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yuradex [85]
3 years ago
12

Use four unit multipliers to convert 170 square yards to square inches The perimeter of a square is 64 inches ​

Mathematics
1 answer:
blsea [12.9K]3 years ago
8 0
Did you know u can use a tutor on this app it’s very helpful :)
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Find the midpoint of A and B where A has coordinates (2, 4)<br> and B has coordinates (-3, -9).
tatuchka [14]

Answer:

(-0.5,-2.5)

Step-by-step explanation:

  • (x1 + x2) / 2 = x midpoint
  • (y1 + y2) / 2 = y midpoint
<h3>X:</h3>

⇒ 2 + -3 = 5

⇒ 5 / 2 = -0.5

<h3>Y:</h3>

⇒ 4 + -9 = -5

⇒ -5 / 2 = -2.5

→ = (-0.5, -2.5)

4 0
3 years ago
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I need this answered please
Basile [38]
I think is gonna be B


3 0
3 years ago
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Write the expression in complete factored form 2a(x-8) + q(x-8)
Mnenie [13.5K]

Answer:

x-8 out of 2a(x-8)+q(x-8)

Step-by-step explanation:


7 0
3 years ago
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Howard is designing a chair swing ride. The swing ropes are 4 44 meters long, and in full swing they tilt in an angle of 2 3 ∘ 2
qaws [65]

Question:

Howard is designing a chair swing ride. The swing ropes are 4 meters long, and in full swing they tilt in an angle of 23°. Howard wants the chairs to be 3.5 meters above the ground in full swing. How tall should the pole of the swing ride be? Round your final answer to the nearest hundredth.

Answer:

7.18 meters

Step-by-step explanation:

Given:

Length of rope, L = 4 m

Angle = 23°

Height of chair, H= 3.5 m

In this question, we are to asked to find the height of the pole of the swing ride.

Let X represent the height of the pole of the swing ride.

Let's first find the length of pole from the top of the swing ride. Thus, we have:

cos \theta = \frac{h}{L}

Substituting figures, we have:

cos(23) = \frac{h}{4}

Let's make h subject of the formula.

h = 4cos(23) = 3.68

The length of pole from the top of the swing ride is 3.68 meters

To find the height of the pole of the swing ride, we have:

X = h + H

X = 3.68 + 3.5

X = 7.18

Height of the pole of the swing ride is 7.18 meters

3 0
3 years ago
please help me, Prove a quadrilateral with vertices G(1,-1), H(5,1), I(4,3) and J(0,1) is a rectangle using the parallelogram me
mestny [16]

Answer:

Step-by-step explanation:

We are given the coordinates of a quadrilateral that is G(1,-1), H(5,1), I(4,3) and J(0,1).

Now, before proving that this quadrilateral is a rectangle, we will prove that it is a parallelogram. For this, we will prove that the mid points of the diagonals of the quadrilateral are  equal, thus

Join JH and GI such that they form the diagonals of the quadrilateral.Now,

JH=\sqrt{(5-0)^{2}+(1-1)^{2}}=5 and

GI=\sqrt{(4-1)^{2}+(3+1)^{2}}=5

Now, mid point of JH=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})

=(\frac{5+0}{2},\frac{1+1}{2})=(\frac{5}{2},1)

Mid point of GI=(\frac{5}{2},1)

Since, mid point point of JH and GI are equal, thus GHIJ is a parallelogram.

Now, to prove that it is a rectangle, it is sufficient to prove that it has a right angle by using the Pythagoras theorem.

Thus, From ΔGIJ, we have

(GI)^{2}=(IJ)^{2}+(JG)^{2}                             (1)

Now, JI=\sqrt{(4-0)^{2}+(3-1)^{2}}=\sqrt{20} and GJ=\sqrt{(0-1)^{2}+(1+1)^{2}}=\sqrt{5}

Substituting these values in (1), we get

5^{2}=(\sqrt{20})^{2}+(\sqrt{5})^{2} }

25=20+5

25=25

Thus, GIJ is a right angles triangle.

Hence, GHIJ is a rectangle.

Also, The diagonals GI=\sqrt{(4-1)^{2}+(3+1)^{2}}=5  and HJ=\sqrt{(0-5)^2+(1-1)^2}=5 are equal, thus, GHIJ is a rectangle.

6 0
3 years ago
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