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3 years ago

Write a real-world problem that could be represented by the following equation. 70 + 5x = 40 + 10x

1 answer:

4 0

Jack wants to paint his house. he collected quotations from two different painters. The first painter's total cost in terms of the number of hours, (x), worked was modelled as f(x) = 70 + 5x, while that of the second painter was g(x) = 40 + 10x. Jack knows that it will take about 6 hours to paint the house. He figured out that the cost quoted by the two painters is equal for 6 hour of work.

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miss Akunina [59]

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3 years ago

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What ides it mean by simplify?

MrRissso [65]

Putting it in simpler terms

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2 years ago

Amelia needs to buy some cat food. At the nearest store, 2 bags of cat food cost $4.50. How much would Amelia spend on 6 bags o

marissa [1.9K]

Answer:

$13.50

Step-by-step explanation:

2 bags of cat food = $4.50

1 bag of cat food = 2/4.50

= 2.25

6 bags of cat food = 2.25*6

=13.50

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2 years ago

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Elijah buys a one-pint bottle of juice for $2.88. What is the unit rate of the cost of the juice per fluid ounce?

mojhsa [17]

Answer: 0.48
Explanation: $2.88 / 16

3 0

3 years ago

Use the fundamental theorem of calculus to find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x

BaLLatris [955]

Answer:

The area of the region is 25,351 $units^2$ .

Step-by-step explanation:

The Fundamental Theorem of Calculus: if $f$ is a continuous function on $[a,b]$ , then

$\int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) | {_a^b}$

where $F$ is an antiderivative of $f$ .

A function $F$ is an antiderivative of the function $f$ if

$F^{'}(x)=f(x)$

The theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

To find the area of the region between the graph of the function $x^5 + 8x^4 + 2x^2 + 5x + 15$ and the x-axis on the interval [-6, 6] you must:

Apply the Fundamental Theorem of Calculus

$\int _{-6}^6(x^5+8x^4+2x^2+5x+15)dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int _{-6}^6x^5dx+\int _{-6}^68x^4dx+\int _{-6}^62x^2dx+\int _{-6}^65xdx+\int _{-6}^615dx$

$\int _{-6}^6x^5dx=0\\\\\int _{-6}^68x^4dx=\frac{124416}{5}\\\\\int _{-6}^62x^2dx=288\\\\\int _{-6}^65xdx=0\\\\\int _{-6}^615dx=180\\\\0+\frac{124416}{5}+288+0+18\\\\\frac{126756}{5}\approx 25351.2$

3 0

3 years ago