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3 years ago

PLS HELP I WILL MARK BRAINLIEST!!

Mathematics

1 answer:

Mandarinka [93]3 years ago

5 0

Answer:

Step-by-step explanation:

I think

A= 1/2 bh

A= 1/2 (14)(4)

A= 56/2

A= 28

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The formula that describes an object's motion is given by s=ut+1/2at^2, where s is the distance traveled, u is the initial veloc

victus00 [196]

None of the offered choices is correct.

The correct equation is
.. a = (2s -2ut)/t^2 . . . . . . . . parentheses are required

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3 years ago

What percent of 12 is 7?<br> Please show work :)

SVETLANKA909090 [29]

It’s 3 hope this helps

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3 years ago

Read 2 more answers

7. If the longer leg of a right triangle is twice as long as the shorter leg, what is the ratio of

kicyunya [14]

Answer: The ratio is 2.39, which means that the larger acute angle is 2.39 times the smaller acute angle.

Step-by-step explanation:

I suppose that the "legs" of a triangle rectangle are the cathati.

if L is the length of the shorter leg, 2*L is the length of the longest leg.

Now you can remember the relation:

Tan(a) = (opposite cathetus)/(adjacent cathetus)

Then there is one acute angle calculated as:

Tan(θ) = (shorter leg)/(longer leg)

Tan(φ) = (longer leg)/(shorter leg)

And we want to find the ratio between the measure of the larger acute angle and the smaller acute angle.

Then we need to find θ and φ.

Tan(θ) = L/(2*L)

Tan(θ) = 1/2

θ = Atan(1/2) = 26.57°

Tan(φ) = (2*L)/L

Tan(φ) = 2

φ = Atan(2) = 63.43°

Then the ratio between the larger acute angle and the smaller acute angle is:

R = (63.43°)/(26.57°) = 2.39

This means that the larger acute angle is 2.39 times the smaller acute angle.

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

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