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3 years ago

If one factor of 56x^4-42x^2y6 is 14x^2y^3 what is the other factor?

Mathematics

1 answer:

nataly862011 [7]3 years ago

7 0

14x^2 y^3( 4x^2-3y^3)

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PLEASE HELP whats the hypotenuse of a right triangle if both its legs are 9 cm round to the nearest tenth of a cm

vivado [14]

Pythagorean theorem: a² + b² = c²

a² and b² are the legs, and c² is the hypotenuse.

Plug in 9 to a² and b² since you know that both the right triangle's legs are 9 cm.

9² + 9² = c²
81 + 81 = c²
162 = c²

Square root both sides.

c = 12.7 cm

6 0

4 years ago

The drama club is selling tickets to the play to raise money for the shows expenses. Each student ticket sales for five dollars

djyliett [7]

Answer:

45 student tickets and 30 adult tickets

Step-by-step explanation:

You can check your work by plugging both answers into both of the original equations. :)

8 0

3 years ago

4 loads of stone weigh23ton. Find the weight of 1 load of stone.

mrs_skeptik [129]

Answer: 1 load of stone weighs 5.75 tons.

Step-by-step explanation: You'll need to divide 23 by 4 leaving you with 5.75. This will help you get a number that is potentially the weight of 1 load. To check to see if this is right, multiply 5.75 by 4 to get 23.

6 0

3 years ago

Which of these statements is true?

Talja [164]

Answer:

B all square are regular polygons

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3 years ago

Identify the x-intercepts of the function below f(x)=x^2+12x+24

damaskus [11]

<u>ANSWER: </u>

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

<u>SOLUTION:</u>

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$

Hence the x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

8 0

3 years ago