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Alex_Xolod [135]
3 years ago
8

I can't figure out how to do this. Please show work

Mathematics
1 answer:
Oksanka [162]3 years ago
7 0
I believe that on the shape is length times the width which is 56
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What is the sum of 8 and 7, doubled
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(8+7) x 2

= 15 x 2

= 30
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Is each line parallel, perpendicular, or neither parallel nor perpendicular to a line whose slope is -6?
babymother [125]

Answer:

Line M - neither

Line N - parallel

Line P - perpendicular

Line Q - neither

Step-by-step explanation:

If a line is perpendicular to another, the slope will be the opposite, e.g. -6, opposite slope, -6.

If a line is parallel to another, the slope will be the exact same, e.g. -6, same slope, 1/6.

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The shipping department must send out 50 cases of phones. Five stores will receive ten cases each. How many cases are left?
scZoUnD [109]
There are none left. 5 x 10 = 50
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3 years ago
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A cross-country team had a total of 10 practices last week, with each practice being in the morning or the afternoon. During eac
GalinKa [24]

x= morning

y = afternoon

 x + y = 10

x=10-y

5x + 6y =57

5(10-y) +6y =57

50-5y +6y =57

y=7

x=10-7 =3

 check 3*5 = 15, 7*6 = 42, 42 + 15 = 57


3 morning practices

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cint%20t%5E2%2B1%20%5C%20dt" id="TexFormula1" title="\frac{d}{dx} \
Kisachek [45]

Answer:

\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2}  t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2

Step-by-step explanation:

\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2}  t^2+1 \ \text{dt} = \ ?

We can use Part I of the Fundamental Theorem of Calculus:

  • \displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

  • \displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

  • \displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

  • \displaystyle\int\limits^b_a \text{f(t) dt}\  = -\int\limits^a_b \text{f(t) dt}

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

  • \displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx}  \int\limits^{x^2}_0 t^2+1 \text{ dt}  

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

  • \displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u]
  • where u represents any function other than a variable

For the first term, replace \text{t} with 2x, and apply the chain rule to the function. Do the same for the second term; replace

  • \displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)  

Simplify the expression by distributing 2 and 2x inside their respective parentheses.

  • [-(8x^2 +2)] + (2x^5 + 2x)
  • -8x^2 -2 + 2x^5 + 2x

Rearrange the terms to be in order from the highest degree to the lowest degree.

  • \displaystyle2x^5-8x^2+2x-2

This is the derivative of the given integral, and thus the solution to the problem.

6 0
2 years ago
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