Problem 1
subtract 3 x
5y=-3x+11
divide 5
y=3/5x+2.2'
Problem 2
add 5x
7y=5x+9
divide 7
y=5/7x+1.3'
'notice'
you can put 2.2 into a fraction (11/5) and 1.3 (9/7) if your teacher allows it, but i put them into decimals
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The general equation of a circle is x^2 + y^2 = r^2. Here we know that the circle passes thru two points: (-3,2) and (1,5). Given that a third point on the circle is (-7, ? ), find the y-coordinate of this third point.
Subst. the known values (of the first point) into this equation: (-3)^2 + (2)^2 = r^2. Then 9 + 4 = 13 = r^2.
Let's check this. Assuming that the equation of this specific circle is
x^2 + y^2 = r^2 = 13, the point (1,5) must satisfy it.
(1)^2 + (5)^2 = 13 is not true, unfortunately.
(1)^2 + (5)^2 = 1 + 25 = 26 (very different from 13).
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Answer:
angle 2 is 55
Step-by-step explanation:
supposing angle 1 and 2 are supplementary angle
since angle 1 is 125
180-125=55
therefore angle 2 is 55
Answer: In the resulting equation: " a² - 12a + 32 = 0 " ;
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The "coefficient" of the "a" term is: " - 12" .
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The "constant" is: " 32 " .
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Explanation:
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Let: "a = x² + 4 " .
Given: (x² + 4)² + 32 = 12x² + 48 ;
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Factor: "12x² + 48" into " (x² + 4) " ;
"12x² + 48" = 12 (x² + 4) " ;
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Given: (x² + 4)² + 32 = 12x² + 48 ;
rewrite as; "a² + 32 = 12a " ;
Subtract "12a" from each side of the equation;
"a² + 32 - 12a = 12a - 12a ;
to get:
" a² - 12a + 32 = 0 " .
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The coefficient of the "a" term; that is:
The "coefficient" of " -12a" ; is: "- 12" .
The constant is: "32<span>" .
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80 c, 80%, 80/100, .80
$1.20, 120%, 120/100, 1.2
$1.60, 160%, 160/100, 1.6
$1.28, 128%, 128/100, 1.28
