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andreev551 [17]
2 years ago
13

35 per cent of a number is what fraction of that number

Mathematics
1 answer:
motikmotik2 years ago
3 0

Answer:

35/100, or 7/20 in simplest form

Step-by-step explanation:

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If tan0=3/4 find sin0
krok68 [10]

Answer:

H2=02+A2

H2=9+16

=25

Step-by-step explanation:

sino =0/H

3/5

5 0
3 years ago
Find the surface area of the triangular prism please.
Kobotan [32]
Surface area =  area of 3 rectangles + 2 triangles

            =  3*(3*7) + 2 * ( 1/2 * 3 * 2.6)

            =   63 +   2* 3.9

            =   70.8  in^2

3 0
3 years ago
The length of a rectangle is 8 inches more than twice its width. The area of the rectangle is 120 square inches. How many inches
Irina-Kira [14]

Answer:

L = 20 inches

Step-by-step explanation:

w = width

L = length

area = 120

W x L = 120

L - 8 = 2W  then L = 2W + 8

substitute for L:

W x (2W + 8) = 120

2W² + 8W -120 = 0

(2W - 12)(W + 10) = 0

2W-12 = 0

2W = 12

W = 6

L = 20

7 0
3 years ago
Zohar is using scissors to cut a rectangle with a length of 5x-2 and a width of 3x+1 out of a larger piece of paper . The perime
Dmitriy789 [7]
The total distance if x=4 is 62
7 0
3 years ago
If a ball is thrown directly upward with a velocity of 50ft/s, it's height (in feet) after t seconds is given by y=50t-16t^2 wha
Tomtit [17]

Answer:

39.0625 ft

Step-by-step explanation:

Since the function is a quadratic representing height, and the coefficient of the t² is negative, the vertex of the parabola will be the maximum height achieved by the ball.  

The general form for a quadratic equation is ax² + bx + c, here a is -16, and b is 50

To find the x coordinate of the vertex, use   x = -b/(2a)

We have x = -50/[2(-16)]

               

                  x = -50/-32

             

                         x= 25/16

Now plug that into the equation to find the y value, which will be the height...

y = 50(25/16) - 16(25/16)²

    y = 1250/16 - 16(625/256)

        y = 1250/16 - 625/16

              y = 625/16

                   y = 39.0625

7 0
3 years ago
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