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3 years ago

Identify each expression as a monomial, binomial, trinomial, or none of the above.

1 answer:

3 0

You can identify an expression by the number of the terms in the expression.

monomial = one term
binomial = two terms
and so on.

$24x^{2}$ y contains only one term, so it is a monomial.

9q + $\frac{4q}{5p}$ - $3p^{2}$ contains three terms, so it is a trinomial.

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PLeAsE sEnD hElP !!!!!!!!!!!!!!!!

Luba_88 [7]

Answer:

5 units

Step-by-step explanation:

If DG, EG and FG are perpendicular bisectors of the sides of triangle ABC, then point G is the circumcenter of the triangle ABC and

BG = AG = CG as radii of the circumcirle.

Consider right triangle BEG. By the Pythagorean theorem,

$BG^2=EG^2+BE^2\\ \\BG^2=4^2+3^2\\ \\BG^2 =16+9\\ \\BG^2=25\\ \\BG=5\ units$

This gives us that

AG = BG = 5 units

3 0

3 years ago

What is the measure of

Helen [10]

Answer:

A = 37 degrees ; B = 82 degrees

Step-by-step explanation:

the angle A plus 143 must measure 180 degrees

angle A = 180 - 143 = 37 degrees

the sum of the inner angles of a triangle is 180 degrees

A + B + C = 180

37 + B + 61 = 180

B = 180 - 98 = 82 degrees

7 0

3 years ago

Read 2 more answers

I need help with this....

Vanyuwa [196]

Answer: x=4

Step-by-step explanation:

because 5x2=10 since there are 5 boxes and you subtract 10 from 30 so you get 20 and then you divide 20 by 5 for each box and get 4

(4+2)x5=30

hope this helps

6 0

2 years ago

Read 2 more answers

(5, -12); m

SOVA2 [1]

This is not the answer but i love ur pfp

6 0

2 years ago

Consider the following function.

Kryger [21]

Answer:

See below

Step-by-step explanation:

I assume the function is $f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, $x=0$ is the only vertical asymptote.

B) Set the first derivative equal to 0 and solve:

$f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

$f'(x)=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-5x+8$

$5x=8$

$x=\frac{8}{5}$

Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:

$f'(2)=-\frac{5}{2^2}+\frac{8}{2^3}=-\frac{5}{4}+\frac{8}{8}=-\frac{5}{4}+1=-\frac{1}{4}$

$f'(1)=-\frac{5}{1^2}+\frac{8}{1^3}=-\frac{5}{1}+\frac{8}{1}=-5+8=3$

Therefore, the function increases on the interval $(0,\frac{8}{5})$ and decreases on the interval $(-\infty,0),(\frac{8}{5},\infty)$ .

C) Since we determined that the slope is 0 when $x=\frac{8}{5}$ from the first derivative, plugging it into the original function tells us where the extrema are. Therefore, $f(\frac{8}{5})=1+\frac{5}{\frac{8}{5}}-\frac{4}{\frac{8}{5}^2 }=\frac{41}{16}$ , meaning there's an extreme at the point $(\frac{8}{5},\frac{41}{16})$ , but is it a maximum or minimum? To answer that, we will plug in $x=\frac{8}{5}$ into the second derivative which is $f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$ . If $f''(x)>0$ , then it's a minimum. If $f''(x)$ , then it's a maximum. If $f''(x)=0$ , the test fails. So, $f''(\frac{8}{5})=\frac{10}{\frac{8}{5}^3}-\frac{24}{\frac{8}{5}^4}=-\frac{625}{512}$ , which means $(\frac{8}{5},\frac{41}{16})$ is a local maximum.

D) Now set the second derivative equal to 0 and solve:

$f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$

$0=\frac{10}{x^3}-\frac{24}{x^4}$

$0=10x-24$

$-10x=-24$

$x=\frac{24}{10}$

$x=\frac{12}{5}$

We then test where $f''(x)$ is negative or positive by plugging in test values. I will use -1 and 3 to test this:

$f''(-1)=\frac{10}{(-1)^3}-\frac{24}{(-1)^4}=-34$ , so the function is concave down on the interval $(-\infty,0)\cup(0,\frac{12}{5})$

$f''(3)=\frac{10}{3^3}-\frac{24}{3^4}=\frac{2}{27}>0$ , so the function is concave up on the interval $(\frac{12}{5},\infty)$

The inflection point is where concavity changes, which can be determined by plugging in $x=\frac{12}{5}$ into the original function, which would be $f(\frac{12}{5})=1+\frac{5}{\frac{12}{5}}+\frac{4}{\frac{12}{5}^2 }=\frac{43}{18}$ , or $(\frac{12}{5},\frac{43}{18})$ .

E) See attached graph

5 0

3 years ago