(-3,-5)(2,4)
slope(m) = (4 - (-5) / (2 - (-3) = (4 + 5) / (2 + 3) = 9/5

y - y1 = m(x - x1)
slope(m) = 9/5
(2,4)...x1 = 2 and y1 = 4
sub
y - 4 = 9/5(x - 1)....this is the correct answer

the student figured the slope wrong....it is 9/5...not -9/5

5 0

3 years ago

The scale factor for a model is 5cm= m model: 9.5cm actual : 30.5m

MaRussiya [10]

We are given dimensions of model and actual figure.

model: 9.5cm actual : 30.5m.

We need to find the length of actual shape for the scale factor 5cm.

Let us assume actual length be x m.

Let us set a proportion now.

Let us convert proportion into fractions.

$\frac{5}{x}=\frac{9.5}{30.5}$

On cross multiplying, we get

9.5x = 5× 30.5

9.5x = 152.5.

On dividing both sides by 9.5, we get

$\frac{9.5x}{9.5}=\frac{152.5}{9.5}$

x=16.05.

<h3>Therefore, the scale factor for a model is 5cm= 16.05 m.</h3>

6 0

3 years ago

Calculus Problem

Roman55 [17]

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

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

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

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

7 0

2 years ago

The grid represents 1 whole. Shade the grid to the model 0.3.

Triss [41]

Answer:

where is the grid? I would help if you have more info

8 0

3 years ago