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2 years ago

If anyone can answer these 4 questions it would be greatly appreciated beyond words can describe

Mathematics

1 answer:

irga5000 [103]2 years ago

8 0

Answer:

answer your own

brainliest me now

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Complete the equation for f(x)

Wewaii [24]

Answer:

-2 would be the answer

Step-by-step explanation:

Can you please mark me brainliest answer

7 0

3 years ago

I don’t know, someone help please.

NeTakaya

Answer:

Radius: 4

Center: (-5,4)

Step-by-step explanation:

To find the radius, you take the square root of what the equation is equal to (in this case that is 16)

To find the center, you set what is in the parenthesis equal to 0 so:

x+5=0 becomes x=-5

y-4=0 becomes y=4

Hope this helped!

3 0

3 years ago

What is the slope of the segment shown for a staircase with 10-inch treads and 7.75-inch risers? As you walk up (or down) the st

s2008m [1.1K]

The formula for slope is

Slope = Rise / Run = Risers / Treads

Slope = 7.75 in /10 in = 0.775

The slope Is 0.775.

The vertical distance is not continuous, thus it is discrete. A discrete function only allows certain points in the interval. When you walk up or down the stairs, the vertical distances are not continuous because of the treads. You cannot stay also mid-air, so it should be discrete. An example of a continuous function would be when you walk up or down an inclined plane.

8 0

3 years ago

Integrala x la a treia ori ln la a doua dx va rog

Studentka2010 [4]

I don't speak Romanian, but the closest translation for this suggests you're trying to compute

$\displaystyle \int x^3 \ln(x)^2 \, dx$

Integrate by parts:

$\displaystyle \int x^3 \ln(x)^2 \, dx = uv - \int v \, du$

where

u = ln(x)² ⇒ du = 2 ln(x)/x dx

dv = x³ dx ⇒ v = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \int x^3 \ln(x) \, dx$

Integrate by parts again:

$\displaystyle \int x^3 \ln(x) \, dx = u'v' - \int v' du'$

where

u' = ln(x) ⇒ du' = dx/x

dv' = x³ dx ⇒ v' = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x) \, dx = \frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx$

So, we have

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \left(\frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx \right)$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \int x^3 \, dx$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \left(\frac14 x^4\right) + C$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac1{32} x^4 + C$

$\boxed{\displaystyle \int x^3 \ln(x)^2 \, dx = \frac1{32} x^4 \left(8\ln(x)^2 - 4\ln(x) + 1\right) + C}$

3 0

2 years ago

Please help quickly (math)

MrMuchimi

Answer:

We are asked for the angle of the translated figure ACB prime
When asked in that manner they want "C'"
Hence A'C'B'= 46 degree

5 0

3 years ago