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3 years ago

Helpppp meeeeee ill mark brainliest

Mathematics

2 answers:

Paraphin [41]3 years ago

7 0

Um... the expression, 10x2 is what? 20 or 20 x

Mazyrski [523]3 years ago

7 0

Answer:

(5x-2)(2x-3)

Step-by-step explanation:

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Which shows two expressions that are equivalent to (Negative 7) (Negative 15) (Negative 5) Yan is climbing down a ladder. Each t

Svet_ta [14]

Answer:

(-8) + (-8) + (-8) + (-8)

Step-by-step explanation:

Since Yan is descending ; we represent his movement as being negative.

Number of rungs descended before stopping = 4

Number of stops = 8

Given that no extra steps was taken ;

Number of ladder rungs descended :

4 rungs per stop for 8 stops

- 4 * 8 = - 32 rungs

Another way to write the product :

Number of stops added in number of rungs descended :

(-8) + (-8) + (-8) + (-8)

3 0

2 years ago

Please answer it now in two minutes

Nikitich [7]

Answer:

3√6

Step-by-step explanation:

tan60=opp/adj

opp(d)=tan60*3√2=√3*3√2=3√6

7 0

3 years ago

Suppose a triangle has sides 3, 4, and 6. Which of the following must be true?

leonid [27]

Answer:

The answer would be B

Step-by-step explanation:

Hope this helps:)....if not then sorry for wasting your time and may God bless you:)

8 0

1 year ago

Read 2 more answers

Determine any asymptotes (Horizontal, vertical or oblique). Find holes, intercepts and state it's domain.

vesna_86 [32]

Factorize the denominator:

$\dfrac{x^2-4}{x^3+x^2-4x-4}=\dfrac{x^2-4}{x^2(x+1)-4(x+1)}=\dfrac{x^2-4}{(x^2-4)(x+1)}$

If $x\neq\pm2$ , we can cancel the factors of $x^2-4$ , which makes $x=-2$ and $x=2$ removable discontinuities that appear as holes in the plot of $g(x)$ .

We're then left with

$\dfrac1{x+1}$

which is undefined when $x=-1$ , so this is the site of a vertical asymptote.

As $x$ gets arbitrarily large in magnitude, we find

$\displaystyle\lim_{x\to-\infty}g(x)=\lim_{x\to+\infty}g(x)=0$

since the degree of the denominator (3) is greater than the degree of the numerator (2). So $y=0$ is a horizontal asymptote.

Intercepts occur where $g(x)=0$ ( $x$ -intercepts) and the value of $g(x)$ when $x=0$ ( $y$ -intercept). There are no $x$ -intercepts because $\dfrac1{x+1}$ is never 0. On the other hand,

$g(0)=\dfrac{0-4}{0+0-0-4}=1$

so there is one $y$ -intercept at (0, 1).

The domain of $g(x)$ is the set of values that $x$ can take on for which $g(x)$ exists. We've already shown that $x$ can't be -2, 2, or -1, so the domain is the set

$\{x\in\mathbb R\mid x\neq-2,x\neq-1,x\neq2\}$

8 0

3 years ago

WHATS THE MEAN?<br> WHATS THE MEDIAN?<br> PLZ HELP ASAP

SCORPION-xisa [38]

Answer:

Step-by-step explanation:

3 0

2 years ago