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iVinArrow [24]
3 years ago
12

When a number is decreased by 3.5%, the result is 91 what is the original number to the nearest tenth

Mathematics
2 answers:
sdas [7]3 years ago
4 0

Answer:

87.8

Step-by-step explanation:

So you have to multiply 91 times 3.5 which equals 318.5

                      91 × 3.5 = 318.5

Then divide 318.5 by 100 which equals 3.185

                      318.5 ÷ 100 = 3.185

Next subtract 91 by 3.185 which equals 87.815

                      91 - 3.185 = 87.815

Last Round 87.815 to 87.8

                       87.8 should be the answer

Hope this helps

rodikova [14]3 years ago
3 0

Answer:

The number is 94.3 (to the nearest tenth)

Step-by-step explanation:

Let the number =x

When x is decreased by 3.5%, we have:

(100% of x)-(3.5% of x)=(100-3.5)% of x =96.5% of x

When a number is decreased by 3.5%, the result is 91.

Therefore:

96.5% of x = 91

Converting percentage to decimal: 96.5%=0.965

0.965x = 91

Divide both sides by 0.965

x=94.3

The number is 94.3 (to the nearest tenth)

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5 0
3 years ago
The green triangle is a dilation of the red triangle with a scale factor of s=1/3 and the center of dilation is at the point (4,
klasskru [66]

Given:

The scale factor is s=\dfrac{1}{3} and the center of dilation is at the point (4,2).

Red is original figure and green is dilated figure.

To find:

The coordinates of point C' and point A.

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Rule of dilation: If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

According to the given information, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let us assume the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

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And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

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