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3 years ago

How do I find the area of this shape (rounded to the nearest 10th place)?

Mathematics

1 answer:

iren2701 [21]3 years ago

3 0

20 times 20 hope this helps

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Rewrite in simplest terms. 4 (-6g - 10h) + 3h - 4 (- 3h - 10g)

kondor19780726 [428]

Answer: 16g - 25h

4 (-6g - 10h) + 3h - 4 (- 3h - 10g)

-24g - 40h +3h +12h +40g

16g - 25h

Step-by-step explanation:

if this is wrong im sorry

8 0

2 years ago

A rocket is launched in the air. The graph below shows the height of the rocket hh in meters after tt seconds.

fgiga [73]

Answer:

The answers are=

(38, 0)
time
in seconds
(19, 1768.9)
Height
in meters

3 0

1 year ago

6^7/6^5 in index form<br>help me

brilliants [131]

Answer:

12^12

Step-by-step explanation:

can u please help me

7 0

2 years ago

LORAN is a long range hyperbolic navigation system. Suppose two LORAN transmitters are located at the coordinates (−60,0) and (6

USPshnik [31]

LORAN follows an hyperbolic path.

The equation of the hyperbola is: $\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$

The coordinates are given as:

$\mathbf{(x,y) = (-60,0)\ (60,0)}$

The center of the hyperbola is

$\mathbf{(h,k) = (0,0)}$

The distance from the center to the focal points is given as:

$\mathbf{c = 60}$

Square both sides

$\mathbf{c^2 = 3600}$

The distance from the receiver to the transmitters is given as:

$\mathbf{2a = 100}$

Divide both sides by 2

$\mathbf{a = 50}$

Square both sides

$\mathbf{a^2 = 2500}$

We have:

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

This gives

$\mathbf{b^2 = 3600 - 2500}$

$\mathbf{b^2 = 1100}$

The equation of an hyperbola is:

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

So, we have:

$\mathbf{\frac{(x - 0)^2}{2500} + \frac{(y - 0)^2}{1100} = 1}$

$\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$

Hence, the equation of the hyperbola is: $\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$