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Naddika [18.5K]

3 years ago

12

a tool set is on sale for $280. the original price of the tool set was $400. what percent of the original price is the sale pric

e

2 answers:

Marianna [84]3 years ago

5 0

70%

Hope this helps! :)

______________

ycow [4]3 years ago

3 0

Answer:

30%

Step-by-step explanation:

400 = 100%

4 = 1%

400 - 280 = 120

120 ÷ 4 = 30

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G=3t^2−5t+6<br><br> P=−8t^2+7t−9<br><br> G+P=

kotykmax [81]

Answer:

-5t^2 +2t -3

Step-by-step explanation:

Collect terms in the sum:

G+P = (3t^2 -5t +6) +(-8t^2 +7t -9) = (3-8)t^2 +(-5+7)t +(6-9)

= -5t^2 +2t -3

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3 years ago

Write 12241.025 how do u read this

charle [14.2K]

Answer:

12241.025

Step-by-step explanation:

Twelve thousand, two hundred and forty one point zero, two, five

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3 years ago

Given the radius is 6 inches, what is the area of the circle?

enyata [817]

Answer:

I wasn't sure how you're supposed to answer, so the answers for each scenario are in the explanation!

Step-by-step explanation:

$A=\pi r^{2}$

$A=\pi (6^{2})$

$A=\pi (36)$

From here, there can be a few different answers:

The exact answer is A=36π inches^2.

If you need to round, round this to wherever you need it: A=113.0973355292326... inches^2

If you're using 3.14 for pi, the answer is A=113.04 inches^2.

8 0

3 years ago

What is (6x^-2)^2 (0.5x)^4 and the steps to do it?

tresset_1 [31]

Square inside the first parentheses,, then square outside both.
Answer: 2.25

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4 years ago

Whats the roots of the following three problems:

klemol [59]

QUESTION 1

The given function is

$f(x)=\frac{(x-3)(x-1)}{(x-1)(x+2)}$

We simplify to get;

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

This function is equal to zero when $x-3=0$

$\Rightarrow x=3$

QUESTION 2

The given function is

$f(x)=\frac{5x^2-10x+5}{2x^2-5x+3}$

$f(x)=\frac{5(x^2-2x+1)}{2x^2-5x+3}$

We factor to get;

$f(x)=\frac{5(x-1)^2}{(2x-3)(x-1)}$

$f(x)=\frac{5(x-1)}{(2x-3)}$

This function equals zero when

$5(x-1)=0$

$x=1$

QUESTION 3

The given function is

$f(x)=\frac{x^3-8}{x^2-6x+8}$

We factor to get,

$f(x)=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+4)}$

$f(x)=\frac{x^2+2x+4}{x+4}$

The function equals zero when $x^2+2x+4=0$

$D=b^2-4ac$

$D=2^2-4(1)(4)$

$D=-12$

Hence the equation has no real roots.

The complex roots are

$x=-\sqrt{3}i-1$ or $x=-1+\sqrt{3}i$

4 0

3 years ago