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valkas [14]
3 years ago
6

The students in Suzanne's school are painting a rectangular mural outside the building that will be 15 feet by 45 feet. Scale dr

awing of the mural is shown. The diagonal is approximately 6.3 units.Write the unit rate for the proportional relationship between lengths on the mural y and lengths in the scale drawing x.
Mathematics
1 answer:
Katarina [22]3 years ago
3 0

Answer:

The dimensions on paper are 1.992ft by 5.976ft with a scale factor of 7.53

Step-by-step explanation:

The first step will be to find the diagonal of the rea life mural.

We can use Pythagoras' Theorem to do this.

Diagonal = \sqrt{15^2 +45^2 } =47.43 feet.

Now we have the real-life diagonal, we will now relate the diagonal of the painting outside with the one on paper. We can do this by dividing the two diagonals.

This will be 47.43 / 6.3 units = 7.53.

The scale factor is 7.53

To get the dimensions of the length and the breadth on paper, we divide the outside dimensions by the scale factor.

This will be

Length = 15/ 7.53 = 1.992

Breadth = 45/7.53 = 5.976

Therefore, the dimensions on paper are 1.992ft by 5.976ft with a scale factor of 7.53

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Need help ASAP! Choose the numbers that are irrational. 1. 0. 87 repeating 2. 3.624 3. √2 4. 0. 6 repeating 5. -5 6. 1 over 3 7.
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Answer:

√2, pi.

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An irrational number cannot be written as a fraction a/b where a and b are integers ( not = 0)..

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Which statement proves that the diagonals of square PQRS are perpendicular bisectors of each other?
scZoUnD [109]
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3 years ago
A ball is kicked 4 feet above the ground with an initial vertical velocity of 55 feet per second. The function h(t)=−16t2+55t+4
kramer

<u>Given</u>:

A ball is kicked 4 feet above the ground with an initial vertical velocity of 55 feet per second.

The function h(t)=-16t^2+55t+4 represents the height h(in feet) of the ball after t seconds.

We need to determine the time of the ball at which it is 30 feet above the ground.

<u>Time:</u>

To determine the time that it takes for the ball to reach a height of 30 feet above the ground, let us substitute h(t) = 30, we get;

30=-16t^2+55t+4

26=-16t^2+55t

Adding both sides of the equation by 16t², we get;

16t^2+26=55t

Subtracting both sides of the equation by 55t, we have;

16t^2-55t+26=0

Let us solve the quadratic equation using the quadratic formula, we get;

t=\frac{-(-55) \pm \sqrt{(-55)^2-4(16)(26)}}{2(16)}

t=\frac{55 \pm \sqrt{3025-1664}}{32}

t=\frac{55 \pm \sqrt{1361}}{32}

t=\frac{55 \pm 36.89}{32}

t=\frac{55 + 36.89}{32} \ or \ t=\frac{55- 36.89}{32}

t=\frac{91.89}{32} \ or \ t=\frac{18.11}{32}

t=2.9 \ or \ t=0.6

The value of t is t = 0.6 because this denotes the time taken by the ball to reach a height of 30 feet from the ground.

Therefore, the time taken by the ball to reach a height of 30 feet above the ground is 0.6 seconds.

5 0
3 years ago
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