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3 years ago

-4=3p-p Show steps how to solve for x

1 answer:

3 0

−4=3p−pStep 1: Simplify both sides of the equation.−4=3p−p−4=3p+−p−4=(3p+−p)(Combine Like Terms)−4=2p−4=2pStep 2: Flip the equation.2p=−4Step 3: Divide both sides by 2.‌2p2‌=‌−42‌p=−2Answer:p=−2

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Find the area of a triangle bounded by the y-axis, the line f(x)=9−4/7x, and the line perpendicular to f(x) that passes through

Setler79 [48]

ANSWER:

The area of the triangle bounded by the y-axis is $\frac{7938}{4225} \sqrt{65} \text { unit }^{2}$

SOLUTION:

Given, $f(x)=9-\frac{-4}{7} x$

Consider f(x) = y. Hence we get

$f(x)=9-\frac{-4}{7} x$ --- eqn 1

$y=9-\frac{4}{7} x$

On rewriting the terms we get

4x + 7y – 63 = 0

As the triangle is bounded by two perpendicular lines, it is an right angle triangle with y-axis as hypotenuse.

Area of right angle triangle = $\frac{1}{ab}$ where a, b are lengths of sides other than hypotenuse.

So, we need find length of f(x) and its perpendicular line.

First let us find perpendicular line equation.

Slope of f(x) = $\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-4}{7}$

So, slope of perpendicular line = $\frac{-1}{\text {slope of } f(x)}=\frac{7}{4}$

Perpendicular line is passing through origin(0,0).So by using point slope formula,

$y-y_{1}=m\left(x-x_{1}\right)$

Where m is the slope and $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$

$y-0=\frac{7}{4}(x-0)$

$y=\frac{7}{4} x$ --- eqn 2

4y = 7x

7x – 4y = 0

now, let us find the vertices of triangle, one of them is origin, second one is point of intersection of y-axis and f(x)

for points on y-axis x will be zero, to get y value, put x =0 int f(x)

0 + 7y – 63 = 0

7y = 63

y = 9

Hence, the point of intersection is (0, 9)

Third vertex is point of intersection of f(x) and its perpendicular line.

So, solve (1) and (2)

$\begin{array}{l}{9-\frac{4}{7} x=\frac{7}{4} x} \\\\ {9 \times 4-\frac{4 \times 4}{7} x=7 x} \\\\ {36 \times 7-16 x=7 \times 7 x} \\\\ {252-16 x=49 x} \\\\ {49 x+16 x=252} \\\\ {65 x=252} \\\\ {x=\frac{252}{65}}\end{array}$

Put x value in (2)

$\begin{array}{l}{y=\frac{7}{4} \times \frac{252}{65}} \\\\ {y=\frac{441}{65}}\end{array}$

So, the point of intersection is $\left(\frac{252}{65}, \frac{441}{65}\right)$

Length of f(x) is distance between $\left(\frac{252}{65}, \frac{441}{65}\right)$ and (0,9)

$\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(9-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+0} \\ &=\frac{252}{65} \end{aligned}$

Now, length of perpendicular of f(x) is distance between $\left(\frac{252}{65}, \frac{441}{65}\right) \text { and }(0,0)$

$\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(0-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+\left(\frac{441}{65}\right)^{2}} \\ &=\frac{\sqrt{(12 \times 21)^{2}+(21 \times 21)^{2}}}{65} \\ &=\frac{63}{65} \sqrt{65} \end{aligned}$

Now, area of right angle triangle = $\frac{1}{2} \times \frac{252}{65} \times \frac{63}{65} \sqrt{65}$

$=\frac{7938}{4225} \sqrt{65} \text { unit }^{2}$

Hence, the area of the triangle is $\frac{7938}{4225} \sqrt{65} \text { unit }^{2}$

8 0

3 years ago

What is the surface area?

Nady [450]

It’s the area of the outside or something

7 0

2 years ago

Read 2 more answers

Here are the 30 best lifetime baseball batting averages of all time, arranged in order from lowest to highest:

hichkok12 [17]

Answer:

Step-by-step explanation:

0.36|6=0.366

8 0

3 years ago

In your biology class, your nal grade is based on several things: a lab score, scores on two major tests, and your score on the

Mariana [72]

Complete Question

In your biology class, your final grade is based on several things: a lab score, scores on two major tests, and your score on the final exam. There are 100 points available for each score. However, the lab score is worth 25% of your total grade, each major test is worth 22:5%, and the final exam is worth 30%. Compute the weighted average for the following scores: 92 on the lab, 81 on the first major test, 93 on the second major test, and 85 on the final exam. (Enter your answer to one decimal place)

Answer:

The weighted average is $W = 87.65$

Step-by-step explanation:

From the question we are told that

The worth of the lab score is 25% of total grade

The worth of each major test is 22.5% of total grade

The worth of final exam is 30% of total grade

The score on lab is 92

The score on first major test 81

The score on second major test 93

The score on final exam is 85

Generally the weighted score is mathematically represented as

$W = 0.25 * 92 + 0.225* 81 + 0.225* 93 +0.30 * 85$

=> $W = 0.25 * 92 + 0.225* 81 + 0.225* 93 +0.30 * 85$

=> $W = 87.65$

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3 years ago

A die is rolled two separate times. What is the probability that the first roll

Masteriza [31]

1/6 times 1/2 is D. 1/12

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2 years ago