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3 years ago

10

Mathematics

2 answers:

natka813 [3]3 years ago

5 0

Answer:

x= -8 , y = 5

x= 25/4 , y = 1/4

Step-by-step explanation:

substitute first eqn into the second eqn:

(7 - 3y)^2 -y^2 = 39

49 - 42y + 9y^2 - y^2 = 39

8y^2 - 42y +10 =0

4y^2 - 21y + 5 = 0

(4y-1) (y-5) = 0

y= 1/4 , 5

when y=1/4

x = 7- 3/4

=25/4

when y= 5

x = 7- 15

= -8

torisob [31]3 years ago

5 0

Answer:

$(x_1 \: ,y_1) = ( \frac{25}{4} \:, \frac{1}{4} )$

$(x_2 , y_2) = ( - 8 , 5)$

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Hey there :)

( 5y + 9 )( 6y - 1 )

We need to use FOIL to expand, that is
First Terms
Outer Terms
Inner Terms
Last Terms

First Outer Inner Last
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Combine, if any, the like-terms
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Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.

Romashka [77]

By definition, the average change of rate is given by:
$AVR = \frac{f(x2)-f(x1)}{x2-x1}$
We will calculate AVR for each of the functions.
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$f(-2) = x^2 + 3x = (-2)^2 + 3(-2) = 4 - 6 = -2

f(3) = x^2 + 3x = (3)^2 + 3(3) = 9 + 9 = 18$
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$AVR = \frac{-20}{-5}$
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f(x) = 3x - 8 interval: [4, 5]:
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$AVR = \frac{7-4}{5-4}$
$AVR = \frac{3}{1}$
$AVR = 3$

f(x) = x^2 - 2x interval: [-3, 4]
$f(-3) = (-3)^2 - 2(-3) = 9 + 6 = 15

f(4) = (4)^2 - 2(4) = 16 - 8 = 8$
$AVR = \frac{8-15}{4+3}$
$AVR = \frac{-7}{7}$
$AVR = -1$

f(x) = x^2 - 5 interval: [-1, 1]
$f(-1) = (-1)^2 - 5 = 1 - 5 = -4

f(1) = (1)^2 - 5 = 1 - 5 = -4$
$AVR = \frac{-4+4}{1+1}$
$AVR = \frac{0}{2}$
$AVR = 0$

Answer:
these functions from the greatest to the least value based on the average rate of change are:
f(x) = x^2 + 3x
f(x) = 3x - 8
f(x) = x^2 - 5
f(x) = x^2 - 2x

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