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3 years ago

Which expression is equivalent to 7a2b + 10a2b2 + 14a2b3?

2 answers:

4 0

The given expression can be simplified in many ways by grouping like terms. The simplest form is obtained by factoring out a²b which gives us the following expression.

a²b(7 + 10b +14b²)

igomit [66]3 years ago

4 0

Answer:

$a^2b(7 + 10b + 14b^2)$

Step-by-step explanation:

Given : $7a^2b + 10a^2b^2 + 14a^2b^3$

To Find: Which expression is equivalent to $7a^2b + 10a^2b^2 + 14a^2b^3$

Solution:

$7a^2b + 10a^2b^2 + 14a^2b^3$

Take $a^2$ as common

$a^2(7b + 10b^2 + 14b^3)$

Now take b as common

$a^2b(7 + 10b + 14b^2)$

Hence $a^2b(7 + 10b + 14b^2)$ is equivalent to $7a^2b + 10a^2b^2 + 14a^2b^3$

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Foil: (x+9)(3x-4)

QveST [7]

QUESTION 1

The given binomial is;

$(x+9)(3x-4)$

First terms are multiplied: $x\times 3x=3x^2$

Outside terms are multiplied: $x\times -4=-4x$

Inside terms are multiplied: $9\times 3x=27x$

Last terms are Multiplied: $9\times -4=-36$

This gives us;

$=3x^2-4x+27x-36$

$=3x^2+23x-36$

QUESTION 2

We want to factor

$6r^5s+4r^4s^2-8r^2s$

The HCF is $2r^2s$

We factor to get;

$2r^2s(3r^3+2r^2s-4)$

QUESTION 3;

$x^2+x-2=0$

Split the middle term;

$x^2+2x-x-2=0$

Factor

$x(x+2)-1(x+2)=0$

$(x+2)(x-1)=0$

$(x+2)=0,(x-1)=0$

$x=-2,x=1$

The solutions are;

$x=-2$ and $x=1$

These are the x-intercepts of the graph of the function

$f(x)=x^2+x-2$

3 0

3 years ago

Find the absolute maximum and minimum values of the following function on the given interval. If there are multiple points in a

Bond [772]

$f(x)=3\sin x+3\cos x\implies f'(x)=3\cos x-3\sin x$

$f$ has critical points where $f'=0$ :

$3\cos x-3\sin x=0\implies\cos x=\sin x\implies\tan x=1\implies x=\pm\dfrac\pi4+2n\pi$

where $n$ is any integer. We get solutions in the interval $\left[0,\frac\pi3\right]$ for $n=0$ , for which $x=\frac\pi4$ .

At this critical point, we have $f\left(\frac\pi4\right)=3\sqrt2\approx4.243$ .

At the endpoints of the given interval, we have $f(0)=3$ and $f\left(\frac\pi3\right)=\frac{3+3\sqrt3}2\approx4.098$ .

So we have the extreme values

$\max\limits_{x\in\left[0,\frac\pi3\right]}f=3\sqrt2$

$\min\limits_{x\in\left[0,\frac\pi3\right]}f=3$

8 0

3 years ago

Find the area of each figure. Use 3.14

grigory [225]

<h3>Answer: 23.13 square meters</h3>

===============================================================

Explanation:

The top and bottom semicircles can be joined to form a full circle. The same can be said about the left and right semicircles. Each circle has diameter 3 and radius 1.5

The area of one circle is

A = pi*r^2

A = 3.14*(1.5)^2

A = 7.065

So two circles doubles to 2*7.065 = 14.13 square meters in area.

Then the last step is to add on the area of the 3 by 3 square (area 3*3 = 9) to get 14.13+9 = 23.13

This area is approximate because pi = 3.14 is approximate. Use more decimal digits in pi to get a more accurate area. However, your teacher wants you to use this specific value so it's best to stick with 23.13

8 0

3 years ago

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Is the following relation a function? Two circles are shown, one labeled x and the other labeled y. The x circle contains the nu

Allushta [10]

Answer:

i believe that it is a function.

Step-by-step explanation:

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