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attashe74 [19]
3 years ago
15

What is this answer

Mathematics
2 answers:
Maurinko [17]3 years ago
6 0
Its 2 my guy my friend
igor_vitrenko [27]3 years ago
5 0
Ok you can find out the answer I can see it right there try also to find it yourself I know you would find it
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What is the rate of change ? <br>y=-5x-3​
KatRina [158]

Answer:

the rate of change is -5.

Step-by-step explanation:

y = -5x - 3

-5 = rate of change

-3 = x intercept

6 0
3 years ago
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How many sides does a circle have
Irina-Kira [14]
A circle doesnt have a sides so 0
4 0
3 years ago
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The circumference of the ellipse approximate. Which equation is the result of solving the formula of the circumference for b?
Serhud [2]

Answer:

b = \sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}

Step-by-step explanation:

Given - The circumference of the ellipse approximated by C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }where 2a and 2b are the lengths of 2 the axes of the ellipse.

To find - Which equation is the result of solving the formula of the circumference for b ?

Solution -

C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }\\\frac{C}{2\pi }  =  \sqrt{\frac{a^{2} + b^{2} }{2} }

Squaring Both sides, we get

[\frac{C}{2\pi }]^{2}   =  [\sqrt{\frac{a^{2} + b^{2} }{2} }]^{2} \\\frac{C^{2} }{(2\pi)^{2}  }   =  {\frac{a^{2} + b^{2} }{2} }\\2\frac{C^{2} }{4(\pi)^{2}  }   =  {{a^{2} + b^{2} }

\frac{C^{2} }{2(\pi )^{2} }  = a^{2} + b^{2} \\\frac{C^{2} }{2(\pi )^{2} }  -  a^{2} = b^{2} \\\sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}  = b

∴ we get

b = \sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}

8 0
3 years ago
A scale of 0.5 inches to 2 feet was used for a model of the goal posts on a football field. If the model was 1 1/2 in high and 5
Paraphin [41]
The actual height is: 1 1/2 * 2 = 3 feet
The actual length is: 5 1/8 * 2 = 10.25 feet

hope this helps you.
6 0
4 years ago
Solve This Using The Quadratic Formula: 3x^2+9x-6=0
aivan3 [116]
<h3>Therefore either  x = \frac{-3+\sqrt{17}}{2}    or,  x = \frac{-3-\sqrt{17}}{2}</h3>

Step-by-step explanation:

3x^2 +9x -6=0                                x = \frac{-b\pm\sqrt{b^2- 4ac} }{2a}

\Leftrightarrow x = \frac{-9\pm\sqrt{9^2- 4.3(-6)} }{2.3}                     here a = 3 ,b = 9 and c= -6

\Leftrightarrow x = \frac{-9\pm\sqrt{153} }{6}

\Leftrightarrow x = \frac{3(-3\pm\sqrt{17}) }{6}

\Leftrightarrow x = \frac{-3\pm\sqrt{17}}{2}

Therefore either  x = \frac{-3+\sqrt{17}}{2}    or,  x = \frac{-3-\sqrt{17}}{2}

4 0
4 years ago
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