What are the zeros of the quadratic function f(x) = 6x2

Nadiya determines that there are no fractions equivalent to 2/10 with a denominator greater than 10, but less than 20, that have

frosja888 [35]

Answer:

Yes, Nadiya is correct. By multiplying the numerator and denominator by 2, the first fraction equivalent to 2/10 is 4/20.

Step-by-step explanation:

To find a fraction equivalent to 2/10, we need to multiply (or divide) the numerator and denominator by the same nonzero whole number.

As we need an equivalent fraction with a denominator greater than 10, we will need to multiply and not divide.

The first nonzero whole number we have is 1. If we multiply the numerator and denominator by 1, we get 2/10. Obviously, 2/10 is equivalent to 2/10 but the denominator is not greater than 10, so it doesn't help us.

The next whole number is 2. When we multiply the numerator and denominator by 2, we get 4/20. The denominator is not less than 20.

If we keep going, we will get 6/30, 8/40 and so on.

Therefore, Nadiya is correct.

The correct answer is the first one.

8 0

2 years ago

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Triangle ABC was translated according to the rule (x, y) → (x + 1.5, y – 3.5) to create the image ΔA'B'C' shown on the coordinat

saveliy_v [14]

Answer:

Graph C

Step-by-step explanation:

Just count out the number of lefts and rights, ups and downs.

6 0

1 year ago

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You are visiting your friend Fabio's house. You find that, as a joke, he filled his swimming pool with Kool-Aid, which dissolved

Rzqust [24]

Answer:

The amount of work required to remove all of the water by pumping it over the side = 3.12 × 10⁶ lbs.ft

Step-by-step explanation:

Work done in moving anything from point A to point B = Fx

For this setup,

If we take an elemental vertical height, dx, The volume would be A.dx

where A = Cross sectional Area = πr² = π(18)² = 1017.88 ft²

dV = 1017.88 dx

The elemental force on that part will be

dF = ρg dV

ρg = 63.8 lbs/ft³

dF = 63.8 × 1017.88 dx = 64940.5 dx

F = ∫dF = 64940.5 dx

W = Fx = (∫dF)x = ∫ 64940.5x dx = 64940.5 ∫ xdx

We'll be integrating from (11 - 6) ft to 11 ft because that's the total height it'll be pumped through

W = 64940.5 (x²/2)¹¹₅ = 64940.5((11² - 5²)/2) = 3.12 × 10⁶ lbs.ft

4 0

2 years ago

Approximate the area under the curve y = x² from x = 2 to x = 5 using a Right Endpoint approximation with 6 subdivisions.

Tanzania [10]

Answer:

$\text{Area}\,=36.75$

Step-by-step explanation:

Using right estimation point simply means to form a bunch of rectangles between the two limits, x =2 and x = 5. and add the areas of all those rectangles.

There must be 6 subdivisions between 2 and 5. so, to do that:

$\Delta{x}=\dfrac{5-2}{6}=0.5$

the length of each subdivision is 0.5 units. That also means that the 6 rectangles in between the limits will each have the base length of 0.5 units.

So the endpoints of each subdivision from 3 to 5 will be:

$\begin{tabular}{|c|c|c|c|c|}3&3.5&4&4.5&5\\\end{tabular}$

By <em>right </em>endpoint approx<em>, </em>we mean that the height of the rectangles will be determined by the right endpoint of each subdivision, that is, it must be equal to the function value of the first limit.

$\begin{tabular}{|c|c|c|}subdivision&$x$&height($y=x^2$)&3 to 3.5&3.5&12.25&3.5 to 4&4&16&4 to 4.5&4.5&20.25&4.5 to 5&5&25\end$

Note that we have used the right-end-point of the subdivision to determine the height the rectangles.

All that's left to do now is to simply calculate the areas of the each of the rectangles. And add them up.

the base of each of the rectangle is $\Delta{x}=0.5$

and the height is determined in the table above.

$\text{Area}\,=(0.5\times12.25)+(0.5\times16)+(0.5\times20.25)+(0.5\times25)$

$\text{Area}\,=0.5(12.25+16+20.25+25)$

$\text{Area}\,=36.75$

3 0

2 years ago

Choose any positive integer. Powers of two here are not very interesting, so choose something else. If the number you have chose

Yakvenalex [24]

Answer:

Step-by-step explanation:

Let the integer be 6 for even and 7 for odd (say)

For 6, we divide by 2, now get 3. Now we multiply by 3 and add 1 to get 10. Now since 10 is even divide by 5, now multiply by 3 and add 1 to get 16. Now divide by 2 again by 2 again by 2 again by 2 till we get rid of even numbers.

The result is 1, so multiply by 3 and add 1 we get 4 now divide 2 times by 2 to get 1, thus this result now again repeats after 2 times.

Say if we select off number 3, multiply by 3 and add 1 to get 10 now divide by 5, now repeat the same process as above for 5 until we get 1 and it gets repeated every third time.

Thus whether odd or even after some processes, we get 1 and the process again and again returns to 1.

5 0

1 year ago

What are the zeros of the quadratic function f(x) = 6x2 – 24x + 1?