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3 years ago

Fuaad solved an absolute value inequality and expressed the solution as -12

1 answer:

5 0

Inequalities require a sign of greater than, less than, greater than or equal to, or less than or equal to. It's never just a single number solution. It's always going to be a range of solutions

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ary Stevens earns $6 an hour at her job and is entitled to time-and-a-half for overtime and double time on holidays. Last week s

Serggg [28]

The answer is D 394.50

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3 years ago

Find the area of the right triangle △DEF with the points D (0, 0), E (1, 1), and F.

Tems11 [23]

Answer:

The area of the triangle is $\sqrt{3}$

Step-by-step explanation:

Given:

Coordinates D (0, 0), E (1, 1)

Angle ∠DEF = 60°

△DEF is a Right triangle

To Find:

The area of the triangle

Solution:

The area of the triangle is = $\frac{1}{2}(base \times height)$

Here the base is Distance between D and E

calculation the distance using the distance formula, we get

DE = $\sqrt{(0-1)^2 + (0-1)^2}$

DE = $\sqrt{(-1) ^2 + (-1)^2$

DE = $\sqrt{1+1}$

DE = $\sqrt{2}$

Base = $\sqrt{2}$

Height is DF

DF = $tan(60^{\circ}) \times DE$

DF = $\sqrt{3} \times DE$

DF = $\sqrt{3} \times\sqrt{2}$

Now, the area of the triangle is

= $\frac{1}{2}({\sqrt{2})(\sqrt{3} \times \sqrt{2})$

= $\frac{1}{2}({\sqrt{2})(\sqrt{3} \sqrt{2})$

= $\frac{1}{2}(2\sqrt{3} )$

= $\sqrt{3}$

6 0

2 years ago

Please help quickly

Crank

Answer:25 m

Step-by-step explanation:

3 0

3 years ago

The value of 63−−√ lies between which two consecutive integers?

Gekata [30.6K]

The square root of 63 is about 7.9, which is between 7 and 8

3 0

3 years ago

1. Find the Least Common Multiple of these two monomials: See picture

gtnhenbr [62]

Answer:

The last choice is correct

$LCM=120a^4b^7c^5$

Step-by-step explanation:

Least Common Multiple (LCM)

To find the LCM we can follow this procedure:

List the prime factors of each monomial.

Multiply each factor the greatest number of times it occurs in either factor.

We have two monomials:

$12a^4b^2c^5$

$40a^3b^7c^1$

The prime factors of the first monomial are:

$2^2,3,a^4,b^2,c^5$

The prime factors of the second monomial are:

$2^3,5,a^3b^7c^1$

LCM = Multiply $2^3*3*5*a^4*b^7*c^5$

These are all the factors the greatest number of times they occur.

Operating:

$LCM=8*15*a^4*b^7*c^5$

$\boxed{LCM=120a^4b^7c^5}$

The last choice is correct

3 0

3 years ago