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Allisa [31]
3 years ago
6

Given the relation y = 7x2 + 3, if the input is 4, what is the output? (4 points)

Mathematics
2 answers:
finlep [7]3 years ago
7 0

Answer:115

Step-by-step explanation:

aivan3 [116]3 years ago
4 0
I'm assuming that you meant: 
y = 7x² + 3 

Remember! Inputs are always x values (unless stated otherwise). Meaning the problem says: 
x = 4 

y = 7(4)² + 3 

The square only applies to the 4. The 7 is not going to be squared! (To be exact it only applies to whatever the value of x is. 

4² = 4·4 = 16

Remember: 
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Follow PEMDAS from left to right (or in the case above from top to bottom).

7·16 + 3 = ? 

16·7 = 10·7 + 6·7 = 70 + 42 = 112
(All I did to multiply was break it up into parts. If it confuses you don't worry about it, and just multiply it out like normal or use a calculator if you are allowed to)

112 + 3 = ? 

Our output is: 
115!
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If p=(-3,-2) and q=(1,6) are the endpoints of the diameter of a circle find the equation of the circle
stira [4]

Answer:

<em>The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²</em>

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

<u>Step-by-step explanation:</u>

Given points end  Points are p(-3,-2) and q( 1,6)

<em>The distance of two points formula</em>

P Q = \sqrt{x_{2} - x_{1})^{2} + ((y_{2} -y_{1})^{2}   }

P Q = \sqrt{1 - (-3)^{2} + ((6 -(-2))^{2}   }

P Q = \sqrt{16+64} = \sqrt{80}

<em>The diameter 'd' = 2 r</em>

                       2 r = √80

                            = \sqrt{16 X 5}

                           = 4 \sqrt{5}

                     <em> r =  2√5</em>

<em>Mid-point of two end points </em>

<em>                        </em>(\frac{x_{1} + x_{2} }{2} , \frac{y_{1} +y_{2} }{2} ) = (\frac{-3+1}{2} ,\frac{-2 +6}{2} )<em></em>

<em>                                               = (-1 ,2)</em>

<em>Mid-point of two end points = center of the circle</em>

<em>                                     (h,k) = (-1 , 2)</em>

                 

The equation of the circle

           (x -h )² +(y-k)² = r²

<em>          (x -(-1) )² +(y-(2))² = (2(√5))²</em>

<em>         x² + 2 x + 1 + y² - 4 y + 4 = 20</em>

<em>          x² + 2 x + y² - 4 y  = 20 -5</em>

<em>           x² + 2 x + y² - 4 y  = 15</em>

<u><em>Final answer</em></u><em>:-</em>

<em>The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²</em>

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

<em>                  </em>

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