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katen-ka-za [31]
3 years ago
12

If g(x) = 3 - 2x find, g(-1)​

Mathematics
1 answer:
pogonyaev3 years ago
3 0

Steps to solve:

g(x) = 3 - 2x when g(-1)

~Substitute

g(-1) = 3 - 2(-1)

~Simplify

g(-1) = 3 - (-2)

~Simplify

g(-1) = 3 + 2

~Add

g(-1) = 5

Best of Luck!

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Answer:

d. Twenty-five cars are randomly selected with the new transmission and 25 with the old transmission.

Step-by-step explanation:

A hypothesis test on the difference between the means must satisfy the following conditions

1) the samples must be randomly selected

2) each sample must have a known standard deviation

3) the samples must be independent

4) the samples must be from normal distribution

Hence only part d gives the right answer.

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The graphs of the equation y = 4x + 1 and y - kx = 10 are perpendicular when k = _______?
Ann [662]

Answer:

k=-\frac{1}{4}

Step-by-step explanation:

we have

Line 1

y=4x+1

Equation in slope intercept form

The slope is equal to

m_1=4

Line 2

y-kx=10

y=kx+10

Equation in slope intercept form

The slope is equal to

m_2=k

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of the slopes is equal to -1)

so

m_1*m_2=-1

substitute

(4)(k)=-1

k=-\frac{1}{4}

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A school with 340 students produces trash that fills 1,190 tall trash cans per week. If each student produces the same amount of
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3.5 trash cans a week
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Given the quadratic function f(x) = 4x^2 - 4x + 3, determine all possible solutions for f(x) = 0
solong [7]

Answer:

The solutions to the quadratic function are:

x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}

Step-by-step explanation:

Given the function

f\left(x\right)\:=\:4x^2\:-\:4x\:+\:3

Let us determine all possible solutions for f(x) = 0

0=4x^2-4x+3

switch both sides

4x^2-4x+3=0

subtract 3 from both sides

4x^2-4x+3-3=0-3

simplify

4x^2-4x=-3

Divide both sides by 4

\frac{4x^2-4x}{4}=\frac{-3}{4}

x^2-x=-\frac{3}{4}

Add (-1/2)² to both sides

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

x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{1}{2}

\left(x-\frac{1}{2}\right)^2=-\frac{1}{2}

\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}

solving

x-\frac{1}{2}=\sqrt{-\frac{1}{2}}

x-\frac{1}{2}=\sqrt{-1}\sqrt{\frac{1}{2}}                 ∵ \sqrt{-\frac{1}{2}}=\sqrt{-1}\sqrt{\frac{1}{2}}

as

\sqrt{-1}=i

so

x-\frac{1}{2}=i\sqrt{\frac{1}{2}}

Add 1/2 to both sides

x-\frac{1}{2}+\frac{1}{2}=i\sqrt{\frac{1}{2}}+\frac{1}{2}

x=i\sqrt{\frac{1}{2}}+\frac{1}{2}

also solving

x-\frac{1}{2}=-\sqrt{-\frac{1}{2}}

x-\frac{1}{2}=-i\sqrt{\frac{1}{2}}

Add 1/2 to both sides

x-\frac{1}{2}+\frac{1}{2}=-i\sqrt{\frac{1}{2}}+\frac{1}{2}

x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}

Therefore, the solutions to the quadratic function are:

x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}

4 0
2 years ago
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