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4 years ago

Which of the following best describes the graph below?

Mathematics

2 answers:

Anni [7]4 years ago

5 0

The answer is a because the graph is slightly curved on the ends and passes the vertical line test

stellarik [79]4 years ago

5 0

Answer:

Step-by-step explanation:

It is a relation that is a function, because it passes vertical line test for a function.

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Can some person help me??????????????

lutik1710 [3]

Answer:

1.2d

198 donors

Step-by-step explanation:

d+0.2d=1.2d

20% more than last year= d+0.2d=1.2d

d=165

1.2(165)=198

Yes, you do get the same number with each of the expression because they are both the same exact things, except one of them has not combined like terms.

7 0

3 years ago

a line segment with one end at C(6,5) has midpoint m(4,2) determine the coordinates of the other endpoint d

meriva

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ C(\stackrel{x_1}{6}~,~\stackrel{y_1}{5})\qquad D(\stackrel{x_2}{x}~,~\stackrel{y_2}{y}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{x+6}{2}~~,~~\cfrac{y+5}{2} \right)~~=~~\stackrel{\stackrel{Midpoint}{M}}{(4~,~2)}\implies \begin{cases} \cfrac{x+6}{2}=4\\[1em] x+6=8\\ \boxed{x= 2}\\ \cline{1-1} \cfrac{y+5}{2}=2\\[1em] y+5=4\\ \boxed{y=-1} \end{cases}$

5 0

3 years ago

Hey <br> Help me please:) :)

zloy xaker [14]

The answer is 68 or B:68

8 0

3 years ago

Find the arc length of the given curve between the specified points.

yuradex [85]

Answer:

4.25

Step-by-step explanation:

Given:

$f(x) = \frac{x^3}{12} + \frac{1}{x} \\\\Arc Length = \int\limits^a_b {\sqrt{1 + (f'(x))^2} } \, dx \\f '(x) = \frac{x^2}{6} - \frac{1}{x^2}\\\\f'(x)^2 = (\frac{x^2}{6} - \frac{1}{x^2})^2 = \frac{x^4}{36} + \frac{1}{x^4} - \frac{1}{3}\\\\1 + f'(x)^2 = \frac{x^4}{36} + \frac{1}{x^4} + \frac{2}{3}\\\\\= \frac{x^8 + 36 + 12x^4}{36x^4}\\\\= \frac{(x^4 + 6)^2}{36x^4}\\\\=\sqrt{1 + f'(x)^2} = \sqrt{ \frac{(x^4 + 6)^2}{36x^4}}\\\\= \frac{x^2}{6} + \frac{1}{x^2} \\\\$

$ArcLength = \int\limits^4_1 {\frac{x^2}{6} + \frac{1}{x^2} } \, dx \\= (\frac{x^3}{18} - \frac{1}{x})\limits^4_1\\\\= (\frac{64}{18} - \frac{1}{4}) - (\frac{1}{18} - 1)\\\\= \frac{17}{4}= 4.25$

8 0

3 years ago

Is my answer correct?

nexus9112 [7]

Answer:

yes

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers