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3 years ago

#4 is 28 it just got cut off

Mathematics

2 answers:

Fudgin [204]3 years ago

5 0

3^2 + (7 - 2)(4) - 6/3

= 9 + (7 - 2)(4) - 6/3

= 9 + (5)(4) - 6/3

= 9 + 20 - 6/3

= 29 - 6/3

= 29 - 2

Answer: 27

Hope that helps!!!! (Answer: Letter Choice, (B3), 27

devlian [24]3 years ago

3 0

Follow PEMDAS. So, first the ( ) needs to be done first. So that now gives us

3^2 + 5 x 4 - 6/3. Next is exponents. So now we have 9 + 5 x 4 -6/3. Next we have to do multiplication. So now we have 9 + 20 - 6/3. Now we have to add. 29-6/3. 29-2 is 27. So, 27 is the answer.

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Please simplify n-2/n^2-4

Doss [256]

Answer:

$\frac{1}{ {n} + {2}}$

Step-by-step explanation:

$\frac{n - 2}{ {n}^{2} - 4} \\ \\ = \frac{n - 2}{ {n}^{2} - {2}^{2} } \\ \\ = \frac{n - 2}{ ({n} + {2})({n} - {2})} \\ \\ = \frac{1}{ {n} + {2}} \\$

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3 years ago

Which expression is equivalent to One-fourth minus three-fourths x? One-fourth (1 + 3 x) One-fourth (1 minus 3 x) One-fourth (ne

Juli2301 [7.4K]

Answer:

1/4(1 - 3x)

Step-by-step explanation:

Step 1: Write out expression

1/4 - 3/4x

Step 2: Factor out GCF

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4 0

3 years ago

Read 2 more answers

Calculate the average mass of 100 marbles that weigh 824g

Lina20 [59]

Answer:

8.24 grams/marble

Step-by-step explanation:

Average mass = (total mass)/(number of marbles)

Here, that would be:

824 g

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7 0

3 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.

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1 year ago

2.8×10 to the -3 power minus 0.00065

Bas_tet [7]

Answer:

the answer is 3.37723077E13

Step-by-step explanation:

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3 years ago