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Tom [10]
3 years ago
13

If a = -9 and b = 4, what is the value of a + b ? A. -5 B. 5 C. -13 D. 13

Mathematics
1 answer:
madreJ [45]3 years ago
7 0

Answer:

A. -5

Step-by-step explanation:

-9 + 4 = -5

hope this helps

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The approximate distance between J and K is units. The approximate distance between K and L is units. The approximate distance b
Dima020 [189]
Count the distance d, which gives you 7.

Use Pythogarean theorem to find c:
{a}^{2}  +  {b}^{2}   =   {c}^{2}
Plug in the numbers:
{4}^{2}  +  {3}^{2}  =  {c}^{2}   \\  \\ = 16 + 9  \\  \\ = 25 \\  \\ c =  \sqrt{25}  = 5
c is 5.

Same thing with g:
{e}^{2}  +  {f}^{2}  =  {g}^{2}  \\  \\    {g}^{2} = {4}^{2}  +  {4}^{2}  \\  \\  = 16 + 16 \\  \\  = 32 \\  \\ g =  \sqrt{32} = 4 \sqrt{2}  = 5.66
g equals 5.66.

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3 years ago
Lisa draws a circle by tracing around the bottom of block. which could be the shape of Lisa's block
Andrej [43]
Let's think of something that one can hold against a page and draw a circle. Some examples are: a cup, a D battery, a can of soda, the tube from the inside of a paper towel roll, a can of beans, etc.

Think of the can of beans. The part that touches the page (and that you trace around with your pencil) is called a face.What these items have in common is that the faces at the ends are circles (they may or may not be the same size).

The name for this 3-D figure is called a cylinder. Her block, therefore, is a cylinder.

Technically, if the ends were ovals we would still call it a cylinder and so to make sure you have the one with the circles at the ends you would say you have a "right circular cylinder" but for most cases people just say "cylinder" and assume the ends are circles. It really depends what level (elementary, middle school, hs, college) of math you are doing whether just cylinder suffices.

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Which BEST represents the height of a home?
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Answer:

D)

Step-by-step explanation:

8 0
2 years ago
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We have the equation:

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By arranging this equation in terms of x and y, we have:


x^2-8x+y^2-6y=-24 \\ \\


By using the method of completing the square, we have:

x^2-8x+\mathbf{\left(\frac{8}{2}\right)^2}+y^2-6y+\mathbf{\left(\frac{6}{2}\right)^2}=-24+\mathbf{\left(\frac{8}{2}\right)^2}+\mathbf{\left(\frac{6}{2}\right)^2} \\ \\ x^2-8x+\mathbf{16}+y^2-6y+\mathbf{9}=-24+\mathbf{16}+\mathbf{9} \\ \\ \boxed{(x-4)^2+(x-3)^2=1}


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Finally, the right answer is c)

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No solution exists for -3x+y=-18 and -3x+y=-6
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