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Setler [38]
3 years ago
10

X^2 + 8x + 7 = 0

Mathematics
1 answer:
dsp733 years ago
7 0

Answer:

x - intercepts: -1, -7

Solutions: [is the x-intercepts] -1,-7

Factors: (x+1)(x+7)

Vertex: -4, -9

Axis of symmetry: x = -4

Step-by-step explanation:

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What does the absolute value of the y coordinate tell you?
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Step-by-step explanation:

in x ,y =0

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3 years ago
Shane purchased a television at a 20% discount. If the sale price is $375.00, what is the original price ?​
Sveta_85 [38]

the answer is going to be 450. all you have to do is add 375+20%

8 0
3 years ago
Can anyone find the answer for this equation?
Lynna [10]
66 hope it is right
5 0
3 years ago
For the function f(x)=x2−24÷3+12<br><br> f(1) <br><br><br> ​f(0)​ <br><br><br> ​f(−1)​
LuckyWell [14K]

Put the values of x to the equation of the function.

f(x)=\dfrac{x^2-24}{3}+12

f(1)=\dfrac{1^2-24}{3}+12=\dfrac{1-24}{3}+12=\dfrac{-23}{3}+12=-7\dfrac{2}{3}+12=4\dfrac{1}{3}\\\\f(0)=\dfrac{0^2-24}{3}+12=\dfrac{-24}{3}+12=-8+12=4\\\\f(-1)=\dfrac{(-1)^2-24}{3}+12=\dfrac{1-24}{3}+12=\dfrac{-23}{3}+12=-7\dfrac{2}{3}+12=4\dfrac{1}{3}

3 0
3 years ago
Assuming that the equations define x and y implicitly as differentiable functions x=f(t),y=g(t) find the slope of the curve x=f(
Mumz [18]
The given equations are
x(t+1)-4t \sqrt{x} =9            (1)
2y+4y^{3/2}=t^{3}+t           (2)

When t=0, obtain
x=9 \\ 2y+4y^{3/2}=0 \,\,=\ \textgreater \ \, y(1+2 \sqrt{y} )=0 \,=\ \textgreater \ \,y=0

Obtain derivatives of (1) and find x'(0).
x' (t+1) + x - 4√x - 4t*[(1/2)*1/√x = 0
x' (t+1) + x - 4√x -27/√x = 0
When t=0, obtain
x'(0) + x(0) - 4√x(0) = 0
x'(0) + 9 - 4*3 = 0
x'(0) = 3
Here, x' means \frac{dx}{dt}.

Obtain the derivative of (2) and find y'(0).
2y' + 4*(3/2)*(√y)*(y') = 3t² + 1
When t=0, obtain
2y'(0) +6√y(0) * y'(0) = 1
2y'(0) = 1 
y'(0) = 1/2.
Here, y' means \frac{dy}{dt}.

Because \frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt}, obtain
\frac{dy}{dx} |_{t=0}\, =  \frac{1/2}{3}= \frac{1}{6}

Answer:
The slope of the curve at t=0 is 1/6.



3 0
3 years ago
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