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3 years ago

What is the area of the trapezoidal window below? 14 inches 6inches 8inches

Mathematics

1 answer:

PSYCHO15rus [73]3 years ago

7 0

Please provide a picture or the given numeric angles/sides/degrees so that i can begin to solve.

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Michael has a total of 15 bills that are either $1 bills or $5 bills. If the total amount of money he has is $47, how many $5 bi

HACTEHA [7]

He has 8 five dollar bills and 2 dollar bills.

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2 years ago

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What is 20 times 20plz tell me

vladimir1956 [14]

Answer:

400

Step-by-step explanation:

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3 years ago

Use the table to identify the values of p and q that should be used to factor2 +3x-10 as (x+ p)(x q).

Nat2105 [25]

we are given

$x^2 +3x -10$

it's factor as (x+p)(x+q)

so, we can write as

$x^2 +3x -10= (x+p)(x+q)$

we can simplify it

$x^2 +3x -10= x^2 +(p+q)x +pq$

now, we can compare both sides

coefficient of x must be equal

so, we get

$3= (p+q)$

$(p+q)=3$

now, we can look at the table where p+q =3

so, we can see that when p+q=3 .... then p=-2 and q=5

so, option-A.......Answer

3 0

3 years ago

0.7 (3s + 4)-1.1s = 7.9

vichka [17]

Answer:

s = 5.1

Step-by-step explanation:

$0.7 (3s + 4)-1.1s = 7.9\\$

Multiply both sides of the equation by 10

$0.7\left(3s+4\right)\cdot \:10-1.1s\cdot \:10=7.9\cdot \:10\\\\= 7\left(3s+4\right)-11s=79\\$

Expand

$\mathrm{Expand\:}7\left(3s+4\right)-11s:\\\quad 10s+28$

Subtract 28 from both sides

$10s+28-28=79-28$

Simplify

$10s=51$

Divide both sides of the equation by 10

$\frac{10s}{10}=\frac{51}{10}$

Simplify

$s=\frac{51}{10}\\\\Decimal = 5.1$

3 0

2 years ago

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Can anybody help me on this please I will be giving 15 points..

oksano4ka [1.4K]

Answer:

BC= $\sqrt{95}$

Angle A (or BAC)=54.31466...

Angle C (or ACB)=35.68533...

Step-by-step explanation:

$\frac{7}{\sin \left(\angle \:ACB\right)}=\frac{12}{\sin \left(90\right)}$

$BC=\sqrt{12^2-7^2}$

$\sqrt{12^2-7^2}=\sqrt{95}$

$\frac{\sqrt{95}}{\sin \left(\angle \:BAC\right)}=\frac{12}{\sin \left(90\right)}\quad :\quad \angle \:BAC=54.31466$

$\frac{7}{\sin \left(\angle \:ACB\right)}=\frac{12}{\sin \left(90\right)}\quad :\quad \angle \:ACB=35.68533$

4 0

3 years ago

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