Question po

If the zeros of a quadratic functions are -2 and 4, which graph could represent the function?

anygoal [31]

Answer:

The Graph having the X-intercepts of (-2,0) and (4,0)

Step-by-step explanation:

The zeroes of the function is the X-intercept.

4 0

3 years ago

The minimum number of rigid transformations required to show that polygon ABCDE is congruent to polygon FGHIJ is 1 2 3 4 . A rot

MaRussiya [10]

I found the corresponding image. Pls. see attachment.

<span>The minimum number of rigid transformations required to show that polygon ABCDE is congruent to polygon FGHIJ is 2 (translation and rotation).

A rotation translation must be used to make the two polygons coincide.

A sequence of transformations of polygon ABCDE such that ABCDE does not coincide with polygon FGHIJ is a translation 2 units down and a 90° counterclockwise rotation about point D </span>

8 0

3 years ago

In the function f(x) = -720h + 10080, the "h" is the number of hours the pool has been draining. Explain in your words how much

Nataly [62]

f(x) = -720h + 10080

10080 is the total amount of water in the pool, 720 is the amount of water you lose each hour.

"h" is the number of hours the pool has been draining, and since you know the pool has been draining for 12 hours, you can plug in 12 for "h".

f(x) = -720h + 10080

f(x) = -720(12) + 10080

f(x) = -8640 + 10080

f(x) = 1440

After 12 hours of draining, 1440 is the amount of water left in the pool. (I don't know the units [ex: gallons, etc.])

3 0

3 years ago

Use the model below to estimate the average annual growth rate of a certain country's population for 1950, 1988, and 2010, where

Morgarella [4.7K]

Answer:

The average annual growth rate of a certain country's population for 1950, 1988, and 2010 are 2.398, 0.9985 and 0.2236 respectively.

Step-by-step explanation:

The given equation is

$Y=-0.0000084x^3+0.00211x^2-0.205x+8.423$

Where Y is the annual growth rate of a certain country's population and x is the number of years after 1900.

Difference between 1950 and 1900 is 50.

Put x=50 in the given equation.

$Y=-0.0000084(50)^3+0.00211(50)^2-0.205(50)+8.423$

$Y=2.398$

Therefore the estimated average annual growth rate of the country's population for 1950 is 2.398.

Difference between 1988 and 1900 is 88.

Put x=88 in the given equation.

$Y=-0.0000084(88)^3+0.00211(88)^2-0.205(88)+8.423$

$Y=0.9984752\approx 0.9985$

Therefore the estimated average annual growth rate of the country's population for 1988 is 0.9985.

Difference between 2010 and 1900 is 110.

Put x=110 in the given equation.

$Y=-0.0000084(110)^3+0.00211(110)^2-0.205(110)+8.423$

$Y=0.2236$

Therefore the estimated average annual growth rate of the country's population for 2010 is 0.2236.

8 0

3 years ago

F(x) = 4√(2x³-1)<br> F'(x) =.....?

goldfiish [28.3K]

Answer:

$\large\boxed{f'(x)=\dfrac{12x^2}{\sqrt{2x^3-1}}}$

Step-by-step explanation:

$f(x)=4\sqrt{2x^3-1}=4\left(2x^3-1\right)^\frac{1}{2}\\\\f'(x)=4\cdot\dfrac{1}{2}(2x^3-1)^{-\frac{1}{2}}\cdot3\cdot2x^2=\dfrac{12x^2}{(2x^3-1)^\frac{1}{2}}=\dfrac{12x^2}{\sqrt{2x^3-1}}\\\\\text{used}\\\\\sqrt{a}=a^\frac{1}{2}\\\\\bigg[f\left(g(x)\right)\bigg]'=f'(g(x))\cdot g'(x)\\\\\bigg[nf(x)\bigg]'=nf'(x)\\\\(x^n)'=nx^{n-1}$

6 0

3 years ago