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AfilCa [17]
3 years ago
11

If you multiply any whole number by a fraction greater than one is your product greater than or less than the whole number

Mathematics
1 answer:
Alik [6]3 years ago
3 0
Your product would be less than because you put a 1 under the whole number and flip the second number
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Solve for <br> 3x + 12= -12
oee [108]

Answer:

x = -8

Step-by-step explanation:

Isolate the variable, x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

First, subtract 12 from both sides:

3x + 12 (-12) = -12 (-12)

3x = -12 - 12

3x = -24

Next, divide 3 from both sides:

(3x)/3 = (-24)/3

x = -24/3

x = -8

x = -8 is your answer.

~

4 0
3 years ago
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How to do number 15?
tatiyna
I think it would be 120 because it is 2 times 60 and they add up to 180 I may be wrong buts that’s what I would do
3 0
3 years ago
I will mark brainliest whoever gives the best answer
Nataliya [291]

Answer:

its a,b,c, or d

Step-by-step explanation:

4 0
3 years ago
How much longer is the hypotenuse of the triangle than its shorter leg?
RUDIKE [14]

This is an incomplete question, here is a complete question and image is also attached below.

How much longer is the hypotenuse of the triangle than its shorter leg?

a. 2 ft

b. 4 ft

c. 8 ft

d. 10 ft

Answer : The correct option is, (b) 4 ft

Step-by-step explanation:

Using Pythagoras theorem in ΔACB :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(AB)^2=(AC)^2+(BC)^2

Given:

Side AC = 6 ft

Side BC = 8 ft

Now put all the values in the above expression, we get the value of side AB.

(AB)^2=(6)^2+(8)^2

AB=\sqrt{(6)^2+(8)^2}

AB=10ft

Now we have to calculate the how much longer is the hypotenuse of the triangle than its shorter leg.

Difference = Side AB - Side AC

Difference = 10 ft - 6 ft

Difference = 4 ft

Therefore, the 4 ft longer is the hypotenuse of the triangle than its shorter leg.

5 0
3 years ago
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F(x) = x^2+3; g(x) = square root of x-2 find f(g(x))
Anettt [7]
F(g(x)) = x + 1 . Just replace square root of x-2 to x in f(x)
5 0
3 years ago
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