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3 years ago

-6h plus 4 - 3plus h equals 11

1 answer:

5 0

$- 6h + 4 - 3 + h = 11$
So we start by combining like terms;
$- 6h + h = - 5h$
$4 - 3 = 1$
Now we have;
$- 5h + 1 = 11$
We then subtract 1 from both sides;
$- 5h + 1 = 11$
$- 5h = 10$
Then we divide both sides by 5;
$- 5h = 10$
$- 5h \div - 5h = h$
$10 \div - 5h = - 2$
Leaving the final answer to be h = -2

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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

3 years ago

I’m stuck on this please

trasher [3.6K]

Answer:

weak negative

Step-by-step explanation:

because its in negative and its in the thousands place.

3 0

2 years ago

Last Question 10 minutes left Middle School Math

kicyunya [14]

Answer:2,2

Step-by-step explanation:

I remember doing this last year

8 0

2 years ago

Read 2 more answers

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makvit [3.9K]

Answer:

2,-8

Step-by-step explanation:

this should be the two answers

6 0

2 years ago

Given the center of a circle as ( 1, -1 ) and a point on the circle at (4 , 2). Write the equation of the circle. *

Andru [333]

Answer:

(x − 1)² + (y + 1)² = 18

Step-by-step explanation:

Equation of a circle is:

(x − h)² + (y − k)² = r²

where (h, k) is the center and r is the radius.

The center is (1, -1), so plugging that in:

(x − 1)² + (y + 1)² = r²

A point on the circle is (4, 2), so plugging that in:

(4 − 1)² + (2 + 1)² = r²

18 = r²

Therefore, the equation is:

(x − 1)² + (y + 1)² = 18

7 0

3 years ago