Answer:
15
Step-by-step explanation:
By solving a linear equation we will see that the initial temperature was 66 degrees.
<h3>
What was the initial temperature?</h3>
Let's define T as the initial temperature.
First, it increases by 18 F, then it decreases 6 F, and finally, it increases another 12 F, such that the end temperature is 94F.
Then we can write:
T + 18F - 6F + 12F = 94F
Now we can solve that linear equation to find T.
T = 94F - 18F + 6F - 12 F = 66 F
In this way, we conclude that the initial temperature was 66 degrees.
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Answer:
35
Step-by-step explanation:
what I did was add GJ together and imaged it in my head to see if it would fit... I'm not sure tho
9. y=-1/4x^2+4x-19
group
y=(-1/4x^2+4x)-19
undistribute -1/4
y=-1/4(x^2-16x)-19
take 1/2 of -16 and squer it to get 64 then add neg and pos inside
y=-1/4(x^2-16x+64-64)-19
factorperfect square
y=-1/4((x-8)^2-64)-19
expand
y=-1/4(x-8)^2+16-19
y=-1/4(x-8)^2-3
vertex is (8,-3)
10.
group
y=(1/4x^2-3x)+18
undistribute
y=1/4(x^2-12x)+18
take 1/2 of -12 and square it and add neg and pos isndie
y=1/4(x^2-12x+36-36)+18
factor
y=1/4((x-6)^2-36)+18
expand
y=1/4(x-6)^2-9+18
y=1/4(x-6)^2+9
get to form (x-h)^2=4p(y-k)
minus 9 both sides and times 4
(x-6)^2=4(y-9)
(x-6)^2=4(1)(y-9)
so 1>0 so opens up and focus is 1 above vertex
vertex is (6,9)
so focus i (6,10)
11.
y=(-1/6x^2+7x)-80
y=(-1/6)(x^2-42x)-80
take 1/2 of linear coefient and squer it and add negative and positive inside
-42/2=-21, (-21)^2=441
y=(-1/6)(x^2-42+441-441)-80
factor perfect square the square
y=(-1/6)((x-21)^2-411)-80
expand
y=(-1/6)(x-21)^2+73.5-80
y=(-1/6)(x-21)^2-6.5
add 6.5 to both sid
y+6.5=(-1/6)(x-21)^2
times both sides by -6
-6(y+6.5)=(x-21)^2
(x-21)^2=-6(y+6.5)
(y-21)^2=4(-3/2)(y-(-6.5))
vertex is
-3/2<0 so directix is above
it is -3/2 or 1.5 units above the vertex
up is y so
-6.5+1.5=-5
the directix is y=-5
11.
in form (y-1)^2=4p(x+3)
opens left or right
(y-1)^2=4(4)(x+3)
vertex is (-3,1)
4>0 so opens right
dirextix is to left
it is 4 units to left
(-3,1)
left right is x
4 left of -3 is -4-3=7
x=-7 is da directix
The value of k is
<h3>How to solve the simultaneous equation?</h3>
Given:
x-y=k.............(eq i)
2x²+y²-15..............(eq ii)
We would make y the subject formula in eq ii
2x²+y²-15= 0
2x² + y²= 15
y²= 15-2x²
y= 
...........(eq iii)
Substitute the value of y into eq i
x-(= k
x- (
= k
k=
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