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4 years ago

From the set (33, 15, 12), use substitution to determine which value of x makes the equation true.

Mathematics

1 answer:

zepelin [54]4 years ago

8 0

X=12 because 3 times 12 equals 36. if you divide by 3 on both sides you would be left with that

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There are four hundred students as Thompson Middle School. If 54% of the students are female, what is the ratio of female to mal

laila [671]

Answer:

1:3

Step-by-step explanation:

I think...

54% of 400= 216

400/216= 1.85

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4 years ago

Mia tried to solve an equation step by step.

natka813 [3]

Answer:

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3 years ago

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What is the value of y in the equation 6.4x+3y=36 when x=3 A 5 B 5.6 C 6 D 8

Vilka [71]

Answer:

5.6

Step-by-step explanation:

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4 years ago

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What are the solutions of the equation x4 + 3x2 + 2 = 0? Use u substitution to solve. x = plus-or-minus i StartRoot 2 EndRoot an

Rom4ik [11]

Answer:

$x=\pm\sqrt{2}i\text{ (or) }x=\pm i$

Step-by-step explanation:

We have been given an equation $x^4 + 3x^2 + 2 = 0$ . We are asked to find the solutions of our given equation using u-substitution.

We can rewrite our given equation as:

$(x^2)^2+3x^2+2 = 0$

Let us assume that $u=x^2$ .

$u^2+3u+2 = 0$

$u^2+2u+u+2 = 0$

$u(u+2)+1(u+2) = 0$

$(u+2)(u+1) = 0$

$(u+2)=0\text{ (or) }(u+1) = 0$

$u=-2\text{ (or) }u=-1$

Upon substituting back the value of u, we will get:

$x^2=-2\text{ (or) }x^2=-1$

$x=\pm\sqrt{-2}\text{ (or) }x=\pm\sqrt{-1}$

$x=\pm\sqrt{-1\cdot 2}\text{ (or) }x=\pm\sqrt{-1}$

Now we will use imaginary unit i. We know that $i^2=-1$ .

$x=\pm\sqrt{i^2\cdot 2}\text{ (or) }x=\pm\sqrt{i^2}$

$x=\pm\sqrt{2}i\text{ (or) }x=\pm i$

Therefore, the solutions for our given equation are $x=\pm\sqrt{2}i\text{ (or) }x=\pm i$ .

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3 years ago

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Find (f/g) (x) for the functions provided: ƒ(x) = x3 − 27, g(x) = 3x − 9

AnnZ [28]

Answer:

$(\frac{f}{g})(x)=\frac{1}{3}(x^2+3x+9)$

Step-by-step explanation:

We have been given that

$f(x)=x^3-27,g(x)=3x-9$

We can use the formula for difference of cubes to simplify the function f(x)

difference of cubes - $a^3-b^3=(a-b)(a^2+ab+b^2)$

$f(x)=x^3-27\\\\=x^3-3^3\\\\=(x-3)(x^2+3x+9)$

And g(x) can be written as

$g(x)=3x-9\\=3(x-3)$

Thus, we have

$(\frac{f}{g})(x)=\frac{(x-3)(x^2+3x+9)}{3(x-3}$

On cancelling the common factors, we get

$(\frac{f}{g})(x)=\frac{1}{3}(x^2+3x+9)$

5 0

3 years ago