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2 years ago

14

Rosa has $175 in her online music account. Each month she buys $15 worth of new songs. Which inequality could be solved

1 answer:

Virty [35]2 years ago

6 0

Answer:

it is C)15x - 175 < 10

Step-by-step explanation:

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Carl jumped 11 5/12 feet in the long jump. Ben jumped 9 7/8 feet. How much farther did Carl jump than Ben?

Wewaii [24]

Answer:

Step-by-step explanation:

1 13/24

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2 years ago

What is the exact value of x? 5⋅21^5x=16

IRINA_888 [86]

Answer:

x= 16 /20420505

Step-by-step explanation:

Let's solve your equation step-by-step.

5(215)x=16

Step 1: Simplify both sides of the equation.

20420505x=16

Step 2: Divide both sides by 20420505.

20420505x /20420505 = 16 /20420505

x= 16 /20420505

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2 years ago

What is the solution to this inequality?

Alika [10]

Answer:

<h2>A</h2>

Step-by-step explanation:

<em><u>e5uiitweyutwwsuoye34t</u></em>

3 0

2 years ago

Read 2 more answers

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

Stells [14]

$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

3 0

2 years ago

if there are more students in one group than the other would it be possible to have equal proportions?

coldgirl [10]

Yes but In the bigger group each person would have less

3 0

3 years ago