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3 years ago

If sin 49° is approximately 34, which is closest to the length of NO

Mathematics

1 answer:

meriva3 years ago

3 0

Answer:

EXPERT ANSWER

Step-by-step explanation:

Correct answer- is 25m

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For every 15 students who run track, there are 7 students who play baseball. There are 112 more students

wolverine [178]

Answer:

There are 210 students running track and 98 students that play baseball.

Step-by-step explanation:

15*14= 210 and 7*14 = 98, 210 - 98 = 112. If this was correct feel free to mark me as brainiest :)

4 0

4 years ago

Read 2 more answers

Carly draws quadrilateral JKLM with vertices J(-3,3), K(3,3), L(2,-1),and M(-2, -1). What is the best way to classify the quadri

Trava [24]

Answer:

Trapezoid

Step-by-step explanation:

The points are shown in the attached graph.

You can clearly see from the picture that it is a "Trapezoid".

Trapezoid is basically a quadrilateral (4 sided-figure) that has 1/2 pair of parallel sides.

On the other hand, a parallelogram has 2 pair of parallel sides.

Thus, a parallelogram is a special type of trapezoid where there are 2 pair of parallel sides. But trapezoid need not be a parallelogram.

The picture shows 1 pair of parallel sides, hence it is a trapezoid.

6 0

3 years ago

How do you solve -4x - 10 < 2

viva [34]

There is a way and you will find it but not from me

5 0

3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$