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3 years ago

3.

Mathematics

2 answers:

katen-ka-za [31]3 years ago

7 0

Answer:

Option B will be your answer

Step-by-step explanation:

B. The balloon starts at a height of 500 ft and rises at a rate of 150 ft.

hope it helps ^_^

d1i1m1o1n [39]3 years ago

6 0

Answer:

the answer is C

Step-by-step explanation:

It shows the line starts at 0.05 a d to get to 3 it has to raise upb150 ft

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30 Points, 5 stars, thanks and Brainliest if answered correctly because I feel generous.

wlad13 [49]

Find the difference between each number:

-24/4 = -6

144/-24 = -6

The next number is multiplied by -6

You have the first 4 terms

5th term = -864 x -6 = 5184

6th term = 5184 x -6 = -31,104

7th term = -31,104 x -6 = 186,624

8th term = 186,624 x -6 = -1,119,744

4 0

3 years ago

Read 2 more answers

Solve −5x = 15. (1 point) −3 3 10 20

Luba_88 [7]

Remember you can do anything to an eqautoin as long as you do oit to both sides

-5x=15
try to get 1x by itself

remember
(ax)/a=x when a=a

so
-5x=15
get x
divide both sides by -5
remember to flip sign
(-5x)/(-5)=15/(-5)
x=-3

answer is first one
x=-3

8 0

2 years ago

Read 2 more answers

I decide to bake some cakes.I need 150g of flour for 12 cakes.How much flour do I need for 30 cakes?

blagie [28]

150 / 12 = 11.6666666667
11.6666666667 x 30 = approximately 350g of flour for 30 cakes

7 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

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

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

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

3 0

3 years ago

How to find the square root and cube root of a number?

tiny-mole [99]

To find the cube root of a number, you want to find some number that when multiplied by itself twice gives you the original number. In other words, to find the cube root of 8, you want to find the number that when multiplied by itself twice gives you 8. The cube root of 8, then, is 2, because 2 × 2 × 2 = 8.

6 0

3 years ago

Read 2 more answers