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Svetradugi [14.3K]
3 years ago
10

Help!!! algebra test - exam

Mathematics
2 answers:
Flauer [41]3 years ago
7 0

Answer:

We cant see the pic

Step-by-step explanation:

aleksandr82 [10.1K]3 years ago
4 0
I can’t see the picture I’m sorry
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In Marta's class, the ratio of students who ate tacos for lunch to the students who ate burritos was 12:18.
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20 30

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Shipping rates for Company A and Company B are shown in the tables below. Which company has shipping rates that you can represen
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Can someone help me get this please???
jarptica [38.1K]

Answer:

With respect to the reference angle, one leg is the adjacent leg, or the leg right next to the reference angle; and the other leg is the opposite leg, or the leg on the opposite side of the triangle. If two triangles are similar, the ratios of corresponding sides are equal.

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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
Slope of (7, -2) and (2, -6)
Stells [14]

Answer:

it's -8/5 because you subtract y2 from y1 and x2 from x1

Step-by-step explanation:

4 0
2 years ago
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