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4 years ago

(2x² - 5) - (3x² + 4).

Mathematics

1 answer:

Goshia [24]4 years ago

3 0

(2x² - 5) - (3x² + 4).

Pretend that there is a -1 in front of :(3x² + 4).

(2x^2-5)-1(3x^2+4)

2x^2-5-3x^-4

2x^2-3x^2-5-4

=-x^2-9

Answer: -x^2-9

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Find the perimeter of quadrilateral PQRS with the vertices P(2,4), Q(2,3), R(-2,-2), and S(-2,3).

storchak [24]

Answer:

$P=16.53\ units$

Step-by-step explanation:

we know that

The perimeter of quadrilateral PQRS is equal to the sum of its four length sides

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

the vertices P(2,4), Q(2,3), R(-2,-2), and S(-2,3)

step 1

Find the distance PQ

P(2,4), Q(2,3)

substitute in the formula

$d=\sqrt{(3-4)^{2}+(2-2)^{2}}$

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

$d=\sqrt{1}$

$dPQ=1\ unit$

step 2

Find the distance QR

Q(2,3), R(-2,-2)

substitute in the formula

$d=\sqrt{(-2-3)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-5)^{2}+(-4)^{2}}$

$dQR=\sqrt{41}\ units$

step 3

Find the distance RS

R(-2,-2), and S(-2,3)

substitute in the formula

$d=\sqrt{(3+2)^{2}+(-2+2)^{2}}$

$d=\sqrt{(5)^{2}+(0)^{2}}$

$dRS=5\ units$

step 4

Find the distance PS

P(2,4), S(-2,3)

substitute in the formula

$d=\sqrt{(3-4)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-1)^{2}+(-4)^{2}}$

$dPS=\sqrt{17}\ units$

step 5

Find the perimeter

$P=PQ+QR+RS+PS$

substitute the values

$P=1+\sqrt{41}+5+\sqrt{17}$

$P=6+\sqrt{41}+\sqrt{17}$

$P=16.53\ units$

5 0

4 years ago

You are considering two job offers. The first job pays $8.50 per hour. The second job pays $7.00 per hour plus $0.25 per unit pr

Sauron [17]

Answer:

Step-by-step explanation:

you just keep counting up and you would get $8.75 will be the money you get but 7 times you have to produce

4 0

3 years ago

Write -72=2y^2-16y-20x in standard form

kifflom [539]

Answer:

-72=-12y-20x

Step-by-step explanation:

-72=2y^2-16y-20x

square the 2 keeping the variable

-72=4y-16y-20x

Then combine like terms in this case the numbers followed by y's

-72=-12y-20x

This is the standard form as there is nothing else you can combine to break this down any further

5 0

3 years ago

Read 2 more answers

Part A

Deffense [45]

Answer: Not safe

Step-by-step explanation:

Let the angle of elevation of the ramp be θ, then:

$\begin{aligned}&\sin \theta=\frac{1.5}{10}=0.15 \\&\theta=\sin ^{-1}(0.15) \approx 8.63^{\circ}\end{aligned}$

But the maximum angle for the ramp cannot exceed 7.1°. The calculated angle of elevation will make it difficult, if not impossible, for motorized wheelchairs to get up the ramp.

Therefore, the ramp at the local library is not safe for motorized wheelchairs

5 0

2 years ago

assume the substance has a half-life of 11 years and the initial amount is 126 grams.How long will it be until only 15 % remains

Mumz [18]

(126) times (1/2 to the x/11 power) = 15% times 126

(1/2) to the x/11 power = 0.15

Take the log of both sides :

(x/11) times log(1/2) = log(0.15)

Multiply both sides by 11 :

'x' times log(1/2) = 11 x log(0.15)

Divide both sides by " log(1/2) " :

x = 11 x log(0.15)/log(1/2) = 30.107 years (rounded)

That's the time it takes for any-size sample of this substance
to decay to 15% of its original size.

5 0

3 years ago