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astraxan [27]
3 years ago
5

Please help with number one and 2 please

Mathematics
1 answer:
sleet_krkn [62]3 years ago
3 0
No. 1 is actually correct. Nice work!

In No. 2, it's much easier to tackle the problem sentence by sentence. In the first sentence, we are given that Val has 6 yards of fabric. In the next sentence, we are given that she uses 1/3 of that piece of fabric to make a shirt. Let's calculate how much she has left now.

6 x (1/3) = 2
Val used up 2 yards, meaning she has 4 yards left.

The next sentence indicates that she uses 3/8 of what she has left, which is 4 yards of fabric. We can apply the same method as the first step.

3/8 x 4 = 3/2

She used up 1.5 yards to make a shirt for her sister, meaning she has 2.5 yards left.

Final Answer: 2.5 yards of fabric


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A bike wheel is 26 inches in diameter. What is the bike wheel's diameter in millimeters (1 inch = 25.4 millimeters)? Determine t
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Answer:

The answer is 660.4 millimeters in diameter.

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3 1
3 years ago
Can someone plz help me asap ​
likoan [24]

Answer:

  • y = 12x - 9

Step-by-step explanation:

<u>Given</u>

  • - 1/3y + 4x = 3

Multiply all terms by -3

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7 0
3 years ago
Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves
Vadim26 [7]

The expression on the left side describes a parabola. Factorize it to determine where it crosses the y-axis (i.e. the line x = 0) :

-3y² + 9y - 6 = -3 (y² - 3y + 2)

… = -3 (y - 1) (y - 2) = 0

⇒   y = 1   or   y = 2

Also, complete the square to determine the vertex of the parabola:

-3y² + 9y - 6 = -3 (y² - 3y) - 6

… = -3 (y² - 3y + 9/4 - 9/4) - 6

… = -3 (y² - 2•3/2 y + (3/2)²) + 27/4 - 6

… = -3 (y - 3/2)² + 3/4

⇒   vertex at (x, y) = (3/4, 3/2)

I've attached a sketch of the curve along with one of the shells that make up the solid. For some value of x in the interval 0 ≤ x ≤ 3/4, each cylindrical shell has

radius = x

height = y⁺ - y⁻

where y⁺ refers to the half of the parabola above the line y = 3/2, and y⁻ is the lower half. These halves are functions of x that we obtain from its equation by solving for y :

x = -3y² + 9y - 6

x = -3 (y - 3/2)² + 3/4

x - 3/4 = -3 (y - 3/2)²

-x/3 + 1/4 = (y -  3/2)²

± √(1/4 - x/3) = y - 3/2

y = 3/2 ± √(1/4 - x/3)

y⁺ and y⁻ are the solutions with the positive and negative square roots, respectively, so each shell has height

(3/2 + √(1/4 - x/3)) - (3/2 - √(1/4 - x/3)) = 2 √(1/4 - x/3)

Now set up the integral and compute the volume.

\displaystyle 2\pi \int_{x=0}^{x=3/4} 2x \sqrt{\frac14 - \frac x3} \, dx

Substitute u = 1/4 - x/3, so x = 3/4 - 3u and dx = -3 du.

\displaystyle 2\pi \int_{u=1/4-0/3}^{u=1/4-(3/4)/3} 2\left(\frac34 - 3u\right) \sqrt{u} \left(-3 \, du\right)

\displaystyle -12\pi \int_{u=1/4}^{u=0} \left(\frac34 - 3u\right) \sqrt{u} \, du

\displaystyle 12\pi \int_{u=0}^{u=1/4} \left(\frac34 u^{1/2} - 3u^{3/2}\right)  \, du

\displaystyle 12\pi \left(\frac34\cdot\frac23 u^{3/2} - 3\cdot\frac25u^{5/2}\right)  \bigg|_{u=0}^{u=1/4}

\displaystyle 12\pi \left(\frac12 u^{3/2} - \frac65u^{5/2}\right)  \bigg|_{u=0}^{u=1/4}

\displaystyle 12\pi \left(\frac12 \left(\frac14\right)^{3/2} - \frac65\left(\frac14\right)^{5/2}\right) - 12\pi (0 - 0)

\displaystyle 12\pi \left(\frac1{16} - \frac3{80}\right) = \frac{12\pi}{40} = \boxed{\frac{3\pi}{10}}

6 0
2 years ago
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