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11111nata11111 [884]

2 years ago

6

Hey I’m confused on this problem help.

1 answer:

PilotLPTM [1.2K]2 years ago

7 0

the answer is b carmella

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50 POINTS AVAILABLE!!

tester [92]

Answer:

they need to be in ratios format

Step-by-step explanation:

6 0

3 years ago

I need help and it’s due today

frez [133]

Answer:

5.63

Step-by-step explanation:

Angle NLM is equal to KLM - KLN.

Since KLM = 137° and KLN = 47°, NLM = °90. (Maybe it doesn't look like a right angle, but the math doesn't lie.)

Since NLM = 16y, then 90 = 16y.

Dividing both sides by 16, we find that:

y = 5.625

5 0

3 years ago

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If log, 7 -3 what is the value of x?<br> A. 21<br> B. 3V7<br> C. 37<br> D. 343

gregori [183]

Answer:

B

Step-by-step explanation:

logₓ 7 = 3

x³ = 7

Take cube root both sides

$x= \sqrt[3]{7}$

4 0

2 years ago

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$tan(\alpha)=cot(\beta)$

<em>Find the value of the length side adjacent to the angle alpha</em>

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

3 years ago

!!20 POINTS!! BRAINLIEST TO CORRECT ANSWER!!

d1i1m1o1n [39]

Answer:

3:9

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers