All categories

3 years ago

Simplify the expression. (7x4)? (7x4)2 =

Mathematics

2 answers:

TEA [102]3 years ago

5 0

Answer:

Step-by-step explanation:

i took the test

NemiM [27]3 years ago

4 0

56 ??? i’m pretty sure that’s the answer

You might be interested in

4 questions 50 points

MAXImum [283]

I hope this helps you

8 0

3 years ago

Solve for x in the triangle.Round your answer to the nearest tenth.

enyata [817]

sin = opposite/hypotenuse

sin 40° = 12/x

Cross multiply.

0.64278760968x = 12

Divide both sides by sin 40°.

x = 18.7 (rounded to nearest tenth).

5 0

3 years ago

Read 2 more answers

Part A

rjkz [21]

Answer:

Step-by-step explanation:

Part A

Solve 4,500 ÷ 90. Enter your answer in the box.

Answer : 50

Part B

Select all the expressions that are equal to 4,500 ÷ 90.

A. 4,500 ÷ 9 tens ✔️

B. 450 ÷ 90 ❌

C. 450 ÷ 9 ❌

D. 4,500 ÷ 9 ❌

E. 4,500 ÷ 90 tens ❌

8 0

2 years ago

Find the area of the striped sector if ∠ A = 45° and the circle has a radius of 10 inches.

Pie

A diagram must be provided along with the question above. Based on my research, here is what I got.

<span>100 pi is the area of the entire circle
That makes :
</span><span>∠ A = 45°/360 = 1/8
</span> 100pi/8 = 25 pi / 2

The answer is option D)25π over 2. I hope this answer helps.

4 0

2 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