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MA_775_DIABLO [31]
3 years ago
10

Find the slope of the line shown on this graph. If your answer is a fraction write the

Mathematics
1 answer:
Lorico [155]3 years ago
7 0

Answer:

1/7

Step-by-step explanation:

This is because, you have to go up one unit and over 7 units before your blue line has a dot present.

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Fittoniya [83]
The anwser is 234 inches
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Plz answer its easy math.
Serga [27]
<h2>Greetings!</h2>

Answer:

B and E

Step-by-step explanation:

The rules of indicies states:

x^{a} ÷x^{b} = x^{a-b}

So 3^{-8} ÷  3^{-4} =  3^{-8 - - 4} =  3^{-4}

So that means B is one of the correct answers.

To find the other correct value, you can divide the two fractions as stated in the question:

3^{-8} by 3^{-4} = \frac{1}{81}

81 is equivalent to 3⁴

So that means E is also correct.


<h2>Hope this helps!</h2>
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Burka [1]

Answer:

Approximately 6.4

Step-by-step explanation:

We can use the pythagorean thereom here, that tells us (a^2)+(b^2)=c^2. C is the hypotenuse, the side opposite from the right angle, while a and b are the other sides. We can insert 5 and 4 as a and b, and solve for c

:(5^2)+(4^2)=c^2

:25+16=c^2

:41=c^2

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What is the Area of the circle in pi units if the diameter is 6 cm?
sergejj [24]

The area of the circle in pi units when the diameter of this circle is 6 centimeters is 9π centimeters.

<h3>What is the area of the circle?</h3>

The area of the circle is the space occupied by it. It is the product of pi and square of its diameter divided by 4. The area of the circle can be given as,

A=\pi\dfrac{d^2}{4}

Here (d) is the diameter of the circle. The diameter of the circle is 6 cm.

d=6

Put this value in the above formula to find the area of the circle as,

A=\pi\dfrac{6^2}{4}\\A=\pi\dfrac{36}{4}\\A=\pi\times9\\A=9\pi\rm\; cm

Thus, the area of the circle in pi units when the diameter of this circle is 6 centimeters is 9π centimeters.

Learn more about the area of the circle here;

brainly.com/question/402655

#SPJ1

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What is the simplest form of 874/942
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Answer:

437/471

Step-by-step explanation:

would consider downloading photomath

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