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3 years ago

Is abc def name the postulates that applies

Mathematics

1 answer:

geniusboy [140]3 years ago

4 0

Answer:

The answer is C. AAA is not a congruency postulate because the side lengths may be different sizes although the angles are the same.

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Evaluate the following expression. y + 9 when y = 12

tino4ka555 [31]

Answer:

Step-by-step explanation:

y+9 = ?

y=12

12+9=21

5 0

2 years ago

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H(x)=-x^2-4 and I(x)=2x+3<br> H(I(s))

ICE Princess25 [194]

Easy peasy
just subsitute I(x) for the x in the h(x) so
h(I(s))=-(2s+3)^2-4
distribute and simplify
h(I(s))=-(4s^2+12s+9)-4
h(I(s))=-4s^2-12s-9-4
h(I(s))=-4s^2-12s-13

3 0

3 years ago

Given the equation 5 + x -12 = 2x - 7. Part A. Solve the equation 5 + x - 12 = 2x - 7. In your final answer, be sure to state th

Nitella [24]

Answer:

srry guys i dont know this

Step-by-step explanation:

5 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

1 year ago

I need someone to explain on how to do this problem.

weqwewe [10]

Answer:

I think it's x=-12 for the buttom one

Step-by-step explanation:

I think when you get open equations like these you would asume 1 is there so rewrite it as 2x + 25 = 1

so take the 25 away from 1. to do that we need to do the opisite of +

2x + 25 = 1

1 - 25 = -24

rewrite

2x = -24

take the 2 away. oppisite of mulitplication is division. so divide 2 by -24

-24 divided by 2 = -12

Hope this helps! Please mark as brainliest! Thanks! If you need me to solve the top one too, I'll do it! Just comment bellow. :D

5 0

3 years ago

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