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2 years ago

HELP PLS!! according to the rational root theorem, -2/5 is a potential rational root of which function?

Mathematics

1 answer:

xeze [42]2 years ago

5 0

Answer:

Step-by-step explanation:

According to the Rational Roots Theorem, the possible rational roots are all the factors of the constant term expressed over the factors of the leading coefficient of the polynomial function. The reason is that the numerator of this rational root is a factor of 4 and the denominator is a factor of 25.

DOES IT HELP?!

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What is the perimeter of triangle JKL (same triangle as question #1)?

Keith_Richards [23]

Answer:

x=30

Step-by-step explanation:

You have to use intercept theorem, also known as Thales's theorem to slvoe this.

6 0

1 year ago

There are 60 girls and 90 boys in the sixth grade at a middle school. What percent of the sixth graders at this middle school ar

KonstantinChe [14]

Answer:

60%

Step-by-step explanation:

First, find the total number of sixth graders

To do this, add the number of girls and boys together

girls+boys

There are 60 girls and 90 boys

60+90

150

Now, find the percent of sixth graders that are boys

To do this, divide the number of boys by the total number of sixth graders

boys/total sixth graders (part/whole)

There are 90 boys, and 150 total sixth graders

90/150

0.6

Convert to a percent. Multiply by 100 or move the decimal place two spots to the right

0.6*100=60

0.6-->6.0-->60.

So, 60% are boys

3 0

2 years ago

Use the double intercept approach to find the graph of -1 = -y + x

NeX [460]

The answer is graph d

3 0

2 years ago

How many cubic inches? Please help!?!

iragen [17]

1575 cubic inches

This is the answer to your question.

7 0

2 years ago

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

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

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

$\sqrt{-1}=i$

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

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