All categories

3 years ago

NEED HELP FAST PLEASE.

Mathematics

1 answer:

Sergio [31]3 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

When 23 is subtracted from the square root of three times a number, the result is 16. find the number?

gizmo_the_mogwai [7]

(√3n)-23=16

Put all of the numbers without a variable on the right side of the equation
(√3n)=39

Simplify
1.732051n=39

Divide 39 by 1.732051
n=22.516658

Round
n=22.5

8 0

3 years ago

Find CG if KG = 41.4

Irina18 [472]

Answer:

13.8

Step-by-step explanation:

CG is 1/3 of KG

so 41.4/3 is 13.8. that is median of CG

5 0

4 years ago

Maria solved the equation 34=x+19. Although her final answer was correct, she could have found it in a simpler way. how could sh

mamaluj [8]

Answer:

She could have subtracted 19 from both sides. Hope this helps :)

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

The diameter of a circle is 22 miles. What is the circle's area? Use 3.14 for pi.. Single choice.

Sonbull [250]

Answer:

379.94 mi²

Step-by-step explanation:

The area of a circle can be figured out with this formula: πr² (3.14 is π here, as it says in the directions).

We know the diameter is 22 miles. Divide that by 2 to get a radius of 11 miles.

The equation now becomes

3.14(11²)

11 squared is 121.

3.14(121)

I recommend splitting the 121 into 100, 20, and 1.

(3.14 * 100) + (3.14 * 20) + (3.14 * 1)

Multiply 3.14 by 100, 20, and 1.

314 + 62.8 + 3.14

Add 314 and 62.8.

376.8 + 3.14

Add 376.8 and 3.14

379.94 mi² is the area of a circle with a 22-mile diameter.

6 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

4 years ago