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STatiana [176]
3 years ago
7

HURRY PLS this is due in 33 minutes pls will give thanks and brainliest

Mathematics
2 answers:
Blizzard [7]3 years ago
7 0
I'm pretty sure it's A.
charle [14.2K]3 years ago
6 0

Answer: B. Draw a chain of bounces to the left, 12 units wide, starting at 0 and ending at –24.

Explanation:

The dividend is obviously -24, so that tells us where we will end. The 12 is the divisor and is the number of groups we need to make (dividing by grouping). It is a positive 12, so we start at zero and move left to stop at the first group or bounce which would be from 0 to -12. The second group or bounce would be from -12 to -24. This will give us our negative amount of groups, because we are moving left, in this case, -2. <em> </em>We can check by using the inverse (or opposite) operation.

12 * -2 =-24

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Find the value of k for x^2 - 2kx+7k -12=0 so that they have two equal roots
Alisiya [41]

Step-by-step explanation:

x^2 -2kx +7k -12 = 0

two equal roots -->Δ = b^2 -4ac = 4k^2 - 4(7k-12)=0 = 4(k^2 - 7k +12)=4(k-4)(k-3)=0

so k =3 and k =4

5 0
3 years ago
Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com
fredd [130]

f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}

f has critical points where the derivative is 0:

2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1

The second derivative is

f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}

and f''(-1)=\frac2e>0, which indicates a local minimum at x=-1 with a value of f(-1)=\frac1e.

At the endpoints of [-2, 2], we have f(-2)=1 and f(2)=e^8, so that f has an absolute minimum of \frac1e and an absolute maximum of e^8 on [-2, 2].

So we have

\dfrac1e\le f(x)\le e^8

\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx

\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}

5 0
3 years ago
In one year, the student enrollment increased from 560 to 588. This was an increase of what percent?
sergiy2304 [10]

Answer:

5% Increase

Step-by-step explanation:

560 * 0.05 = 28

560 + 28 = 588

8 0
3 years ago
Read 2 more answers
A group of students volunteered to finish a task in 25 days .1 Q of students did not come n the work could b completed in 35 day
SpyIntel [72]

Answer:

p=7Q/2

Step-by-step explanation:

Original number of students:

p students to do 1 job in 25 days.

Let r= the rate for 1 student.

pr*25=1

pr*25=1 is the work rate equation for p students.

Lesser number of students:

p-Q students came to do the job and time required was 35 days.

(P-Q)*r*35=1.

The unknowns are p, Q and r

Equate the original number of students and the lesser number of students

pr*25=(P-Q)*r*35

25rp=35rp - 35Qr

Collect like terms

25rp-35rp = -35Qr

Divide both sides by -5

-5rp+7rp=- 7rp

It can be re written as

7rp-5rp=-7Qr

2rp=7Qr

Make p the subject of the formula

p=7Qr/2r

p=7Q/2

p=7Q/2 is the original number of students

-10rp = -35Qr

The system of these two equations can be solved for p. See the THREE unknown

variables, p, r, and Q. You might assume that either r or Q would be a constant.

5 0
3 years ago
In the equation below, t is the time in hours it toon to drive to the beach. 63t=315
Tju [1.3M]

Answer:

5 hours

Step-by-step explanation:

63t=315

t = 315/63

t=5

6 0
3 years ago
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