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2 years ago

True or false? The graph represents a function

Mathematics

2 answers:

Cloud [144]2 years ago

8 0

Answer:

False. This is not a function.

Step-by-step explanation:

By using the vertical line test, it shows that this graph is not a function. The imaginary vertical line would go through the graph twice. To be a function, the imaginary vertical line can only go through once.

Viefleur [7K]2 years ago

8 0

The statement is false because if the red line had a vertical line going through it the line would hit it twice as shown in the picture below

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What equation represents the relationship between p and q shown in the table

mash [69]

q = 10p because all of them are multiplied by ten

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2 years ago

Help!!!!! Please!!!!!!!! Math!!!!!! Quickly!!!!!! Brainiest answer to correct answer!!!!!!!!

muminat

The answer should be 103. Because the interior angles of all triangles add up to 180 degrees, you can just add together the two known interior angles; subtract them from 180; and then find the supplement.

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2 years ago

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In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

2 years ago

In the diagram B is the midpoint of a A and C. B is also the midpoint or E and F. C is the midpoint of A and D.

aivan3 [116]

Answer:

AD = 84

Step-by-step explanation:

Since B is the midpoint of AC then AB = BC = 2x - 5

Since C is the midpoint of AD then AC = CD, thus

AB + BC = CD, that is

2x - 5 + 2x - 5 = x + 29

4x - 10 = x + 29 ( subtract x from both sides )

3x - 10 = 29 ( add 10 to both sides )

3x = 39 ( divide both sides by 3 )

x = 13

Hence

AD = AB + BC + CD

= 2x - 5 + 2x - 5 + x + 29 = 5x + 19, thus

AD = (5 × 13) + 19 = 65 + 19 = 84

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3 years ago

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garri49 [273]

The answer is 0.2 hope it helps

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