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3 years ago

Can you please help on this problem

Mathematics

1 answer:

ArbitrLikvidat [17]3 years ago

4 0

This will be 15/8 because you cancel the common factors

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Brainliest fir CORRECT answer

viktelen [127]

Answer:

y ≈ 5.2

Step-by-step explanation:

∠ F = 180° - (42 + 48)° ← sum of angles in a triangle

∠ F = 180° - 90° = 90°

Thus Δ DEF is right at F

Using the sine ratio in the right triangle

sin48° = $\frac{opposite}{hypotenuse}$ = $\frac{DF}{DE}$ = $\frac{y}{7}$ ( multiply both sides by 7 )

7 × sin48° = y , then

y ≈ 5.2 ( to the nearest tenth )

4 0

3 years ago

A ball is kicked straight up into the air from a height of 48ft with an initial velocity of 88 ft/s. After how many seconds does

AlladinOne [14]

Answer:

Step-by-step explanation:

5 0

3 years ago

What is the probability that 2 randomly selected months have 31 days?

aniked [119]

Answer:

31.8%

Step-by-step explanation:There are 7 months that have 31 days so,

The odds the first selected month has 31 days is 7/12

The next month there are only 6 months left with 31 days of only 11 months to choose from

so the odds for the second are 6/11

To calculate the odds they BOTH have 31 months you multiply the two odds:

(7/12) x (6/11) = 42/132 = .318 so 31.8%

6 0

3 years ago

Read 2 more answers

Convert from radians to degrees. Show all work. You can type pi for the symbol <br> 7/12 pi

mariarad [96]

Answer:

The answer is $\frac{7\pi }{12}^{c}=105^{0}$

Step-by-step explanation:

The expression is

$\frac{7\pi }{12}^{c}\\= \frac{7\pi }{12}*\frac{180}{\pi }^{0} \,;[1^{c}=\frac{180}{\pi}^{0}]\\=7*15\\=105^{0}$

Hope you have understood this....

pls mark my answer as the brainliest

8 0

3 years ago

In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

7 0

3 years ago