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3 years ago

Giving brainlest to the first right answer

Mathematics

1 answer:

Neko [114]3 years ago

5 0

Answer:

Sorry man I could answer it I could find the step

But I hope you have good look on this!

But if You are using trygonometry here

a , 4

m 3 , 16

m 2, 20

m 2

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If DE≅EF, CD = v + 45, and CF = 4v, what is the value of v?

irinina [24]

Answer:

<h2>since DE≅EF</h2>

CD=CF (cpct)

$v + 45 = 4v \\ 4v - v = 45 \\ 3v = 45 \\ v = \frac{45}{3} \\ v = 15$

<h2>hope it helps you<3</h2>

3 0

2 years ago

For A I somehow got 41.05 but its 2.95cm can someone show me the working and what I did wrong?

777dan777 [17]

Answer:

Step-by-step explanation:

a). tan(75°) = $\frac{\text{Opposite side}}{\text{Adjacent side}}$

= $\frac{11}{k}$

k = $\frac{11}{\text{tan}(75)}$

k = 2.947

k = 2.95 cm

b). cos(52°) = $\frac{16}{s}$

s = $\frac{16}{\text{cos}(52)}$

s = 25.988

s ≈ 25.99 cm

c). sin(5°) = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

= $\frac{16.1}{q}$

q = $\frac{16.1}{\text{sin}(5)}$

q = 184.727

q ≈ 184.73 cm

4 0

2 years ago

98 POINTS!!!!! PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Maslowich

Hello!

I believe the answer is C.

I hope it helps!

3 0

3 years ago

Read 2 more answers

What is 5 divided by 5

Triss [41]

Answer:

Step-by-step explanation:

You have 5 jelly beans and 4 imaginary friends

You have to split the jellybeans between yourself and your pretend friends.

Each of you gets 1 jellybean

(Except you get all the jellybeans because your friends don't exist)

Hope this helps :)

6 0

2 years ago

Divide. (18x^3 + 12x^2 - 3x) ÷ 6x^2

nlexa [21]

$\bold{[ \ Answer \ ]}$

$\boxed{\bold{\frac{x^3\left(6x^2+4x-1\right)}{2}}}$

$\bold{[ \ Explanation \ ]}$

$\bold{Divide: \ \left(18x^3\:+\:12x^2\:-\:3x\right)\:\div \:6x^2}$

$\bold{-------------------}$

$\bold{Rewrite}$

$\bold{18x^3+12x^2-3x \ = \ x^2\frac{x\left(6x^2+4x-1\right)}{2}}$

$\bold{Rewrite}$

$\bold{x^2\frac{x\left(6x^2+4x-1\right)}{2}}$

$\bold{Multiply \ Fractions \ (a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c})}$

$\bold{\frac{x\left(6x^2+4x-1\right)x^2}{2}}$

$\bold{Rewrite}$

$\bold{x\left(6x^2+4x-1\right)x^2 \ = \ x^3\left(6x^2+4x-1\right)}$

$\bold{Simplify}$

$\bold{\frac{x^3\left(6x^2+4x-1\right)}{2}}$

$\boxed{\bold{[] \ Eclipsed \ []}}$

3 0

3 years ago

Read 2 more answers