All categories

3 years ago

Please help KaikersCupcake!!

Mathematics

1 answer:

lina2011 [118]3 years ago

4 0

The third one is correct because y>-then you have to do 3x1= and 3x8=24 so that will be y>-3/24.

You might be interested in

In isosceles pqr with base , pq = 2x + 3, and pr = 9x - 11. what is the value of x?

Anon25 [30]

First, draw and label a rough sketch of an isosceles triangle. Now put in the values of pq=2x+3 and pr=9x-11

You know that the triangle is isosceles, therefore, the sides pq and pr are identical.

Now, make the equations equal to each other to find the value of x.

2x+3=9x-11
3+11=9x-2x
14=7x
14/7=x
2=x

Therefore the value of x=2.

Hope I helped :)

8 0

2 years ago

Sarah rolls a die. What is the chance of her rolling a 4?!

Mariulka [41]

depends on the type of dice, some have many sides . how many rolls, how many seconds until it stops rolling, which quantiative set of dice rolls do you have that you use to verify the accuricy of predicitons.

In short, not sure.

3 0

3 years ago

Read 2 more answers

How do i solve that question?

yawa3891 [41]

a) The solution of this ordinary differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ .

b) The integrating factor for the ordinary differential equation is $-\frac{1}{x}$ .

The particular solution of the ordinary differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ .

<h3>How to solve ordinary differential equations</h3>

a) In this case we need to separate each variable ( $y, t$ ) in each side of the identity:

$6\cdot \frac{dy}{dt} = y^{4}\cdot \sin^{4} t$ (1)

$6\int {\frac{dy}{y^{4}} } = \int {\sin^{4}t} \, dt + C$

Where $C$ is the integration constant.

By table of integrals we find the solution for each integral:

$-\frac{2}{y^{3}} = \frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32} + C$

If we know that $x = 0$ and $y = 1$ , then the integration constant is $C = -2$ .

The solution of this ordinary differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ . $\blacksquare$

b) In this case we need to solve a first order ordinary differential equation of the following form:

$\frac{dy}{dx} + p(x) \cdot y = q(x)$ (2)

Where:

$p(x)$ - Integrating factor
$q(x)$ - Particular function

Hence, the ordinary differential equation is equivalent to this form:

$\frac{dy}{dx} -\frac{1}{x}\cdot y = x^{2}+\frac{1}{x}$ (3)

The integrating factor for the ordinary differential equation is $-\frac{1}{x}$ . $\blacksquare$

The solution for (2) is presented below:

$y = e^{-\int {p(x)} \, dx }\cdot \int {e^{\int {p(x)} \, dx }}\cdot q(x) \, dx + C$ (4)

Where $C$ is the integration constant.

If we know that $p(x) = -\frac{1}{x}$ and $q(x) = x^{2} + \frac{1}{x}$ , then the solution of the ordinary differential equation is:

$y = x \int {x^{-1}\cdot \left(x^{2}+\frac{1}{x} \right)} \, dx + C$

$y = x\int {x} \, dx + x\int\, dx + C$

$y = \frac{x^{3}}{2}+x^{2}+C$

If we know that $x = 1$ and $y = -1$ , then the particular solution is:

$y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$

The particular solution of the ordinary differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ . $\blacksquare$

To learn more on ordinary differential equations, we kindly invite to check this verified question: brainly.com/question/25731911

3 0

2 years ago

Plz help !! Me with it

astraxan [27]

Answer:

x=31

Step-by-step explanation:

4+ sqrt (x+5) = 10

Subtract 4 from each side

4 -4+ sqrt (x+5) = 10-4

sqrt(x+5) = 6

Square each side

(sqrt(x+5))^2 = 6^2

x+5 = 36

Subtract 5 from each side

x+5-5 = 36-5

x = 31

7 0

2 years ago

Read 2 more answers

What is the average rate of change of f over the interval -1 on a separate sheet of paper and upload the work as a photo. Show a

Anarel [89]

Answer:

Average rate of change for the function $f(x)= x^2-x-1$ over the interval -1<x<1 is -1

Step-by-step explanation:

We need to find average rate of change of f over the interval -1 < x < 1

The function given is: $f(x)= x^2-x-1$

The formula used to find average rate of change is:

$Average\:rate\:of\:change=\frac{f(b)-f(a)}{b-a}$

We have, a = -1 and b = 1

Finding f(b) when b=1

$f(x)=x^2-x-1\\f(1)=(1)^2-(1)-1\\f(1)=1-1-1\\f(1)=-1$

Now, finding f(a), when a= -1

$f(x)=x^2-x-1\\f(-1)=(-1)^2-(-1)-1\\f(1)=1+1-1\\f(-1)=2-1\\f(-1)=1$

Now, putting values and finding average rate of change

$Average\:rate\:of\:change=\frac{f(b)-f(a)}{b-a}\\Average\:rate\:of\:change=\frac{f(1)-f(-1)}{1-(-1)}\\Average\:rate\:of\:change=\frac{-1-(1)}{1-(-1)}\\Average\:rate\:of\:change=\frac{-1-1}{1+1}\\Average\:rate\:of\:change=\frac{-2}{2}\\Average\:rate\:of\:change=-1$

So, average rate of change for the function $f(x)= x^2-x-1$ over the interval -1<x<1 is -1

4 0

3 years ago