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dybincka [34]
3 years ago
12

Need help

Mathematics
2 answers:
Alika [10]3 years ago
4 0

Answer:

16

Step-by-step explanation:

2*8= 16

emmasim [6.3K]3 years ago
3 0

Answer:

Step-by-step explanation:

128

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I will mark brainliest !
harina [27]

Answer:

90 degrees

Step-by-step explanation:

A circle circumscribed over a right triangle is the same as a circle circumscribed over a rectangle built from the extension of the right triangle.

That tells us that the hypotenuse is the diameter of the circle in that case.

Here, we can see a triangle that has the hypotenuse as the diameter of the circumscribed circle. We can infer that the triangle is a right triangle.

4 0
3 years ago
Which stands for 250 feet below sea level?
Lubov Fominskaja [6]
-250 feet, because you are 250 feet below sea level.
3 0
3 years ago
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Lunna [17]

Answer:

3rd one

Step-by-step explanation:

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3 years ago
Need help with 12 plz
Flura [38]

Answer:

see below    PLEASE GIVE BRAINLIEST

Step-by-step explanation:

HOUSE  - LIBRARY                            = 1 3/5 MILES

LIBRARY - MALL - KYLE'S HOUSE   =  1 1/3 + 4/5 = 4/3 + 4/5 =

                                                                    20/15 + 12/15 = 32/15 = 2 2/15 miles



8 0
3 years ago
PLEASE HELP
lisabon 2012 [21]

Step-by-step explanation:

The figure below shows a portion of the graph of the function j\left(x\right) \ = \ 4^{x-2}, hence the average rate of change (slope of the blue line) between the x and x+h is

                     \text{Average rate of change} \ = \ \displaystyle\frac{\Delta y}{\Delta x} \\ \\ \rule{3.7cm}{0cm} = \dsiplaystyle\frac{f\left(x+h\right) \ - \ f\left(x\right)}{\left(x \ + \ h \right) \ - \ x} \\ \\ \\  \rule{3.7cm}{0cm} = \displaystyle\frac{f\left(x + h\right) \ - \ f\left(x\right)}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x+h-2} \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x-2+h} \ - \ 4^{x-2}}{h}

                                                            \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h}\right) \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h} \ - \ 1 \right)}{h}

7 0
2 years ago
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