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Kryger [21]
3 years ago
11

Question 16 of 25

Mathematics
2 answers:
yKpoI14uk [10]3 years ago
6 0
I think the answer is a but I am not sure
lawyer [7]3 years ago
4 0

Answer:

To get the vertex of the parabola we proceed as follows;

y=-7(x-4)^2-5

The above can be written as:

y=-7x^2+56x-117

The values of a,b and c are:

a=-7, b=56 and c=-117

x=-b/(2a)

x=-56/(-7*2)=4

but;

y=-7x^2+56x-117

y=-7(4)^2+56(4)-117

y=-5

Thus;

x=4 and y=-5

The vertex will be at point (4,-5)

Hope it helped... Plz mark as BRAINLIEST... And follow me..

Thanks

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Josef's closet is 5 feet wide and 3 feet deep. What is the area of his closet floor?
choli [55]

Answer:

15f^2

Step-by-step explanation:

multiply lenghtxwidth

always make sure to square your answer when doing area

6 0
3 years ago
Every weekend, Morgan buys interesting clothes at her local thrift store and then resells them on an auction website. If she bri
Artyom0805 [142]
150 - s <== ur expression

when s = 27.13
150 - 27.13 = 122.87

when s = 55.14
150 - 55.14 = 94.86
6 0
3 years ago
Read 2 more answers
Find the integral using substitution or a formula.
Nadusha1986 [10]
\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx

Derivative of the denominator:
\rm (x^2+2x+5)'=2x+2

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx

and then split the fraction,

\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx

Based on our previous test, we know that a simple substitution will work for the first integral: \rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx

So the first integral changes,

\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx

integrating to a log,

\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2

So we have this going on,

\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx

Bringing all that stuff together as a single square,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx

Making the substitution: \rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx

giving us,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du

simplying a lil bit,

\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du

and hopefully from this point you recognize your arctangent integral,

\rm ln|x^2+2x+5|+\frac52arctan(u)

undo your substitution as a final step,
and include a constant of integration,

\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c

Hope that helps!
Lemme know if any steps were too confusing.

8 0
3 years ago
What is m pls help i will mark brainly
-Dominant- [34]

Answer:

x=18

Step-by-step explanation:

Hey There!

So if you did not know all of the angles in a triangle will add up to equal 180

So we use the equation

180=2x+3x+5x

and solve for x

step 1 combine like terms

2x+3x=5x

5x+5x=10x

so now we have

180=10x

step 2 divide each side by 10

180/10=18

therefore x = 18

hope this helps and if you have any more questions feel free to ask :)

6 0
3 years ago
Read 2 more answers
Based on a kc value of 0.210 and the data table given, what are the equilibrium concentrations of xy, x, and y, respectively?
a_sh-v [17]

The activation energy for the forward and the one for the reverse reaction are similar because they attained chemical equilibrium. A chemical equilibrium happens when both of the reactant and products achieve the same concentration.

4 0
3 years ago
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