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Pavlova-9 [17]
4 years ago
13

[50 POINTS] SHOW ON PAPER, OR TAKE A SCREENSHOT ON HOW YOU DID IT.

Mathematics
2 answers:
dusya [7]4 years ago
8 0

Answer:

1.)  6.48 - 3. 50 = 2.98

Mia needs $2.98 more dollars to buy the frame.

2.)  7.65 - 5.40 = 2.25  

An adult ticket costs $2.25 more than a child's ticket

3.)  13.75 + 5.40 + 9 = 28.15  

William earned $28.15 all together

Step-by-step explanation:

I hope this helps!

snow_tiger [21]4 years ago
6 0

Answer:

Step-by-step explanation:

1. 6.48 - 5.50 = 0.98

2. 7.65 - 5.40 = 2.25

3. 13.79 + 5.40 + 9= 28.19

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6x³

Step-by-step explanation:

X x X = x² + 4x + 2x = 6x³

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A warning/caution sign?

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4x-4x-5=-5 which statement is true for the equation
Tasya [4]
4x - 4x - 5 = -5
4x - 4x = -5 + 5
0 = 0

this equation has infinite solutions
3 0
4 years ago
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There are 765 marbles in a box. There are twice as many blue marbles as green marbles. There are three times as many red marbles
Digiron [165]
B=blue
g=green
r=red
2 times as many blue as green
b=2g
3 times as many red as blue
r=3b
subsitute 2g for b
r=3(2g)=6g
r=6g

r+g+b=765
so we subsitute
r=6g
g=g
b=2g

6g+g+2g=765
add like terms
9g=765
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g=85
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5 0
3 years ago
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
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