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Strike441 [17]
3 years ago
5

Quadrilateral A and B are scaled copies of each other. Quadrilateral A has side lengths of 2, 3, 5, and 6. Quadrilateral B's lon

gest side is 21. What is the perimeter of Quadrilateral B?
Mathematics
1 answer:
harkovskaia [24]3 years ago
4 0

Answer:

Quadrilateral A has side lengths 2, 3, 5, and 6. Quadrilateral B has side lengths 4, 5, 8, and 10. Could one of the quadrilaterals be a scaled copy of the other ...

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You are able to express numbers in like halves 
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Delray is starting a small computer business he apply to his bank for a loan of $12000. If the interest rate is 11% find the int
DENIUS [597]

I believe the answer would be $13,320.

Hope this helps ;}

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3 years ago
PLEASE HELP<br> A) 8<br> C) 1<br> B) 6<br> D) 7<br><br> What is the value of x?
SVEN [57.7K]

Answer: 7

Step-by-step explanation:

As tangents drawn from a common external point to a circle are congruent,

  • 8x - 7 = 6x + 7
  • 2x - 7 = 7 [subtract 2x from both sides]
  • 2x = 14 [add 7 to both sides]
  • x = 7 [divide both sides by 2]
5 0
2 years ago
F (n) = 65 - 100n<br> f (39) =
yan [13]

f(39)=65-100(39)

65-3900

= -3835

6 0
2 years ago
A post is supported by two wires (one on each side going in oppositedirections) creating an angle of 80° between the wires. The
Vladimir [108]

Using the Sine rule,

\frac{\sin A}{A}=\frac{\sin B}{B}=\frac{\sin C}{C}\begin{gathered} \text{Let A = 14m,} \\ Substituting the variables into the formula,Where the length of the wires are, AP = xm and BP = ym[tex]\begin{gathered} \frac{\sin80^0}{14}=\frac{\sin40^0}{x} \\ \text{Crossmultiply,} \\ x\times\sin 80^0=14\times\sin 40^0 \\ Divide\text{ both sides by }\sin 80^0 \\ x=\frac{14\sin40^0}{\sin80^0} \\ x=9.14m \end{gathered}

Hence, the length of wire AP (x) is 9.14m.

For wire BP (y)m,

Sum of angles in a triangle is 180 degrees,

A^0+P^0+B^0=180^0\begin{gathered} \text{Where A}^0=\text{ unknown,} \\ P^0=80^0\text{ and,} \\ B^0=40^0 \\ A^0+80^0+40^0=180^0 \\ A^0+120^0=180^0 \\ A^0=180^0-120^0 \\ A^0=60^0 \end{gathered}

Using the side rule to find the length of wire BP,

\begin{gathered} \frac{\sin 60^0}{y}=\frac{\sin 80^0}{14} \\ \text{Crossmultiply,} \\ y\times\sin 80^0=14\times\sin 60^0 \\ \text{Didive both sides by }\sin 80^0 \\ y=\frac{14\times\sin 60^0}{\sin 80^0} \\ y=12.31m \end{gathered}

Hence, the length of wire BP (y) is 12.31m

Therefore, the length of the wires are (9.14m and 12.31m).

4 0
1 year ago
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