What is the answer to 9*20^15-23/5

100 points! Mhanifa can you please help? Look at the picture attached. I will mark brainliest!

dimulka [17.4K]

Answer:

1 and 2.

Midpoints calculated, plotted and connected to make the triangle DEF, see the attached.

D= (-2, 2), E = (-1, -2), F = (-4, -1)

3.

As per definition, midsegment is parallel to a side.

Parallel lines have same slope.

Find slopes of FD and CB and compare.

m(FD) = (2 - (-1))/(-2 -(-4)) = 3/2
m(CB) = (1 - (-5))/(1 - (-3)) = 6/4 = 3/2
As we see the slopes are same

Find the slopes of FE and AB and compare.

m(FE) = (-2 - (- 1))/(-1 - (-4)) = -1/3
m(AB) = (1 - 3)/(1 - (-5)) = -2/6 = -1/3
Slopes are same

Find the slopes of DE and AC and compare.

m(DE) = (-2 - 2)/(-1 - (-2)) = -4/1 = -4
m(AC) = (-5 - 3)/(-3 - (-5)) = -8/2 = -4
Slopes are same

4.

As per definition, midsegment is half the parallel side.

We'll show that FD = 1/2CB

FD = $\sqrt{(2+1)^2+(-2+4)^2}$ = $\sqrt{3^2+2^2}$ = $\sqrt{13}$
CB = $\sqrt{(1 + 5)^2+(1+3)^2}$ = $\sqrt{6^2+4^2}$ = 2 $\sqrt{13}$
As we see FD = 1/2CB

FE = 1/2AB

FE = $\sqrt{(-4+1)^2+(-1+2)^2}$ = $\sqrt{3^2+1^2}$ = $\sqrt{10}$
AB = $\sqrt{(-5 -1)^2+(3-1)^2}$ = $\sqrt{6^2+2^2}$ = 2 $\sqrt{10}$
As we see FE = 1/2AB

DE = 1/2AC

DE = $\sqrt{(-2+1)^2+(2+2)^2}$ = $\sqrt{1^2+4^2}$ = $\sqrt{17}$
AC = $\sqrt{(-5 +3)^2+(3+5)^2}$ = $\sqrt{2^2+8^2}$ = 2 $\sqrt{17}$
As we see DE = 1/2AC

3 0

2 years ago

Read 2 more answers

In Jamie's pocket there are 45 coins.

AVprozaik [17]

D:N
2 :3
Add them together to get the total which = 5
5 = 45
2 ?
= 18
No of dimes is 18

7 0

2 years ago

Read 2 more answers

What is the distance from point D to point C

In-s [12.5K]

Answer:

-4to 2

Step-by-step explanation:

it lies in the x axis

4 0

2 years ago

g A manufacturer is making cylindrical cans that hold 300 cm3. The dimensions of the can are not mandated, so to save manufactur

sdas [7]

Answer:

The dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

Step-by-step explanation:

A cylindrical can holds 300 cubic centimeters, and we want to find the dimensions that minimize the cost for materials: that is, the dimensions that minimize the surface area.

Recall that the volume for a cylinder is given by:

$\displaystyle V = \pi r^2h$

Substitute:

$\displaystyle (300) = \pi r^2 h$

Solve for h:

$\displaystyle \frac{300}{\pi r^2} = h$

Recall that the surface area of a cylinder is given by:

$\displaystyle A = 2\pi r^2 + 2\pi rh$

We want to minimize this equation. To do so, we can find its critical points, since extrema (minima and maxima) occur at critical points.

First, substitute for h.

$\displaystyle \begin{aligned} A &= 2\pi r^2 + 2\pi r\left(\frac{300}{\pi r^2}\right) \\ \\ &=2\pi r^2 + \frac{600}{ r} \end{aligned}$

Find its derivative:

$\displaystyle A' = 4\pi r - \frac{600}{r^2}$

Solve for its zero(s):

$\displaystyle \begin{aligned} (0) &= 4\pi r - \frac{600}{r^2} \\ \\ 4\pi r - \frac{600}{r^2} &= 0 \\ \\ 4\pi r^3 - 600 &= 0 \\ \\ \pi r^3 &= 150 \\ \\ r &= \sqrt[3]{\frac{150}{\pi}} \approx 3.628\text{ cm}\end{aligned}$

Hence, the radius that minimizes the surface area will be about 3.628 centimeters.

Then the height will be:

$\displaystyle \begin{aligned} h&= \frac{300}{\pi\left( \sqrt[3]{\dfrac{150}{\pi}}\right)^2} \\ \\ &= \frac{60}{\pi \sqrt[3]{\dfrac{180}{\pi^2}}}\approx 7.25 6\text{ cm} \end{aligned}$

In conclusion, the dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

7 0

2 years ago

Plz answer this and explain how

Dahasolnce [82]

Answer:

Option H, π/3 and 2π/3

Step-by-step explanation:

sin x = sqrt(3) / 2

It happens at 60 degrees and 120 degrees. In radians, it is π/3 and 2π/3

Answer: Option H, π/3 and 2π/3

8 0

2 years ago