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A software designer is mapping the streets for a new racing game. all of the streets are depicted as either perpendicular or par

ElenaW [278]

The equation of the central street PQ is -1.5x - 3.5y = -31.5 option (b) is correct.

<h3>What is a straight line?</h3>

A straight line is a combination of endless points joined on both sides of the point.

We have a straight line:

$-7x+3y=-21.5$

Convert it to the general form given below:

$\rm y=mx+c$

$\rm 3y=-21.5+7x\\\\ \rm y = \frac{-21.5}{3}+\frac{7}{3}x$ or

$\rm y = \frac{7}{3}x-\frac{21.5}{3}$

$m = \frac{7}{3}$ (Slope of AB line)

For the slope(m') of the PQ line:

$\rm m'=-\frac{1}{m}$ ( because AB and PQ are perpendicular to each other)

$\rm m' = -\frac{3}{7}$

We know the:

$\rm (y-y')=m'(x-x')$

Where (x', y') = (7, 6), we get:

$\rm (y-6)=-\frac{3}{7} (x-7)$

$\rm 7(y-6)=-3 (x-7)\\\\\rm 7y-42= -3x+21\\\\\rm 7y= -3x+21+42\\\\\rm 3x+7y=63$

$\rm -1.5x-3.5y=-31.5$ (multiply by -1/2 on both sides)

Thus, the equation of the central street PQ is -1.5x - 3.5y = -31.5

Learn more about the straight line.

brainly.com/question/3493733

6 0

1 year ago

Find the tangent line approximation for 10+x−−−−−√ near x=0. Do not approximate any of the values in your formula when entering

Svetllana [295]

Answer:

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Step-by-step explanation:

We are asked to find the tangent line approximation for $f(x)=\sqrt{10+x}$ near $x=0$ .

We will use linear approximation formula for a tangent line $L(x)$ of a function $f(x)$ at $x=a$ to solve our given problem.

$L(x)=f(a)+f'(a)(x-a)$

Let us find value of function at $x=0$ as:

$f(0)=\sqrt{10+x}=\sqrt{10+0}=\sqrt{10}$

Now, we will find derivative of given function as:

$f(x)=\sqrt{10+x}=(10+x)^{\frac{1}{2}}$

$f'(x)=\frac{d}{dx}((10+x)^{\frac{1}{2}})\cdot \frac{d}{dx}(10+x)$

$f'(x)=\frac{1}{2}(10+x)^{-\frac{1}{2}}\cdot 1$

$f'(x)=\frac{1}{2\sqrt{10+x}}$

Let us find derivative at $x=0$

$f'(0)=\frac{1}{2\sqrt{10+0}}=\frac{1}{2\sqrt{10}}$

Upon substituting our given values in linear approximation formula, we will get:

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}(x-0)$

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}x-0$

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Therefore, our required tangent line for approximation would be $L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$ .

8 0

2 years ago

a rectangular gift box is 10 inches long, 7 inches wide, 1.25 inches high, and had a surface area of 182.5 square inches. gift w

nalin [4]

1) 2 * (10 * 7 + 7 * 1,25 + 10 * 1,25) = 2 * (70 + 8,75 + 12,5) = 2 * 91,25 = 182,5 square inches - a surface area
2) 182,5 * 0,012 = <span>$</span>2,19 - answer.

6 0

2 years ago

Please i need help. show work

Nadya [2.5K]

Just download the app photomath it helps a lot with equations plus it shows u how to do them

7 0

3 years ago

Problem PageQuestion Working together, two pumps can drain a certain pool in hours. If it takes the older pump hours to drain th

gizmo_the_mogwai [7]

Answer:

10.5 hours.

Step-by-step explanation:

Please consider the complete question.

Working together, two pumps can drain a certain pool in 6 hours. If it takes the older pump 14 hours to drain the pool by itself, how long will it take the newer pump to drain the pool on its own?

Let t represent time taken by newer pump in hours to drain the pool on its own.

So part of pool drained by newer pump in one hour would be $\frac{1}{t}$ .

We have been given that it takes the older pump 14 hours to drain the pool by itself, so part of pool drained by older pump in one hour would be $\frac{1}{14}$ .