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vladimir1956 [14]
3 years ago
14

What percent of 70 is 28?

Mathematics
2 answers:
Finger [1]3 years ago
6 0

Answer:

40%

Step-by-step explanation:

28/70*100=40

Mumz [18]3 years ago
5 0

Answer: 40%

Step-by-step explanation:

This represents this equation:

70*x =28

x=28/70=0,4

Multiply that with 100 to get the percentage.

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Line n is parallel to Line P. Find the measure of each angle.<br> Please help
Margaret [11]

Answer:

m<1=140

m<2= 40

m<3= 140

m<4= 40

m<5=140

m<6= 40

m<7= 140

4 0
2 years ago
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OMG PLEASE HELP ME PLESE
slava [35]

Answer:

use demos

Step-by-step explanation:

7 0
3 years ago
The number of pieces of popcorn in a large movie theatre popcorn bucket is normally distributed, with a mean of 1515 and a stand
Murrr4er [49]

Answer: 99.7%, Im pretty confident about that. Im taking the test right now.

Step-by-step explanation:

3 0
3 years ago
17
Aleksandr-060686 [28]

Answer:

The expressions that give the value of y are A - 3B and (1/3)A - B

The solution is (27/13, -60/13)

Step-by-step explanation:

We can see both equation A and equation B.

Equation A: 2x + (1/4)y = 3

Equation B: (2/3)x - y = 6

To find the value of y, we have to solve both equations A and equation B simultaneously. This is done by multiplying equation B by 3 and subtracting from equation A (A - 3B) to get:

(13/4)y = -15

y = -60/13

you can also get y by dividing equation A by 3 and subtracting equation B (1/3A - B)

Put y = -60/13 in equation A to get x:

2x + (1/4)(-60/13) = 3

2x = 3 + 15/13

2x = 54/13

x = 27/13

The solution is (27/13, -60/13)

7 0
3 years ago
Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 2t, 1 + 6t, 1 + 14t . Find the points at which their p
rusak2 [61]

Answer:

DNE

Step-by-step explanation:

Given that two particles travel along the space curves

r_1(t) = (t, t^2, t^3)\\ r_2(t) = (1 + 2t, 1 + 6t, 1 + 14t )

To find the points of intersection:

At points of intersection both coordinates should be equal.

i.e. r1 =r2

Equate corresponding coordinates

t=1+2t\\t^2=1+6t\\t^3=1+14t

I equation gives t =-1

Substitute in II equation to get t^2 = -5

i.e. t cannot be real

Hence no point of intersection

DNE

8 0
3 years ago
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