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3 years ago

PLEASE HELP! I WILL MAKE YOU BRAINLIST!

Mathematics

1 answer:

Mice21 [21]3 years ago

5 0

Answer: I think it is C

Step-by-step explanation: 32/4=8 so 1 cm = 8km

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Choose the correct slope of the line that passes through the points (1, −3) and (3, −5).

Lelu [443]

I believe it is -1.

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2 years ago

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Simplify the following expression: -2/10(1 - 2x + 2 )

const2013 [10]

Answer:

2/10-4/10x

Step-by-step explanation:

8 0

2 years ago

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In 2002, a certain town recorded 15,300 acres of underdeveloped land. Each year

Katyanochek1 [597]

The approximate amount of underdeveloped land in 2013 is A = 15300 X e^-0.07(11).

<h3>How do you determine the future value when there is continuous compounding? </h3>

The formula for calculating future value of an investment when there is continuous compounding is A = P x $e^{r}$ x N

Where:

P = population in 2002
r = rate of decrease = 0.07
n = number of years = 2013 - 2002 = 11 years

A = 15300 X e^-0.07(11)

To learn more about future value, please check: brainly.com/question/25760893

6 0

1 year ago

Write an equation for an ellipse centered at the origin, which has foci at (\pm5,0)(±5,0)(, plus minus, 5, comma, 0, )and vertic

Virty [35]

Answer:

The equation of ellipse is

$\frac{x^2}{41}+\frac{y^2}{16}=1$

Step-by-step explanation:

The equation of an ellipse is

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

where(h,k) is the center and c is distance from the center to the foci is given by $a^2-b^2=c^2$ . a is the distance from the center to the vertices and b is the distance from the center to the co-vertices.

The center of the ellipse is the mid-point of the vertices.

The mid point of the vertices $(\pm\sqrt{41},0)$ is

= $(\frac{\sqrt{41}+(-\sqr{41})}2,\frac{0+0}2)$

=(0,0)

a is the distance between the center and the vertices.

So, $a=\sqrt{(0-\sqrt{41})^2+(0-0)^2}$

$=\sqrt{41}$

c is the distance between the center and the foci.

So, $c=\sqrt{(0-5)^2+(0-0)^2$

$a^2-b^2=c^2$

$\Rightarrow (\sqrt41)^2-b^2=5^2$

$\Rightarrow b^2=(\sqrt41)^2- 5^2$

$\Rightarrow b^2=41-25$

$\Rightarrow b^2=16$

The equation of ellipse is

$\frac{(x-0)^2}{(\sqrt{41})^2}+\frac{(y-0)^2}{16}=1$

$\Rightarrow \frac{x^2}{41}+\frac{y^2}{16}=1$

7 0

3 years ago

Domain f(x)=5^(2x-1)

notka56 [123]

The attached image may have what you are looking for.

4 0

2 years ago