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3 years ago

What equation goes through the points (4,-9) and is perpendicular to the x axis

Mathematics

1 answer:

alekssr [168]3 years ago

3 0

X=4 is your equation

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ANSWER ASAP!! Find the area of a triangle with legs that are: 12 m, 15 m, and 9 m.

Setler79 [48]

Answer:

Step-by-step explanation:

<h3>METHOD 1:</h3>

You can use the Heron's formula:

$A=\sqrt{p(p-a)(p-b)(p-c)}$

where

<em>p</em><em> - half of perimeter</em>

<em>a, b, c</em><em> - lengths of sides</em>

We have

$a=12m;\ b=15m;\ c=9m$

Calculate:

$p=\dfrac{12+15+9}{2}=\dfrac{36}{2}=18\ (m)\\\\A=\sqrt{18(18-12)+(18-15)(18-9)}\\\\A=\sqrt{(18)(6)(3)(9)}\\\\A=\sqrt{2916}\\\\A=54\ (m^2)$

<h3>METHOD 2:</h3>

Let's check that it is not a right triangle.

If the sum of the squares of the two shorter sides is equal to the square of the longest side, then this triangle is rectangular.

We have

$9m < 12m$

Check:

$9^2+12^2=81+144=225\\15^2=225$

This is a right trianglr wherew 9m and 12m are legs and 15m is a hypotenuse.

The formula of an area of a right triangle is:

$A=\dfrac{ab}{2}$

Substitute:

$A=\dfrac{(9)(12)}{2}=\dfrac{108}{2}=54\ (m^2)$

6 0

3 years ago

On a 66-sided number cube, which is more likely: rolling an odd number or rolling an even number?

statuscvo [17]

Hello, how have you been?

I believe that the answer to this question is exactly %55. There is 33 even numbers, and 33 odd numbers :)

Hope i helped :)

5 0

3 years ago

Read 2 more answers

Determine the numerical length of JK.

Zarrin [17]

Answer:

So the two sectors add to the full length:

JK + IJ = IK

x+6 + 9 = 2x

Rearrange:

x=15

So the numerical length:

2*15

5 0

3 years ago

Read 2 more answers

What is the simplest form for 15:25

vazorg [7]

Answer:

3/5

Step-by-step explanation:

Both can be divided by 5. So, 15 divided by 5 is 3, and 25 divided by 5 is 5! Hoped this helped :)

3 0

2 years ago

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