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3 years ago

What is the factored form of 36x4 – 25?

Mathematics

1 answer:

dalvyx [7]3 years ago

5 0

(6x2 + 5) • (5 - 6x2)

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Find VW. Please help BRAINLIEST.....

In-s [12.5K]

Answer:

sin50 =wx÷wv = 0.766÷1 = 6÷wv = 0.766wv = 6 => wv = 6÷0.766 => wv = 7.8328 ^-^

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4 years ago

Student Council is selling candy grams. They are using the formula

Luden [163]

Answer:

Our equation is:

y = f(x) = 2.5*x

The domain is the set of the values of x.

The range is the set of the values of y.

They must sell between 55 and 60 candy grams to meet their goal.

If we assume these as restrictions for the domain, then the minimum value of x is 55, and the maximum value of x is 60.

Then the domain is:

D: 55 ≤ x ≤ 60.

Now that we know the domain, we can find the range.

As our equation is linear with a positive coefficient, the minimum in the range will coincide with the minimum in the domain, then we have:

y = 2.5*55 = 137.5

And the maximum will coincide with the maximum in the domain:

y = 2.5*60 = 150.

Then the range is:

y = 137.5 ≤ y ≤ 150

7 0

3 years ago

Is (-3,4) a solution to the equation y=3x+107<br> O A solution<br> O Not a solution

Ira Lisetskai [31]

Answer:

Step-by-step explanation:

its is not a solution, however if the equal sign actually looks like

$\leqslant$

then it would be yes

4 0

3 years ago

Read 2 more answers

Plot the ordered paired (4,-5) state which quadrant or on which axis the point lies

Rudiy27

Answer:

The point is in the 4th quadrant.

Explanation:

Given the ordered pair(4, -5), we have that x = 4 and y = -5. Plotting this point, we'll have;

Note that quadrants are labelled in an anti-clockwise direction with the top right portion of the graph as the 1st quadrant. Looking at the plotted point, we can see that it i in the 4th quadrant.

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1 year ago

Explain why the function is discontinuous at the given number

Blizzard [7]

For $f$ to be continuous at $x=5$ , we need to have

$\displaystyle\lim_{x\to5}f(x)=f(5)$

Note that $x\to5$ means that $x\neq5$ , but that $x$ is *approaching* 5. We're told that for $x\neq5$ , we have

$f(x)=\dfrac{x^2-5x}{x^2-25}$

We can write

$\dfrac{x^2-5x}{x^2-25}=\dfrac{x(x-5)}{(x+5)(x-5)}=\dfrac x{x+5}$

and the limit would be

$\displaystyle\lim_{x\to5}\frac x{x+5}=\dfrac5{5+5}=\dfrac5{10}=\dfrac12\neq1=f(5)$

and so $f$ is discontinuous.

5 0

3 years ago