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Licemer1 [7]
4 years ago
15

CHRISTINE MAKES HOUSE CALLS. FOR EACH

Mathematics
1 answer:
kkurt [141]4 years ago
6 0

I'm sorry but what is your question?? I would like to help.

You might be interested in
2x + 5y + 30 = 0 The point ( 5 2 , y) is a solution to the equation shown. What is the value of y? A) -7 B) -5 C) 0 D) 5
Korolek [52]

2x + 5y + 30 = 0

2(52) + 5y + 30 = 0

104 + 5y + 30 = 0

134 + 5y = 0

         5y = -134

            y = -\frac{134}{5}

Neither A, B, C, or D are the the correct answer.  Did you mistype the problem?

5 0
3 years ago
Find −30÷(6a)-30÷(6a) when a=−5a=-5 . Write the answer in simplest form
Scorpion4ik [409]
Instead of waiting for someone to answer you try mathway
6 0
3 years ago
Using suitable identity find 97 ×103​
Ket [755]

Answer:

9,991

Step-by-step explanation:

97 \times 103 \\  \\  = (100 - 3)(100 + 3) \\  \\  {(100)}^{2}  -  {(3)}^{2}  \\  \\  = 10000 - 9 \\  \\  = 9,991

4 0
3 years ago
The area of a rectangular ceiling tile is 756 square inches. The perimeter is 114 inches. What are the dimensions of the tile?
kogti [31]

Here we need to use what we know about rectangles to make a system of equations.

By solving that system we found that the tile has a length of 36 inches and a width of 21 inches.

Remember that for a rectangle of length L and width W, the perimeter is:

P = 2*(L + W)

And the area is:

A = W*L

Here we know that the perimeter is 114 inches, then we can write:

114in = 2*(L + W)

We also know that the area is 756 in^2, then we can write:

756 in^2 = L*W

So we found two equations, which means that we have a system of two equations with two variables:

114in = 2*(L + W)

756 in^2 = L*W

To solve this, the first step is to isolate one of the variables in one of the equations, we can isolate L in the first equation:

114in = 2*(L + W)

114in/2 = (L + W)

57in = L + W

57in - W = L

Now that we have an expression equivalent to L, we can replace it in the other equation to get:

756 in^2 = L*W

756 in^2 = (57in - W)*W

Now we can solve this for W.

756 in^2 = W*57in - W^2

W^2 - W*57in + 756 in^2 = 0

The solutions are given by the Bhaskara's formula:

W = \frac{57in \pm \sqrt{(-57in)^2 - 4*1*(756in^2)} }{2*1} = \frac{57in \pm 15in}{2}

Then the two possible values of the width will be:

W = (57in + 15in)/2 =  36 in

W = (57in - 15in)/2 = 21 in

Suppose that we choose the second solution, W = 21in

Now using the equation 57in - W = L we can find the value of L

L = 57in - W = 57in - 21in = 36in

L = 36in

Then we found that the tile has a length of 36 inches and a width of 21 inches.

If you want to learn more, you can read:

brainly.com/question/11137975

4 0
3 years ago
The length of a rectangle is 22 centimeters and the width is
nataly862011 [7]

Answer:

78cm

Step-by-step explanation:

L₁/W₁ = L₂/W₂

22/4 = L₂/6

5.5*6=L₂

L₂=33cm

Perimeter of rectangle = 2L+2W

=2(33)+2(6)

=66+12

=78cm

7 0
3 years ago
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