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4 years ago

Plzzzz help

Mathematics

2 answers:

timofeeve [1]4 years ago

4 0

Answer:

1) x=4

2) x<5

3) x=3

4) x ≤ -5

5) x=16.1

Step-by-step explanation:

andrew-mc [135]4 years ago

3 0

Question 1) x=4
the rest I rly dk

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Assalaamu alaikum...,

aleksandr82 [10.1K]

Answer:

see below

Step-by-step explanation:

alaikum salam.

2. find cos Ф, sin Ф, tan Ф

cos Ф = 5 / 13

Ф = 67.38

sin Ф = 12 / 13

Ф = 67.38

tan Ф = 12 / 5

Ф = 67.38

-----------------------

3. find sin β, cos β, tan β

sin β = 4 / 5

β = 53.13

cos β = 3 / 5

β = 53.13

tan β = 4 / 3

β = 53.13

4 0

4 years ago

Read 2 more answers

Suppose y varies jointly as x and z. What is y when x = 2 and z = 3, if y = 20 when x = 4 and z = 3?

Julli [10]

Answer:

y = 10

Step-by-step explanation:

Given y varies jointly as x and z then the equation relating them is

y = kxz ← k is the constant of variation

To find k use the condition y = 20 when x = 4 and z = 3 , then

20 = k × 4 × 3 = 12k ( divide both sides by 12 )

$\frac{20}{12}$ = k , that is

k = $\frac{5}{3}$

y = $\frac{5}{3}$ xz ← equation of variation

When x = 2 and z = 3 , then

y = $\frac{5}{3}$ × 2 × 3 = 5 × 2 = 10

7 0

2 years ago

What is less 2 over 5 or 7 over 15

Veseljchak [2.6K]

Answer:

2 over 5

Step-by-step explanation:

What you need to do is find a common denominator. In this case it is 15. 2 over 5 = 6 over 15 so since 6/15 < 7/15 and 6/15= 2/5, 2/5 < 7/15

8 0

4 years ago

Read 2 more answers

The harmonic motion of a particle is given by f(t) = 2 cos(3t) + 3 sin(2t), 0 ≤ t ≤ 8. (a) When is the position function decreas

iren [92.7K]

For the last part, you have to find where $f'(t)$ attains its maximum over $0\le t\le8$ . We have

$f'(t)=-6\sin3t+6\cos2t$

so that

$f''(t)=-18\cos3t-12\sin2t$

with critical points at $t$ such that

$-18\cos3t-12\sin2t=0$

$3\cos3t+2\sin2t=0$

$3(\cos^3t-3\cos t\sin^2t)+4\sin t\cos t=0$

$\cos t(3\cos^2t-9\sin^2t+4\sin t)=0$

$\cos t(12\sin^2t-4\sin t-3)=0$

So either

$\cos t=0\implies t=\dfrac{(2n+1)\pi}2$

$12\sin^2t-4\sin t-3=0\implies\sin t=\dfrac{1\pm\sqrt{10}}6\implies t=\sin^{-1}\dfrac{1\pm\sqrt{10}}6+2n\pi$

where $n$ is any integer. We get 8 solutions over the given interval with $n=0,1,2$ from the first set of solutions, $n=0,1$ from the set of solutions where $\sin t=\dfrac{1+\sqrt{10}}6$ , and $n=1$ from the set of solutions where $\sin t=\dfrac{1-\sqrt{10}}6$ . They are approximately