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Mars2501 [29]
3 years ago
6

Can someone please help me I really need this

Mathematics
1 answer:
svp [43]3 years ago
7 0
Honestly i don’t really know but maybe a bar graph? you could put the monthly sales in the x axis and the earnings in the y axis. sorry
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PLEASEEEEEEEEE HELPPPPPPPPPP 70-100 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!! I NEED ONE MORE ANSWER BC I WANNA SEE WHICH ANSWER IS THE
Zina [86]

Answer and Explanation:

Gina wanted to swim at the pool. She is allowed to spend at most $30. The cost to swim in the pool is 3 dollars per hour plus a flat fee of $3. How many hours can Gina swim without going over her spending limit?

This problem can be represented by  3x+3\leq 30.

'x' would be the number of hours. "At most" means less than or equal to. So, the value of x would have to be less than or equal to 30.

3x+3\leq 30-3\\\\3x\leq 27\\\\3x/3\leq 27/3\\\\\boxed{x\leq9}

So, Gina would have a maximum of 9 hours to swim in the pool.

Hope this helps.

3 0
3 years ago
PLEASEEE HELP AND SHOW WORK​
Verdich [7]
I mean it should be 7 cause 2 angles are congruent.
6 0
3 years ago
|x+2|-3=9 please help!!! Solve twice!!! Two answers!!!
Over [174]
×= 10 and ×= -14

Find x by simplifying both sides of the equation then isolate the variable.
6 0
3 years ago
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Please I need help with differential equation. Thank you
Inga [223]

1. I suppose the ODE is supposed to be

\mathrm dt\dfrac{y+y^{1/2}}{1-t}=\mathrm dy(t+1)

Solving for \dfrac{\mathrm dy}{\mathrm dt} gives

\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{y+y^{1/2}}{1-t^2}

which is undefined when t=\pm1. The interval of validity depends on what your initial value is. In this case, it's t=-\dfrac12, so the largest interval on which a solution can exist is -1\le t\le1.

2. Separating the variables gives

\dfrac{\mathrm dy}{y+y^{1/2}}=\dfrac{\mathrm dt}{1-t^2}

Integrate both sides. On the left, we have

\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\int\frac{\mathrm dz}{z+1}

where we substituted z=y^{1/2} - or z^2=y - and 2z\,\mathrm dz=\mathrm dy - or \mathrm dz=\dfrac{\mathrm dy}{2y^{1/2}}.

\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\ln|z+1|=2\ln(y^{1/2}+1)

On the right, we have

\dfrac1{1-t^2}=\dfrac12\left(\dfrac1{1-t}+\dfrac1{1+t}\right)

\displaystyle\int\frac{\mathrm dt}{1-t^2}=\dfrac12(\ln|1-t|+\ln|1+t|)+C=\ln(1-t^2)^{1/2}+C

So

2\ln(y^{1/2}+1)=\ln(1-t^2)^{1/2}+C

\ln(y^{1/2}+1)=\dfrac12\ln(1-t^2)^{1/2}+C

y^{1/2}+1=e^{\ln(1-t^2)^{1/4}+C}

y^{1/2}=C(1-t^2)^{1/4}-1

I'll leave the solution in this form for now to make solving for C easier. Given that y\left(-\dfrac12\right)=1, we get

1^{1/2}=C\left(1-\left(-\dfrac12\right)^2\right))^{1/4}-1

2=C\left(\dfrac54\right)^{1/4}

C=2\left(\dfrac45\right)^{1/4}

and so our solution is

\boxed{y(t)=\left(2\left(\dfrac45-\dfrac45t^2\right)^{1/4}-1\right)^2}

3 0
3 years ago
Which rule describes the translation ABCD - A’B’C’D
monitta

Answer:

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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