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PilotLPTM [1.2K]
3 years ago
12

Jordan has some music books. he will buy 9 new music books each year. he will have 52 music books in 5 years.

Mathematics
1 answer:
natita [175]3 years ago
8 0

Answer:

760 music books per each 5 years

Step-by-step explanation:

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A craft show has an admission fee of $1.50 for children and $4.00 for adults. On Saturday 155 people came to the carnival and $5
rjkz [21]
X+y=155. Y= -x+155

1.50x +4(-x+155)= 520
1.50x-4x+620=520
-2.5x =-100
X= 40. Y=110
8 0
3 years ago
Mrs. Davis went to a produce market to buy bananas and strawberries. She spent $8.00. If the bananas were $0.50 per pound, and t
9966 [12]

9514 1404 393

Answer:

  4 pounds

Step-by-step explanation:

Let b represent the pounds of bananas Mrs. Davis bought. The weight of strawberries would be (7-b) and their cost would be 4×0.50 = 2.00 per pound. Her total purchase was ...

  0.50b + 2.00(7 -b) = 8.00

  -1.50b +14 = 8 . . . . simplify

  6 = 1.50b . . . . . . add 1.50b -8

  4 = b . . . . . . . . divide by 1.50

Mrs. Davis bought 4 pounds of bananas.

7 0
2 years ago
0.06 is 10 times as much as
kenny6666 [7]
0.06 is 10 times as much as 0.6
8 0
3 years ago
Read 2 more answers
PLEASE HELP ASAPPPPPPPPPPP
mixas84 [53]

\boxed{\boxed{\sf Area\:of\:the\:figure=10x^4-50x^3+5x^2+246x-196}}

Refer to the attachment for calculations.

6 0
3 years ago
Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )
DedPeter [7]
(a) First find the intersections of y=e^{2x-x^2} and y=2:

2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}

So the area of R is given by

\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx

If you're not familiar with the error function \mathrm{erf}(x), then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line y=1 with y=e^{2x-x^2}.

1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2

So the area of S is given by

\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx
\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve y=e^{2x-x^2} and the line y=1, or e^{2x-x^2}-1. The area of any such circle is \pi times the square of its radius. Since the curve intersects the axis of revolution at x=0 and x=2, the volume would be given by

\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx
5 0
3 years ago
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