solutions are [tex](-\sqrt{\frac{7}{4}},0.5)[/tex], [tex](\sqrt{\frac{7}{4}},0.5)[/tex], [tex](-1,-1)[/tex], and [tex](1,-1)[/tex].

Step-by-step explanation:

I'm assuming the system is [tex]\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.[/tex]:

[tex]x^2+y^2=2[/tex]

[tex]x^2+(2x^2-3)^2=2[/tex]

[tex]x^2+(4x^4-12x^2+9)=2[/tex]

[tex]x^2+4x^4-12x^2+9=2[/tex]

[tex]4x^4-11x^2+9=2[/tex]

[tex]4x^4-11x^2+7=0[/tex]

[tex]x^4-11x^2+28=0[/tex]

[tex](x^2-7)(x^2-4)=0[/tex]

[tex](4x^2-7)(x^2-1)=0[/tex]

[tex]4x^2-7=0[/tex]

[tex]4x^2=7[/tex]

[tex]x^2=\frac{7}{4}[/tex]

[tex]x=\pm\sqrt{\frac{7}{4}}[/tex]

[tex]x^2-1=0[/tex]

[tex]x^2=1[/tex]

[tex]x=\pm1[/tex]

[tex]y=2x^2-3[/tex]

[tex]y=2(\pm\sqrt{\frac{7}{4}})^2-3[/tex]

[tex]y=2({\frac{7}{4}})-3[/tex]

[tex]y=\frac{7}{2}-3[/tex]

[tex]y=\frac{1}{2}[/tex]

[tex]y=2x^2-3[/tex]

[tex]y=2(\pm1)^2-3[/tex]

[tex]y=2(1)-3[/tex]

[tex]y=2-3[/tex]

[tex]y=-1[/tex]

Therefore, [tex]x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}[/tex] and [tex]y=-1,\frac{1}{2}[/tex]

The ordered pair solutions are [tex](-\sqrt{\frac{7}{4}},0.5)[/tex], [tex](\sqrt{\frac{7}{4}},0.5)[/tex], [tex](-1,-1)[/tex], and [tex](1,-1)[/tex].