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natta225 [31]
3 years ago
10

If 40 percent of a number is 32 what is 25 percent of that number

Mathematics
1 answer:
gulaghasi [49]3 years ago
6 0
Your answer is 31.25
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Cesar, Carmen, and Dalila raised $95.34 for their tennis team. Carmen raised $12.12 less than Cesar, and Cesar raised $35 more t
NeTakaya

Answer:

Cesar's amount can be expressed by x + 12.12

Dalila's amount can be expressed as  x - 22.88

Step-by-step explanation:

If Carmen raised 12.12.less than Cesar, that means his amount is 12.12 more than hers so that is <u>x + 12.12</u> for <u>Cesar</u>.

If  Cesar raised 35 more than Dalila, her amount is Cesar's minus 35. Calculate: x+12.12 - 35. the difference is 22.88 less than x  so the amount for <u>Dalila is x- 22.88</u>

5 0
3 years ago
Can someone help me please?
alekssr [168]

Answer:

SHEEESH

Step-by-step explanation:

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3 years ago
D;the craft store discussed above, the clay cone is $12, the clay cylinder is $30, and the clay sphere is $28. Which is the best
777dan777 [17]

Answer:

the cylinder

Step-by-step explanation:

if you look at it the cylinder is the biggest and largest and take must of the space

3 0
1 year ago
The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2
maxonik [38]

Given:

The red figure dilated with a scale factor of s=\dfrac{1}{3} and the center of dilation is at the point (4,2) to get the green figure.

To find:

The coordinates of C' and A.

Solution:

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)

In given problem, 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 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)

m=12+4

m=16

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).

4 0
3 years ago
How do you solve 3k+2(5k-3)=7
slamgirl [31]

3k+2(5k-3)=7

=>3k+10k-6=7

=>13k=13

=>k=13/13

Therefore, k=1

8 0
3 years ago
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