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2 years ago

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A 13 -ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft from the house, t

he base is moving away at the rate of 15 ft/sec.
a. What is the rate of change of the height of the top of the ladder?
b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then?
c. At what rate is the angle between the ladder and the ground changing then?

1 answer:

VARVARA [1.3K]2 years ago

8 0

I think the answer is B because all the air other answers are incorrect

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624 feet to 702 feet state whether the percent of the change is an increase or decrease

Ghella [55]

Answer:

The percent of change is an increase.

Step-by-step explanation:

624 has been changed to 702.

Percent change = [(Value after - Value before ) / Value before]*100

= $\frac{702-624}{624}*100$

= $\frac{78}{624}*100$

So we can clearly see that percentage change is a positive change. This means that the Value after change is higher than value before the change. So the percent of change is an increase.

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3 years ago

Plot the following points on a xy-plane.<br> (5,2) , (-2, 1) , (-1,-3)

aliya0001 [1]

Answer: See below

Step-by-step explanation:

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3 years ago

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1. What is an equation for the line with slope 2/3 and y-intercept 9?

Monica [59]

Question 1:

The generic equation of the line is given by:
$y = mx + b
$
Where,
m: slope of the line
b: cutting point with the y axis
Substituting values we have:
$y = \frac{2}3}x + 9
$
Answer:
an equation for the line with slope 2/3 and y-intercept is:
$y = \frac{2}3}x + 9$

Question 2:

The standard equation of the line is given by:
$y-yo = m (x-xo)
$
Where,
m: slope of the line
(xo, yo): ordered pair that belongs to the line
The slope of the line is:
$m = \frac{y2-y1}{x2-x1}$
Substituting values:
$m = \frac{1-(-3)}{3-1}$
$m = \frac{1+3}{3-1}$
$m = \frac{4}{2}$
$m = 2
$
We choose an ordered pair:
$(xo, yo) = (3, 1)
$
Substituting values:
$y-1 = 2 (x-3)
$
Rewriting:
$y = 2x - 6 + 1

y = 2x - 5$
Answer:
An equation in slope-intercept form for the line that passes through the points (1, -3) and (3,1) is:
$y = 2x - 5
$

Question 3:

For this case, since the variation is direct, then we have an equation of the form:
$y = kx
$ Where,
k: constant of variation.
We then have the following equation:
$-4y = 8x
$ Rewriting we have:
$y = \frac{8}{-4}x
$
$y = -2x
$ Therefore, the constate of variation is given by:
$k = -2
$
Answer:
the constant of variation is:
$k = -2$

Question 4:

For this case, since the variation is direct, then we have an equation of the form:
$y = kx
$ Where,
k: constant of variation.
We must find the constant k, for this we use the following data:
y = 24 when x = 8
Substituting values:
$24 = k8
$ Clearing k we have:
$k = \frac{24}{8} 
$
$k = 3
$
Therefore, the equation is:
$y = 3x
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$y = 3 (10)

y = 30$ Answer:
the value of y when x = 10 is:
$y = 30$

6 0

3 years ago

Read 2 more answers

PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS

Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

$a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\$

With respect to reference angle A, we have:

opposite side = a = 4
adjacent side = b = $\sqrt{33}$
hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

$\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\$

Then cosine which is adjacent over hypotenuse

$\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\$

Tangent is the ratio of opposite over adjacent

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\$

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

cosecant, abbreviated as csc, is the reciprocal of sine
secant, abbreviated as sec, is the reciprocal of cosine
cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

$\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{ ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

------------------------------------------------------

Summary:

The missing side is $b = \sqrt{33}$

The 6 trig functions have these results

$\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0

1 year ago

Plzzzzzz help!! I will give brainliest to whoever answers

jarptica [38.1K]

Answer:

4 and -3 i think

Step-by-step explanation:

the dot is on 4 and -3

5 0

3 years ago

Read 2 more answers