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3 years ago

My answer is B do you agree

Mathematics

2 answers:

insens350 [35]3 years ago

8 0

Answer:

Area is increased by 4 i have my question and i need you to answer plz thxxxx

Step-by-step explanation:

Nastasia [14]3 years ago

6 0

Answer:

The area increased by a factor of 4

Step-by-step explanation:

If the scale factor is 4 and only the height changes

A = 1/2 bh

A = 1/2 b(4h)

A = 1/2 bh(4)

A = A4

The new area is the old area increased by 4

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On a coordinate plane, line P Q goes through (negative 3, negative 3) and (3, negative 1). Point R is at (1, 1).

olga55 [171]

Answer:

0, 2/3

Step-by-step explanation:

Its the only one that make sense.

hope thats right

edge 2020

7 0

3 years ago

Read 2 more answers

Simplify the expression -4^2(3x - 7)

maxonik [38]

Answer:

−48x+112

Step-by-step explanation:

evatulate: −16 (3−7)

-48+112

3 0

3 years ago

Find the volume of the solid.

dmitriy555 [2]

In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

In the plane $z=2$ , we have

$2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2$

which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$

3 0

1 year ago

Find the tangent of each angle that is not the right angle. Drag and drop the numbers into the boxes to show the tangent of each

omeli [17]

Answer:

tan D = $\frac{6}{11}$ or 0.545

tan F = $\frac{11}{6}$ or 1.833

Step-by-step explanation:

To find tangent is always $\frac{opposite}{adjacent}$

so the opposite of angle D is line EF which is 4.8 and the adjacent side is line ED which is 8.8 so $\frac{4.8}{8.8}$ which is $\frac{6}{11}$ or 0.545

The opposite of angle f is line ED which is 8.8 and the adjacent is line EF which is 4.8 so $\frac{8.8}{4.8}$ is $\frac{11}{6}$ or 1.833

3 0

3 years ago

A beach has to enclose a rectangular area, because some endangered species are nesting there. They have 200 feet of rope to rope

kkurt [141]

Area is equal to length times width. The perimeter (the amount of rope) has to equal twice the length added to twice the width so we're left with:
A = l * w
200 = 2l + 2w
solve for either l or w
l = 100 - w
plug into the area equation to get one equation with two variables
A = w(100 - w)
A = -w^2 + 100w
take the derivative
A' = -2w + 100
set the derivative equal to zero
0 = -2w + 100
2w = 100
w = 50
This is the width that maximizes the area
with a width of 50, the length must also be 50 to have a perimeter of 200
therefore, they can rope up to 50 * 50 = 2500 ft^2

6 0

3 years ago