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andrew11 [14]
3 years ago
14

Find the vertex of y = x^2 – 2x – 15

Mathematics
2 answers:
svet-max [94.6K]3 years ago
7 0
The vertex is ( 1, -16 ).
solong [7]3 years ago
7 0

Answer:

The vertex is at (h, k), which here is (1, -16).

Step-by-step explanation:

One of the quicker ways to approach this proplem is to "complete the square," which puts the equation into "vertex form."

Starting with y = x^2 – 2x – 15, add 1 to, and then subtract 1 from,  x^2 – 2x :

y = x^2 – 2x + 1 - 1– 15

Now re-write   x^2 – 2x   as (x - 1)^2 - 1:

y = (x - 1)^2 - 1 – 15. or  y = (x - 1)^2 - 16, which has the form y = (x - h)^2 + k.  Comparing these two equations, we see that h = 1 and k = -16.

The vertex is at (h, k), which here is (1, -16).

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Find the surface area of each figure?Please help
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6(2.5 x 2.5) = 37.5m²

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2(11 x 7) + 2(9 x 7) + 2(9 x 11) = 478m²

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Answer: 503m²
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4 0
2 years ago
Naomi has earned $54 mowing lawns the past 2 days. She worded 21/2 hours yesterday and 41/4 hourse today. If Naomi is paid the s
Klio2033 [76]
She worked 2 and 1/2 hours yesterday, and today she did 4 and 1/4 hours.

first off, let's check how many hours she worked total, and divide 54 by it.

so, first off, let's convert the mixed fractions to "improper" and add them up.

\bf \stackrel{mixed}{2\frac{1}{2}}\implies \cfrac{2\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{5}{2}}&#10;\\\\\\&#10;\stackrel{mixed}{4\frac{1}{4}}\implies \cfrac{4\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{17}{4}}\\\\&#10;-------------------------------

\bf \cfrac{5}{2}+\cfrac{17}{4}\impliedby \textit{our LCD will be 4}\implies \cfrac{(2\cdot 5)+(1\cdot 17)}{4}&#10;\\\\\\&#10;\cfrac{10+17}{4}\implies \cfrac{27}{4}\implies \stackrel{total~hours}{6\frac{3}{4}}&#10;\\\\\\&#10;\cfrac{\stackrel{wages}{54}}{\stackrel{hours}{6\frac{3}{4}}}\implies \cfrac{54}{\frac{27}{4}}\implies \cfrac{\frac{54}{1}}{\frac{27}{4}}\implies \cfrac{54}{1}\cdot \cfrac{4}{27}\implies \cfrac{216}{27}\implies 8~\frac{\$}{hr}
6 0
3 years ago
Solve the initial value problem y'=2 cos 2x/(3+2y),y(0)=−1 and determine where the solution attains its maximum value.
zloy xaker [14]

Answer:

y=\frac{-3\pm\sqrt{4sin2x+1}}{2}

x={\pi}{4}

Step-by-step explanation:

We are given that

y'=\frac{2cos2x}{3+2y}

y(0)=-1

\frac{dy}{dx}=\frac{2cos2x}{3+2y}

(3+2y)dy=2cos2x dx

Taking integration on both sides then we get

\int (3+2y)dy=2\int cos 2xdx

3y+y^2=sin2x+C

Using formula

\int x^n=\frac{x^{n+1}}{n+1}+C

\int cosx dx=sinx

Substitute x=0 and y=-1

-3+1=sin0+C

-2=C

sin0=0

Substitute the value of C

y^2+3y=sin2x-2

y^2+3y-sin 2x+2=0

y=\frac{-3\pm\sqrt{(3)^2-4(1)(-sin2x+2)}}{2}

By using quadratic formula

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

y=\frac{-3\pm\sqrt{9+4sin2x-8}}{2}=\frac{-3\pm\sqrt{4sin2x+1}}{2}

Hence, the solution y=\frac{-3\pm\sqrt{4sin2x+1}}{2}

When the solution is maximum then y'=0

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cos\frac{\pi}{2}=0

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3 0
3 years ago
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Vaselesa [24]

Answer:

A = 80   S =70

Step-by-step explanation:

#Adults tickets = A

#Students tickets = S

A + S = 150

5A (means the price is $5 times the number of Adult tickets) equals the total amount for all of the Adult tickets altogether

3S (means the price is $3 times the number of Student tickets) equals the total amount for all of the students tickets altogether

5A + 3S = 610    (Means adding the total of all the student tickets plus adult tickets will equal $610 for all tickets sold)

Using both equations now, you can use substitution or elimination to solve for one of the variables. Then you can use the variable to substitute to solve the remaining one.

A + S = 150

5A + 3S = 610

Substitution:

A = 150 - S  (Rearrange the first equation by moving S to the other side)

Substitute into the other equation

5 (150 - S) + 3S = 610

750 - 5S + 3S = 610      Combine like terms :    -2S = -140

Solve for S = 70

Substitute into A + S = 150        A + (70) = 150

A = 80

3 0
3 years ago
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