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3 years ago

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An Algebra I test contains 30 questions. Some of the questions are worth 4 points each, and some are worth 2 points each. The to

tal test is worth 100 points. Let f represent the number of questions worth 4 points and t, the number of questions worth 2 points.

1 answer:

Roman55 [17]3 years ago

6 0

Answer:

Step-by-step explanation:

f+t=40 and 4f+2t=100

4f+2t=120

4f+2t=100

2t=20

t=10

So f=20

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Answer:

(-6, 6).

Step-by-step explanation:

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Describe several methods you could use to determine the rational zeros of a polynomial function. Which would you choose to use f

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3 years ago

Calculus Problem

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The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

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We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

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2 years ago

Mr. Walker asked his students to use the associative property to find an expression that is equivalent to (13 + 15 + 20) + (20 +

pychu [463]

I got it as 4 when calculating all have the same number which is 133

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3 years ago

A penguin walks 15 feet in 9 seconds.

shtirl [24]

Answer:

12 seconds.

Step-by-step explanation:

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3 years ago

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