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2 years ago

Explain how you know which terms to combine when combining like terms.

Mathematics

1 answer:

joja [24]2 years ago

8 0

Answer:

See like terms have the same variables

E.g 2x+4y-x+7y

We should take the same variable together with the same signs so here

2x-x will be 1x

4y+7y will be 11 y

So x+11y

Hope it helps!

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Write as a Fraction 65%

Licemer1 [7]

Answer:

13/20

Step-by-step explanation:

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2 years ago

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Drag the expressions into the boxes to correctly complete the table.

lora16 [44]

Answer:

SUMMARY:

$x^4+\frac{5}{x^3}-\sqrt{x}+8$ → Not a Polynomial

$-x^5+7x-\frac{1}{2}x^2+9$ → A Polynomial

$x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi$ → A Polynomial

$\left|x\right|^2+4\sqrt{x}-2$ → Not a Polynomial

$x^3-4x-3$ → A Polynomial

$\frac{4}{x^2-4x+3}$ → Not a Polynomial

Step-by-step explanation:

The algebraic expressions are said to be the polynomials in one variable which consist of terms in the form $ax^n$ .

Here:

$n$ = non-negative integer

$a$ = is a real number (also the the coefficient of the term).

Lets check whether the Algebraic Expression are polynomials or not.

Given the expression

$x^4+\frac{5}{x^3}-\sqrt{x}+8$

If an algebraic expression contains a radical in it then it isn’t a polynomial. In the given algebraic expression contains $\sqrt{x}$ , so it is not a polynomial.

Also it contains the term $\frac{5}{x^3}$ which can be written as $5x^{-3}$ , meaning this algebraic expression really has a negative exponent in it which is not allowed. Therefore, the expression $x^4+\frac{5}{x^3}-\sqrt{x}+8$ is not a polynomial.

Given the expression

$-x^5+7x-\frac{1}{2}x^2+9$

This algebraic expression is a polynomial. The degree of a polynomial in one variable is considered to be the largest power in the polynomial. Therefore, the algebraic expression is a polynomial is a polynomial with degree 5.

Given the expression

$x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi$

in a polynomial with a degree 4. Notice, the coefficient of the term can be in radical. No issue!

Given the expression

$\left|x\right|^2+4\sqrt{x}-2$

is not a polynomial because algebraic expression contains a radical in it.

Given the expression

$x^3-4x-3$

a polynomial with a degree 3. As it does not violate any condition as mentioned above.

Given the expression

$\frac{4}{x^2-4x+3}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

Therefore, is not a polynomial because algebraic expression really has a negative exponent in it which is not allowed.

SUMMARY:

$x^4+\frac{5}{x^3}-\sqrt{x}+8$ → Not a Polynomial

$-x^5+7x-\frac{1}{2}x^2+9$ → A Polynomial

$x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi$ → A Polynomial

$\left|x\right|^2+4\sqrt{x}-2$ → Not a Polynomial

$x^3-4x-3$ → A Polynomial

$\frac{4}{x^2-4x+3}$ → Not a Polynomial

3 0

2 years ago

Alex rents a car for one day. The charge is $18 plus $0.12 per mile. Alex wants to spend exactly $30. How many miles can he driv

katen-ka-za [31]

Answer:

Alex can drive for 100 miles.

Step-by-step explanation:

First, you do 30 - 18, which is 12. Divide 12 by 0.12 to get your answer, which is 100 miles.

Hope this answer helped!

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What is Ava's pay each month

jolli1 [7]

The answer to your question is $1978

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2 years ago

Point A is the midpoint of side XZ and point B is the midpoint of side YZ. Triangle X Y Z is cut by line segment A B. Point A is

AveGali [126]

Answer:

<h3>B. 4units</h3>

Step-by-step explanation:

Find the diagram attached. The diagram is a similar triangle.

From the diagram, XZ/XY = CA/AB

Given XZ = 2x-2+2x-2 = 4x-4

XY =5x-7

CA = 2x-2

AB= x+1

On substituting this parameters into the formula to get x first

4x-4/5x-7 = 2x-2/x+1

cross multiply

(4x-4)(x+1) = 5x-7(2x-2)

open the parenthesis

4x²+4x-4x-4 = 10x²-10x-14x+14

4x²-4 = 10x²-24x+14

10x²-4x²-24x+14+4 = 0

6x²-24x+18 = 0

x²-4x+3 = 0

x²-3x-x+3 = 0

x(x-3)-1(x-3) =0

(x-3)(x-1) = 0

x = 3 and 1

Next is to get length AX.

Given AX = 2x-2

Substitute x = 3 into the expression

AX = 2(3)-2

AX = 6-2

AX = 4 units

Hence the measure of length AX is 4 units

4 0

2 years ago