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2 years ago

I need help finding Sin B, Cos B, and Tan A for a geometry problem. please help meee

Mathematics

1 answer:

Scilla [17]2 years ago

6 0

Answer:

Sin B = 5/13

Cos B = 12/13

tan A = 12/5

Step-by-step explanation:

Sin B = opposite side/ hypotenuse

Sin B = 5/13

Cos B = adjacent side / hypotenuse

= 12/13

tan A = opposite side /adjacent side

= 12/5

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Sophia has three stickers for every two stickers that Ethan has. If Sophia gives Ethan 16 of her stickers, then Ethan will have

olga_2 [115]

Answer:

Step-by-step explanation:

Let s represent the number of stickers Sophia has. Then the number Ethan has is (2/3)s. The transfer of 16 stickers can be described by the equation ...

(s -16)/(2/3s +16) = 1/2 . . . . . sophia/ethan = 1/2 after the transfer

2s -32 = 2/3s +16 . . . . . . . . .cross multiply

4/3s = 48 . . . . . . . . . . . . . . . add 32-2/3s

s = (48)(3/4) = 36 . . . . . . . . .multiply by 3/4

Together, the total number of stickers held is s +(2/3)s = (5/3)s.

5/3s = (5/3)(36) = 60 . . . . . total of Sophia's and Ethan's stickers

Together, Sophia and Ethan have 60 stickers.

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2 years ago

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morpeh [17]

4-2/3x=x-6/5

Answer: X=6

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2 years ago

About how many more website hits were there on Saturday than on Friday?

Ainat [17]

Answer:

1,000 hits

Step-by-step explanation:

Saturday: 2,000 hits

Friday: 1,000 hits

2,000 - 1,000 = 1,000

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2 years ago

Read 2 more answers

37 Ifx - 7 = 1, then x + 19

Virty [35]

Answer:

152

Step-by-step explanation:

x =7+1 = 8

8 x 19 =152

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2 years ago

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Write f(x)=8x^2-4x+11 in vertex form

Vlad [161]

Answer:

$f(x)=8(x-\frac{1}{4})^{2}+\frac{21}{2}$

$f(x)=8(x-0.25)^{2}+10.5$

Step-by-step explanation:

we have

$f(x)=8x^{2}-4x+11$

This is a vertical parabola open upward (because the leading coefficient is positive)

The vertex is a minimum

Convert to vertex form

Factor the leading coefficient

$f(x)=8(x^{2}-\frac{1}{2}x)+11$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)=8(x^{2}-\frac{1}{2}x+\frac{1}{16})+11-\frac{1}{2}$

$f(x)=8(x^{2}-\frac{1}{2}x+\frac{1}{16})+\frac{21}{2}$

Rewrite as perfect squares

$f(x)=8(x-\frac{1}{4})^{2}+\frac{21}{2}$ ----> equation in vertex form

$f(x)=8(x-0.25)^{2}+10.5$

The vertex is the point (0.25,10.5)

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3 years ago