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3 years ago

Can someone help me with this? Preferably with an explanation on how to actually figure this out. Thank you!

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

5 0

Answer:

2^-4, 1/16, (1/2)^4

There is a rule with exponents, if you are dividing two exponents you subtract the first exponent from the other to simplify it, also meaning it's equivalent.

So in this case 5-9 = 2^-4

The others were found by evaluation.

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PLEASE HELP ME ASAP 20 POINTS SHOW WORK

ZanzabumX [31]

Answer:

i got it wrong i don't want to give the wrong answer

question 4

smaller answer is -9

larger answer is 2

Step-by-step explanation:

i did the test

8 0

3 years ago

Read 2 more answers

2/3=x/6=9/y to find the value of y in this proportion you must divide 54 by

scZoUnD [109]

Answer:

You must divide $54$ by $4$

Step-by-step explanation:

we have

$\frac{2}{3}=\frac{x}{6}=\frac{9}{y}$

Step 1

Solve for x

$\frac{2}{3}=\frac{x}{6}$

$3x=6*2\\x=12/3\\x=4$

Step 2

Find the value of y

$\frac{x}{6}=\frac{9}{y}$

Substitute the value of x

$\frac{4}{6}=\frac{9}{y}$

$4y=9*6\\y=54/4\\y= 13.5$

therefore

You must divide $54$ by $4$

6 0

3 years ago

Read 2 more answers

If g(x) = 4 square root of x then g(45) is

AlladinOne [14]

Hi,

So we have g(x) = 4√(x). We're looking for g(45). Think of it like this: whatever number is in the place of x in g(x), just place that number AS x.

Therefore, we have :
g(45) = 4√(45) ⇒ (4) × (√(45)) ⇒ (4)(3√(5)) = 12√(5).

-Hope this helps!

5 0

3 years ago

I need help finding y

kolezko [41]

to solve for y, we must multiply by t on both side to get ride of the t in the denominator. By doing this, we will get:

${t}^{3}$

on the right side.

Subtracting 3, and we have successfully isolated y.

It would be impossible to get a quantitative value for y if we don't know the value of t.

4 0

3 years ago

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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

2 years ago