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3 years ago

What is 56x2 Just a regular normal question from me Your welcome in advance

Mathematics

1 answer:

guapka [62]3 years ago

4 0

Answer:

112

Step-by-step explanation:

hope this helps

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A box stall is a square or rectangular enclosure that can house one or more animals. Derek wants to construct a square box stall

stepladder [879]

Your answer is 550000!!!

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3 years ago

The temperature dropped 12F in 8 hours .if the final temperature was -7,what was the starting temperature

bekas [8.4K]

Just add 12 to -7, -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12

6 0

3 years ago

Need answer within 15 min

nikdorinn [45]

Answer:

Step-by-step explanation:

12-6+4

6 0

2 years ago

I need help bruv. For the function y=x^2 - 5x - 6 .

lara [203]

Answer:

a. Vertex = (2.5,-12.25)

b. y-intercept = (0,-6)

c. x-intercept = (6,0) ; (-1,0)

Step-by-step explanation:

Your equation is written as:

$y = ax^{2} + bx +c$

The easiest way to find the vertex is writing the equation this way:

$y = a(x-h)^{2}+k$

Being the vertex (h,k)

So first complete the square

$y = x^{2} -5x -6 \\y = x^{2} - 2(2.5x) -6 \\y = x^{2} - 2(2.5x) +2.5^{2} - 2.5^{2} -6 \\y = (x^{2} -2(2.5x) + 2.5^{2}) -6.25 - 6 \\y = (x-2.5)^{2} -12.25$

Vertex : (2.5,-12.25)

To find the y-intercept you need to replace the equation when x = 0 and get y

$y = (0)^{2} - 5(0) -6\\y = - 6$

To find the x-intercept you need to replace the equation when y = 0 and get x

$0 = x^{2} - 5x -6 \\0 = (x-6)(x+1)\\x = 6\\x = -1$

4 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

2 years ago