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3 years ago

EXPERTS/ACE/GENIUSES/TRUSTED HELPERS HELPP ME

Mathematics

1 answer:

Bad White [126]3 years ago

8 0

Answer is B, the graph increases by about 0.28 per year

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21π<br> Rational<br> Irrational

lesantik [10]

Answer: Rational

Step-by-step explanation:

21 $\pi$ is rational because it is $\pi$ times 21 which is 65.9734457254 and numbers that terminate or repeat are rational.

4 0

3 years ago

A package of breadcrumbs to feed the ducks at the duck pond costs $0.25. How many packages of bread crumbs could be bought with

musickatia [10]

23.50 divided by .25 = 94

3 0

4 years ago

Read 2 more answers

Quadrilateral WXYZ has interior angles that meet the following criteria: Angles W and X are congruent . The smallest angle's mea

Reptile [31]

Smallest angle is $50$ °

<u>Step-by-step explanation:</u>

Here we have , Quadrilateral WXYZ has interior angles that meet the following criteria:

Angles W and X are congruent.

According to question , angles are congruent i.e. Both angle measure same

⇒ $X=W$

Now ,

• The smallest angle's measure is 100° less than twice the measure of W.

According to above statement we have ,

Smallest angle = $2W-100$

• The largest angle's measure is 10º more than twice the measure of X. ( correction , in question given as 2X )

According to above statement we have ,

Largest angle = $10+2X=10+2X$

Now , We know that sum of all 4 angles in 360 i.e.

⇒ $X+W+10+2X+2W-100=360$ { X=W }

⇒ $6W-90=360$

⇒ $6W=450$

⇒ $W=75$

Smallest angle is : $2W-100 = 2(75)-100=50$

Therefore , Smallest angle is $50$ ° .

4 0

3 years ago

3600 divided by 1760?

balu736 [363]

Answer:

$3600 \div 1760 \\ \\ = 2.04545$

5 0

3 years 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