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Paul [167]
3 years ago
15

A worker cut pieces of string that are each 3/5 yard in length how many pieces of string can the worker cut from a piece of stri

ng that is 1/4 yard in length
Mathematics
1 answer:
Jet001 [13]3 years ago
6 0

We cannot cut any pieces from the 1/4yard string as it is smaller than the 3/5 yard pieces we want to cut.

<u>SOLUTION: </u>

It is given to us that a worker cut pieces of string that are each 3/5 yard in length.

We have been asked to find out how many pieces of string can the worker cut from a piece of string that is 1/4 yard in length.

The total length of the string available for cutting is \frac{1}{4} yard.

We shall convert \frac{1}{4} into decimal form for the simplicity of the calculation. Which gives us:

= \frac{1}{4}

= 0.25 yard

Now the small pieces that want to be cut of 0.25 yard are given as \frac{3}{5} yards. So, now we convert \frac{3}{5} yard into decimal as well for uniformity.

= \frac{3}{5}

= 0.6 yard

So from the calculations we can see that 0.25 yard is smaller than 0.6 yard.

Therefore, we cannot cut any pieces from the \frac{1}{4} yard string as it is smaller than the pieces we want to cut.

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The measure of angle 1 is 65 degrees. What are the measures of the other seven angles? (Enter the
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Answer:

See solution below

Step-by-step explanation:

According to the diagram shown

m<1 = m<5 = 5=65 degrees (corresponding angle)

m<5 = m<4 - 65 degrees (alternate interior angle)

m<9 = m<8 = 65degrees (corresponding angle)

m<5 = m<8 = 65dgrees (vertically opposite angles)

m<6+m<8 = 180

m<6 + 65 = 180

m<6 = 180 - 65

m<6 = 115degrees

m<2 = m<6 = 115degrees (corresponding angles)

m<6 = m<7 = 115degrees (vertically opposite angles)

m<3 = m<7 = 115degrees(corresponding angle)

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3 years ago
The total stopping distance T for a vehicle is T = 2.5x + 0.5x 2 , where T is in feet and x is the speed in miles per hour. Appr
Kobotan [32]

Answer:

%  change in stopping distance = 7.34 %

Step-by-step explanation:

The stooping distance is given by

T = 2.5 x + 0.5 x^{2}

We will approximate this distance  using the relation

f (x + dx) = f (x)+ f' (x)dx

dx = 26 - 25 = 1

T' =  2.5 + x

Therefore

f(x)+f'(x)dx = 2.5x+ 0.5x^{2} + 2.5 +x

This is the stopping distance at x = 25

Put x = 25 in above equation

2.5 × (25) + 0.5× 25^{2} + 2.5 + 25 = 402.5 ft

Stopping distance at x = 25

T(25) = 2.5 × (25) + 0.5 × 25^{2}

T(25) = 375 ft

Therefore approximate change in stopping distance = 402.5 - 375 = 27.5 ft

%  change in stopping distance = \frac{27.5}{375} × 100

%  change in stopping distance = 7.34 %

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Answer:

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2:D(6,-3) E(2,-3) F(3, -4.5) G(5, -4.5)

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Step-by-step explanation:

Before we begin, lets define the transformations:

(x, y)

T(a, b) = (x+a, y +b)

Rxaxis = (x, -y)

Ryaxis = (-x, y)

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D.5(1/2x, 1/2y)

R(-90, O) = (-y, x)

1. T. A(6,-1) B(14,0) C(11,4)

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2. Rx D(-12,-6) E(-4,-6) F(-6, -9) G(-10, -9)

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3. Rx=1 H(-1,2) I(4, 2) J(4,-2) K(-1,-2)

   T H(1,4) I(6, 4) J(6,0) K(1,0)

   r-90 H(-4,1) I(-4, 6) J(0,6) K(0,1)

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