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Dmitriy789 [7]
3 years ago
8

Judy earns $65 for working five hours. find the constant rate of variation

Mathematics
2 answers:
d1i1m1o1n [39]3 years ago
7 0

Answer:

which principle prevents a brach from abusing  its power

Step-by-step explanation:

DanielleElmas [232]3 years ago
4 0

Answer:.

Step-by-step explanation:.

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What is the value of x is the rational expression below equal to zero? <br><br> 2-x/2+x
garri49 [273]

Answer:

X=-4

Step-by-step explanation:

Write the polynomial as an equation.

y

=

2

−

x

÷

2

+

x

Find the x-intercepts.

Tap for more steps...

x-intercept(s):  

(

−

4

,

0

)

Find the y-intercepts.

Tap for more steps...

y-intercept(s):  

(

0

,

2

)

List the intersections.

x-intercept(s):  

(

−

4

,

0

)

y-intercept(s):  

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2

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7 0
3 years ago
Dennis is cutting construction paper into rectangles for a project he needs to cut one rectangle that is 12 inches times 15 1/4
Nataly_w [17]

Answer:

Dennis need 288.91 square inches of construction paper for his project.

Step-by-step explanation:

Dennis is cutting construction paper into rectangles for a project.

He needs to cut one rectangle that is 12 inches times 15 1/4 inches

First convert the mixed fraction to improper fraction

15\frac{1}{4} = \frac{(15\times4) + 1}{4} = \frac{61}{4}

So the area of 1st rectangle is

A_1 = 12\times \frac{61}{4} = 183 \: in^{2}

He needs to cut another rectangle that is 10 1/3"" x 10 1/4""

The symbol " means inches

First convert the mixed fraction to improper fraction

10\frac{1}{3} = \frac{(10\times3) + 1}{3} = \frac{31}{3}

10\frac{1}{4} = \frac{(10\times4) + 1}{4} = \frac{41}{4}

So the area of 2nd rectangle is

A_2 = \frac{31}{3}\times \frac{41}{4} = 105.91 \: in^{2}

The total construction paper needed for this project is

A = A_1 + A_2\\\\A = 183 + 105.91\\\\A = 288.91 \: in^{2}

Therefore, Dennis need 288.91 square inches of construction paper for his project.

6 0
4 years ago
Look at the picture help please
Fofino [41]
X = 39ft
It’s the same units as house A it’s just a reflection
3 0
3 years ago
If you have 36 ft of fencing what are the area of the different rectangles you could enclose with the fencing?consider only whol
n200080 [17]

The fence you have must fit the perimeter of the rectangle.

With 36 feet of fence, these are the rectangles that you can enclose:

1-ft x 17-ft . . . Area = 17 ft²

2 x 16 . . . Area =  32 ft²
3 x 15 . . . Area =  45 ft²
4 x 14 . . . Area =  56 ft²
5 x 13 . . . Area =  65 ft²
6 x 12 . . . Area =  72 ft²
7 x 11 . . . Area =  77 ft²
8 x 10 . . . Area =  80 ft²
9 x 9 . . . Area =  81 ft²


6 0
4 years ago
Read 2 more answers
Factor completly: x^2-2x-24
melisa1 [442]
<span>x^2-2x-24 = (x - 6)(x + 4)

cause
</span>(x - 6)(x + 4)
= x^2 - 6x + 4x - 24
= x^2 - 2x - 24

answer
(x - 6)(x + 4)

hope it helps
4 0
4 years ago
Read 2 more answers
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