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3 years ago

What is is distance between (0,2) and (5,2)

Mathematics

1 answer:

likoan [24]3 years ago

3 0

5 units. The only variable changing is x so it's just 5 units up from the second coordinate

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Find the greatest common monomial factor : 4a^5b^2 and ab

3241004551 [841]

4a^5b^2=2*2*a*a*a*a*a*b*b
ab=a*b
greatest factor is ab

8 0

2 years ago

What is the answe ?????

patriot [66]

1)-9,-6,-3,7,11
2)8
3)1

3 0

2 years ago

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Mrs. Walters drove 360 miles on 25 gallons of gas. Mr. Walters drove 480 miles on 32 gallons of gas.

Temka [501]

Answer:

Mr.Walters car can go 15 miles on one gallon of gas.

Step-by-step explanation:

To find this, you first want to put it into a ratio.

480:32

Next, you want to divide.

480/32

After you divide, you get 15. That tells you that you get 15 miles off of one gallon of gas.

8 0

3 years ago

Read 2 more answers

Let x be a positive integer. If (-3)ºx (-3)* = (-3)14,<br> what is x?

Nezavi [6.7K]

Answer:

$-3^{\circ \:}x\left(-3\right)=\left(-3\right)\cdot \:14 : x = \frac{840}{180^{\circ \:}}$

Decimal Form:

$x = -267.38030...$

Step-by-step explanation:

$-3^{\circ \:}x\left(-3\right)=\left(-3\right)\cdot \:14$

Remove Parentheses: $(-a) = -a$

$-3^{\circ \:}x\left(-3\right)=-3\cdot \:14$

Multiply the numbers: $3 * 14 = 42$

$-3^{\circ \:}x\left(-3\right)=-42$

Divide both sides by: $-3^{\circ \:}\left(-3\right)$

$\frac{-3^{\circ \:}x\left(-3\right)}{-3^{\circ \:}\left(-3\right)}=\frac{-42}{-3^{\circ \:}\left(-3\right)}$

Simplify:

$x=-\frac{840}{180^{\circ \:}}$

Hope I helped. If so, may I get brainliest and a thanks?

Thank you, have a good day! =)

3 0

3 years ago

Write the standard form equation of the ellipse shown in the graph, and identify the foci.

vlada-n [284]

Answer option A

From the given graph is a Vertical ellipse

Center of ellipse = (-2,-3)

Vertices are (-2,3) and (-2,-9)

Co vertices are (-6,-3) and (2,-3)

The distance between center and vertices = 6, so a= 6

The distance between center and covertices = 4 , so b= 4

The general equation of vertical ellipse is

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

(h,k) is the center

we know center is (-2,-3)

h= -2, k = -3 , a= 6 and b = 4

The standard equation becomes

$\frac{(x+2)^2}{4^2} + \frac{(y+3)^2}{6^2}=1$

$\frac{(x+2)^2}{16} + \frac{(y+3)^2}{36}=1$

Foci are (h,k+c) and (h,k-c)

$c=\sqrt{a^2-b^2}$

Plug in the a=6 and b=4

$c=\sqrt{6^2-4^2}$

$c=\sqrt{20}$

$c=2\sqrt{5}$ , we know h=-2 and k=-3

Foci are $(-2,-3+2\sqrt{5})$ and $(-2,-3-2\sqrt{5})$

Option A is correct

6 0

3 years ago