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vlada-n [284]
2 years ago
12

F(x)=3x-2 h(x)=3x g(x)=x squared find the value of x when f(x)=19

Mathematics
1 answer:
Yuliya22 [10]2 years ago
4 0

The value of f(19) in the given equation is 55.

<h3>What is the function of a linear equation?</h3>

The function of a linear equation illustrates a straight line graph and for a given set of input into the function, there is usually a specific output.

Given that:

  • f(x) = 3x - 2

We are to find the f(x) = 19.

So;

f(19) = 3(19) - 2

f(19) = 57 - 2

f(19) = 55

Learn more about the function of  a linear equation here:

brainly.com/question/14323743

#SPJ1

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Equation for this problem:            Total 3 cubed / (L x W)

So 18.5cm x 15cm = 277.5 is the product of L and W. Now you need to devide the total by 277.5 which is 14. So the aquarium is 14 cm deep.

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2 years ago
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Anit [1.1K]

Answer:

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2 years ago
Solve the equation for all values of x by completing the square.<br> x2 + 8.0 + 15 = 0
Aleonysh [2.5K]

Answer:

x= -3

x= -5

Step-by-step explanation:

x^2 + 8x + 15 = 0

find which 2 numbers when multiplied equal 15 and when added equal 8

they are 3 and 5

(x+3)(x+5)

x+3 =0

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x= -5

6 0
2 years ago
I need help please and thank you
Ira Lisetskai [31]

This is a false statement.

When you are solving using square roots, you need to be aware that answers can be both positive and negative. When we solve this, you see there are two possible answers.

x^2 - 9 = 0

x^2 = 9

x = +/- 3

While 3 is an answer, so is -3. If we square either of those numbers, we get 9, which will satisfy the equation.

5 0
3 years ago
Evalute costheta if sintheta = (sqrt5)/3
torisob [31]

Answer:

\large\boxed{\cos\theta=\pm\dfrac{2}{3}}

Step-by-step explanation:

Use \sin^2x+\cos^2x=1.

We have

\sin\theta=\dfrac{\sqrt5}{3}

Substitute:

\left(\dfrac{\sqrt5}{3}\right)^2+\cos^2\theta=1\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(\sqrt5)^2}{3^2}+\cos^2\theta=1\qquad\text{use}\ (\sqrt{a})^2=a\\\\\dfrac{5}{9}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{5}{9}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{9}{9}-\dfrac{5}{9}\\\\\cos^2\theta=\dfrac{4}{9}\to \cos\theta=\pm\sqrt{\dfrac{4}{9}}\\\\\cos\theta=\pm\dfrac{\sqrt4}{\sqrt9}\\\\\cos\theta=\pm\dfrac{2}{3}

3 0
3 years ago
Read 2 more answers
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