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3 years ago

8

Find cos(0) Please help

1 answer:

Kisachek [45]3 years ago

6 0

Answer:

cos o=base /hypotenuse=EF/DF

c.is your answer

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Is 10.78778777 a rational or irrational number

3241004551 [841]

Answer:

Im pretty sure its irrational

Step-by-step explanation:

5 0

3 years ago

Create a function that has a derivative of f(x) = 2x + 1 at a given input value of your choice.

Taya2010 [7]

The opposite operation to taking a derivative is an integral. Integrate to find original function.

f(x) = x^2 + x + C

C is constant because we didn't have bounds on the integral. it lets you choose the input value let's say is just (1,0)

0 = 1 + 1 + C
- 2 = C

function at input value (1,0)
f(x) = x^2 + x - 2

now if you check by taking derivative is :
f' = 2x + 1

6 0

2 years ago

If you shift the quadratic parent function, f(x) = x2, left 12 units, what is the equation of the new function?

Shtirlitz [24]

C is your answer i believe

3 0

3 years ago

Hitman42 [59]

Answer:

this is an equilateral triangle

by using distance formula

5 0

3 years ago

Find the quotient using polynomial long division:

Snowcat [4.5K]

(1) $50y^3=5y\cdot 10y^2$ , and

$10y^2\cdot(5y-4)=50y^3-40y^2$

Subtract this from $50y^3+10y^2-35y-7$ to get a remainder of

$(50y^3+10y^2-35y-7)-(50y^3-40y^2)=50y^2-35y-7$

What we've done here is show that

$\dfrac{50y^3+10y^2-35y-7}{5y-4}=10y^2+\dfrac{50y^2-35y-7}{5y-4}$

We can keep going as long as the degree of the remainder's numerator is at least the same as the degree of its denominator.

Next, $50y^2=5y\cdot 10y$ , and

$10y\cdot(5y-4)=50y^2-40y$

Subtract this from the previous remainder to get a new remainder of

$(50y^2-35y-7)-(50y^2-40y)=5y-7$

which means

$\dfrac{50y^3+10y^2-35y-7}{5y-4}=10y^2+10y+\dfrac{5y-7}{5y-4}$

Finally, $5y=5y\cdot1$ , and

$1\cdot(5y-4)=5y-4$

Subtract this from the previous remainder to get

$(5y-7)-(5y-4)=-3$

So we end up with

$\dfrac{50y^3+10y^2-35y-7}{5y-4}=\boxed{10y^2+10y+1}-\dfrac3{5y-4}$

(the quotient is the expression in the box)

(2) Using the same process as before:

$8m^4=2m\cdot4m^3$

$4m^3\cdot(2m+1)=8m^4+4m^3$

$(8m^4-4m^2+m+4)-(8m^4+4m^3)=-4m^3-4m^2+m+4$

$\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-\dfrac{4m^3+4m^2+m+4}{2m+1}$

$-4m^3=2m\cdot(-2m^2)$

$-2m^2\cdot(2m+1)=-4m^3-2m^2$

$(-4m^3-4m^2+m+4)-(-4m^3-2m^2)=-2m^2+m+4$

$\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-2m^2-\dfrac{2m^2-m-4}{2m+1}$

$-2m^2=2m\cdot(-m)$

$-m\cdot(2m+1)=-2m^2-m$

$(-2m^2+m+4)-(-2m^2-m)=2m+4$

$\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=4m^3-2m^2-m+\dfrac{2m+4}{2m+1}$

$2m=2m\cdot1$

$1\cdot(2m+1)=2m+1$

$(2m+4)-(2m+1)=3$

$\implies\dfrac{8m^4-4m^2+m+4}{2m+1}=\boxed{4m^3-2m^2-m+1}+\dfrac3{2m+1}$

8 0

3 years ago