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seraphim [82]
3 years ago
10

The area of the kite is 60in^2. Find the value of x.

Mathematics
1 answer:
Mashcka [7]3 years ago
6 0

Answer:

8 in

Step-by-step explanation:

Area = (d1 × d2)/2

d1 = 15 in

d2 = x in

60 = ( 15 * x)/2

120 = 15x

x = 120/15

= 8 in

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alexandr1967 [171]
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x= 7.5

I hope this helps!
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3 years ago
What is the interquartile range
motikmotik

<em>Answer:</em>

<em></em>

<em>Step-by-step explanation:</em>

<em>Khan academy huh?</em>

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Alex_Xolod [135]
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4 0
3 years ago
Find the angle between the given vectors to the nearest tenth of a degree. u = , v = (2 points)
Liono4ka [1.6K]

Answer:

<h2>3.6°</h2>

Step-by-step explanation:

The question is incomplete. Here is the complete question.

Find the angle between the given vectors to the nearest tenth of a degree.

u = <8, 7>, v = <9, 7>

we will be using the formula below to calculate the angle between the two vectors;

u*v = |u||v| cos \theta

\theta is the angle between the two vectors.

u = 8i + 7j and v = 9i+7j

u*v = (8i + 7j )*(9i + 7j )

u*v = 8(9) + 7(7)

u*v = 72+49

u*v = 121

|u| = √8²+7²

|u| = √64+49

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|v| = √9²+7²

|v| = √81+49

|v| = √130

Substituting the values into the formula;

121= √113*√130 cos θ

cos θ = 121/121.20

cos θ = 0.998

θ = cos⁻¹0.998

θ = 3.6° (to nearest tenth)

Hence, the angle between the given vectors is 3.6°

5 0
3 years ago
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prisoha [69]

Answer:

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