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Naya [18.7K]
2 years ago
6

Two friends, Parker and David, took summer jobs. The graph below represents David's earnings in dollars and cents, yy, for worki

ng xx hours.

Mathematics
1 answer:
siniylev [52]2 years ago
8 0

Answer:

WORK For Exercises 44 and 45, use the information below and in the graph. Andrea works in a restaurant and is paid every two weeks. If Andrea earns $ 6.50 an hour, illustrate the Distributive Property by writing two expressions representing Andrea's pay last week.

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How do you do rotation of the origin ?
anzhelika [568]
The rotation rule would be (-y, x)

Write/mark all your coordinates down. Now plot all your prime points and draw a line connecting them.

Hope this helps!~
6 0
4 years ago
Help please
exis [7]

Answer:

You can put this solution on YOUR website! 5X - 20 = 10 * X 5X - 20 = 10X Subtract 5x from each side-20 = 5X Divide each side by 5-4 = X. Let's make sure it works... 20 less than five times a number (-20-20) is equal to the product of ten and that number (10 * -4)

Step-by-step explanation:

8 0
3 years ago
Which choice represents the result when 4 + 6m is subtracted from 15m? Select ALL that apply.
Amiraneli [1.4K]
Answer = 9m + 4

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7 0
3 years ago
Read 2 more answers
Kelly has 50% more money than Luis. Luis has $40. If Luis gets $8 more, will Kelly still have more money than Luis? What percent
OverLord2011 [107]

Answer:

Kelly will still have more money than Luis

Kelly will have 25% more

Step-by-step explanation:

Let x represent the amount of money Kelly has and y, Luis

Luis has $40

and Kelly has 50% more money than Luis

Therefore, Kelly has;

50% of $40 + $40

= 0.5*40 + 40

= $60

If Luis gets $8 more, Luis will have;

$8 + $40 = $48

since Luis' $48 is less than Kelly's $60, Kelly will still have more money than Luis

Difference in their money = $60 - $48 = $12

Therefore, Kelly will have;

12/48 * 100 more than Luis

= 25% more

8 0
3 years ago
Find the length of the curve y = integral from 1 to x of sqrt(t^3-1)
Arlecino [84]
y=\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt

By the fundamental theorem of calculus,

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt=\sqrt{x^3-1}

Now the arc length over an arbitrary interval (a,b) is

\displaystyle\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=\int_a^b\sqrt{1+x^3-1}\,\mathrm dx=\int_a^bx^{3/2}\,\mathrm dx

But before we compute the integral, first we need to make sure the integrand exists over it. x^{3/2} is undefined if x, so we assume a\ge0 and for convenience that a. Then

\displaystyle\int_a^bx^{3/2}\,\mathrm dx=\frac25x^{5/2}\bigg|_{x=a}^{x=b}=\frac25\left(b^{5/2}-a^{5/2}\right)
6 0
3 years ago
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