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DerKrebs [107]
3 years ago
7

Allison has a poster that is 15 in by 18 in. What will the dimensions of the poster be if she scales it down by a factor of one-

third ? 15 in by 18 in 45 in by 54 in 2.5 in by 2 in 5 in by 6 in
Mathematics
2 answers:
erastovalidia [21]3 years ago
8 0
15×13=5
18×13=6
so 5in by 6in
nadya68 [22]3 years ago
4 0

Answer:

5 in by 6 in

Step-by-step explanation:

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Rewrite 8^(9/5) as a radical expression. I think it might be D, I'm unsure though.
AnnZ [28]
It is D, Nancy Reynolds from Chicago.

The denominator of the exponent is the index of the radical, the base of the exponent is the base of the radical, and the numerator of the exponent is the exponent of the radicand.

8^{\frac{9}{5}}= \sqrt[5]{8^9}
7 0
2 years ago
Geometry question help
ololo11 [35]
The answer is b because
7 0
2 years ago
Please help with this question. I would appreciate it!
Veronika [31]

Answer:

Mass of A = 5760 grams

Step-by-step explanation:

Surface area is given, so we can set-up a similarity equation to solve for k, proportionality constant. Here, we will relate both of them through k^2, because they are 2 dimensional (area). Thus

Surface Area of A = k^2 * Surface Area of B

28 = k^2 * 40.32

k^2 = 28/40.32

k^2 = 25/36

k = Sqrt (25/36)

k = 5/6

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Mass of A = 5760 grams

8 0
3 years ago
Read 2 more answers
Write a definite integral that represents the area of the region. (Do not evaluate the integral.) y1 = x2 + 2x + 3 y2 = 2x + 12F
Svet_ta [14]

Answer:

A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36

Step-by-step explanation:

The equations are:

y = x^{2} + 2x + 3

y = 2x + 12

The two graphs intersect when:

x^{2} + 2x + 3 = 2x + 12

x^{2} = 0

x_{1}  = 3\\x_{2}  = -3

To find the area under the curve for the first equation:

A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx

To find the area under the curve for the second equation:

A_{2} = \int\limits^3__-3}{2x + 12} \, dx

To find the total area:

A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx

Simplifying the equation:

A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx

Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).  

3 0
2 years ago
Graph the following inequality. Click on the graph until the correct one appears.
exis [7]

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Note that if instead we hade y≥3x+5 the line would be solid.

If you draw a line parallel to the y-axis through any x-value the value of y can be any on that vertical line that is above but not on the line y=3x+5

. So you end up with a 'feasible' region of solutions for y

5 0
3 years ago
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