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pychu [463]
3 years ago
11

on field day, 1/10 of the students in Mrs. Brown's class competes in jumping events, 3/5 of the student competes in running even

ts, and 1/10 competed in throwing events. what part of mrs. brown's class did not compete in jumping, running or throwing events?
Mathematics
2 answers:
hjlf3 years ago
6 0
1/10 + 1/10 = 2/10
10 / 5 = 2
3/5 = 6/10
2/10 + 6/10 = 8/10
10/10 - 8/10= 2/10
Vikentia [17]3 years ago
6 0
I think the answer is 2/10
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Let S = {A, B, C, D} be some set of four elements of a vector space. Suppose that D = 2A + B + 3C and C = A − B. Is {A, B, D} li
Olin [163]

Answer:

As explained

Step-by-step explanation:

given that D = 2A + B + 3C and S = {A, B, C, D}, C = A − B

for {A, B, D}  to be linearly dependent implies that A B and D will have values that not dependent on the sample space C.

if C = A -B is substituted in the original equation, the generated equation, 5A -2B will have values each as against C which is linearly independent.

Also, D will have values that are linearly dependent.

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3 years ago
A flare is launched from a life raft with an initial upward velocity of 192 feet per second. How many seconds will it take for t
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7 0
3 years ago
Find the area of the region which is inside the polar curve r=5sin(θ) but outside r=4. Round your answer to four decimal places
natka813 [3]

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be 3.75 square units.

<h3>What is an area bounded by the curve?</h3>

When the two curves intersect then they bound the region is known as the area bounded by the curve.

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be

Then the intersection point will be given as

\rm 5 \sin \theta  = 4\\\\\theta = 0.927 , 2.214

Then by the integration, we have

\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214}[ (5 \sin \theta)^2 - 4^2] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [25\sin ^2 \theta - 16] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [ \dfrac{25}{2}(1 - \cos 2\theta ) - 16] d\theta \\

\rightarrow \dfrac{1}{2} [\dfrac{25 \theta }{5} - \dfrac{25 \cos 2\theta }{2} - 16\theta]_{0.927}^{2.214} \\\\\\\rightarrow \dfrac{1}{2} [\dfrac{25(2.214 - 0.927) }{5} - \dfrac{25 (\cos 2\times 2.214 - \cos 2\times 0.927) }{2} - 16(2.214 - 0.927]\\

On solving, we have

\rightarrow \dfrac{1}{2} \times 7.499\\\\\rightarrow 3.75

Thus, the area of the region is 3.75 square units.

More about the area bounded by the curve link is given below.

brainly.com/question/24563834

#SPJ4

5 0
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