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yaroslaw [1]
3 years ago
6

What the answer now

Mathematics
1 answer:
stiks02 [169]3 years ago
7 0

Answer:

35.6 yd²

Step-by-step explanation:

Area of ∆UVW can be solved if we know the lengths of 2 sides and their included angle.

We are Given just 1 side, UV (w). Use the law of sines to find UW (v).

Thus:

\frac{v}{sin(V)} = \frac{w}{sin(W)}

W = 137°

w = 19 yd

V = 180 - (137 + 22) = 21°  => sum of triangle

v = ??

Plug in the values and solve for v

\frac{v}{sin(21)} = \frac{19}{sin(137)}

Multiply both sides by sin(21)

\frac{v}{sin(21)}*sin(21) = \frac{19}{sin(137)}*sin(21)

v = \frac{19*sin(21)}{sin(137)}

v = 10 yd (approximated)

Find area of ∆UVW:

Area = ½*UV*UW*sin(U)

Area = ½*v*w*sin(U)

= ½*10*19*sin(22)

Area = 35.6 yd² (to nearest tenth)

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After 3.5 hours ,pasha had traveled 217 miles if she travels at the constant speed how far will she have travelled after 4 hours
vesna_86 [32]

Answer:

248 miles

Step-by-step explanation:

we know that

After 3.5 hours ,Pasha had traveled 217 miles

so

using proportion

Find out how many miles she will have traveled after 4 hours

\frac{3.5}{217}\ \frac{hours}{miles}=\frac{4}{x}\ \frac{hours}{miles}\\\\x=217(4)/3.5\\\\x=248\ miles

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3 years ago
Suppose j varies jointly with g and​ v, and j=2 when g=4 and v=3. <br> Find j when g=8 and v=9.
vladimir2022 [97]

Answer:

j=12

Step-by-step explanation:

we know that

In a join variation, If j varies jointly with respect to g and v, the equation will be of the form j = kgv

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step 1

Find the value of k

we have

j=2,g=4,v=3

substitute and solve for k

2 = k(4)(3)

2 = 12k

k=\frac{1}{6}

The equation is equal to

j=\frac{1}{6}gv

step 2

Find the value of j when g=8,v=9

substitute the values in the equation and solve for j

j=\frac{1}{6}(8)(9)

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5 0
3 years ago
Read 2 more answers
If E F = 7 , A C = 16 , and A F = 9
Vilka [71]

Answers:

CB = 14

GF = 8

FB = 9

EF is parallel to CB

====================================

Explanations:

Points E and F are midpoints of their respective sides. They form the midsegment EF. Because EF is a midsegment, A midsegment is half the length of its parallel counterpart, so CB is two times longer than EF. If EF is 7 units long, then CB = 2*EF = 2*7 = 14

For similar reasons, GF is parallel to AC. If AC = 16, then half of that is GF = (1/2)*AC = 0.5*16 = 8.

FB = FA = 9 as these segments have the same single tickmark to indicate they are the same length

EF is parallel to CB because EF is a midsegment, and this is one of the properties of being a midsegment. We can show that quadrilateral EGBF is a parallelogram to help prove this.

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