Given that3x:8=7:2Calculate the value ofx.Give your answer in its simplest form

Mathematics

1 answer:

laila [671]3 years ago

5 0

The answer is 28/3, and in simplest form its 9 1/3

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Find the value of x. (6x-3) 75°

Alenkinab [10]

the value of x is 13

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H(t)=(t+3)

Anarel [89]

Answer:

Step-by-step explanation:

h(t) = (t + 3)² + 5, -5 ≤ t ≤ -1

The average rate of change is the change in h over the change in t.

(h(-1) − h(-5)) / (-1 − (-5))

= ((-1 + 3)² + 5 − ((-5 + 3)² + 5)) / (-1 + 5)

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A family wants to remodel a house and spend at most $135,000 total for the entire project. a)If the builder wants to make a 15%

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) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

$\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}$

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$