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lina2011 [118]
3 years ago
10

secant AC and BD intersect at Point e inside F what is the measure of AED if measurements CD equals 30 a b is 50 and CB is 170

Mathematics
1 answer:
andrey2020 [161]3 years ago
6 0

Check the picture below.

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Cual es el resultado?/ who is the answer? <br> -m+4-(-m)=
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Answer:

-m+4-(-m) = -m+4+m \\                  = 4

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3 years ago
Which of the following are true for the number 8.5? Select all that apply
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The answer should be A and D
7 0
3 years ago
NEED ANSWER ASAP:
andriy [413]

Answer:

14/27 feet

Step-by-step explanation:

2 1/3 feet : 4 1/2 feet

= 7/3 : 9/2

= 7/3 ÷ 9/2

= 7/3 × 2/9

= (7*2)/(3*9)

= 14/27 feet

2 1/3 feet : 4 1/2 feet = 14/27 feet

8 0
3 years ago
Ms. Pacheco, Mr Edwards, and Mr Richards are three math teachers at turner middle school. Ms. Pacheco is three years older than
zubka84 [21]
Mr edwards = Mr r x 2
Mr r = Mr p - 3 = Mr e ÷ 2.                       Mr p = 30
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Mr p = Mr r + 3.                                         Mr r = 27

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Mr e = 2(mr p) - 6

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5 0
3 years ago
Read 2 more answers
A piece of wire 28 ft long is cut into two pieces. One piece is used to form a square, and the remaining piece is used to form a
marusya05 [52]

Answer:

Wire should be cut in two parts with the length = 15.69 ft and 12.31 ft

Step-by-step explanation:

Length of the wire = 28 ft has been cut in two pieces.

One piece is used to form a square and remaining piece to form a circle.

Let the length of the wire which forms the square is 'l' ft.

Area of the square = (side)²

Perimeter of the square = 4(side) = l

Length of one side = \frac{l}{4}

So, the area of the square = \frac{l^{2}}{16} ft²

Now length of the remaining part = perimeter of the circle = (28 - l) ft

2πr = (28 - l)

r = \frac{28-l}{2\pi }

Area of the circle formed = πr²

= \frac{\pi(28-l)^{2} }{4(\pi )^{2} }

= \frac{(28-l)^{2}}{4\pi }

Combined area of the square and circle

= \frac{l^{2}}{16} + \frac{(28-l)^{2}}{4\pi }

= \frac{l^{2}}{16} + \frac{(28-l)^{2}}{4\pi }

Now to maximize the area we will find the derivative of the area with respect to l.

\frac{dA}{dl}=\frac{d}{dl}[\frac{l^{2}}{16}+\frac{(28-l)^{2}}{4\pi}]

= \frac{d}{dl}[\frac{l^{2}}{16}+\frac{(28)^{2}+l^{2}-56l }{4\pi }]

= \frac{2l}{16}+\frac{(2l-56)}{4\pi }

Now equate the derivative to zero.

\frac{2l}{16}+\frac{(2l-56)}{4\pi } = 0

(2l-56)=-\frac{2l}{16}\times 4\pi

(2l-56)=-\frac{\pi l}{2}

2l+\frac{\pi l}{2}=56

\frac{(4l+l\pi )}{2}=56

l(4 + π) = 112

l(4 + 3.14) = 112

l = \frac{112}{7.14}

l = 15.69 ft

Length of the other part = 28 - 15.69 = 12.31 ft

Therefore, wire should be cut in two parts with the length = 15.69 ft and 12.31 ft

8 0
3 years ago
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