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3 years ago

Use the vertical line test to determine if the relation {(–6, –2), (–2, 6), (0, 3), (3, 5)} is a function

Mathematics

1 answer:

hammer [34]3 years ago

6 0

Answer: Its a function

Step-by-step explanation:

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Every day, Kylie's orange juice stand uses 1/2 of a bag of oranges. How many days will

nordsb [41]

Answer:

9 Days

Step-by-step explanation:

1/2x10=5

1/2x8=4

Logic tells you that 1/2x9=4 1/2

4 1/2 divided by 9= 1/2

So this answer is reasonable!!!

Please mark me brainlist!!! UwU

3 0

4 years ago

Please don't give me a link as an answer

tino4ka555 [31]

Answer:

x=2

Step-by-step explanation:

x+2=4

x=2

6 0

3 years ago

Read 2 more answers

Given f(x) = x − 7 and g(x) = x2 . Find g(f(−1)).

ivann1987 [24]

Answer:

Step-by-step explanation:

$g(f(-1))$ can be found by first finding $f(-1)$ then taking that result and putting it into the function $g$ .

$f(-1)$ means we are going to take the expression $x-7$ and evaluate it for $x=-1$ :

$-1-7$

$-8$

So $f(-1)=-8$ .

So we have this so far:

$g(f(-1))=g(-8)=(-8)^2=64$ .

$g(-8)$ was found by replacing $x$ in $x^2$ with $-8$ .

8 0

3 years ago

Write an equation of the line in slope-intercept form: y-intercept of 2, x-intercept of 4

zvonat [6]

Answer:

$\sf y=-\dfrac12x+2$

Step-by-step explanation:

y-intercept is when x = 0, so (0, 2)

x-intercept is when y = 0, so (4, 0)

$\sf slope=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{0-2}{4-0}=-\dfrac12$

Slope-intercept form of linear equation: $\sf y=mx+b$

(where m is the slope and b is the y-intercept)

Given:

$\sf m=-\dfrac12$
b = 2

$\sf \implies y=-\dfrac12x+2$

3 0

3 years ago

likoan [24]

It's easy to show that $7\tan(4x)$ is strictly increasing on $x\in\left[0,\frac\pi8\right]$ . This means

$M = \max \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/12} = 7\sqrt3$

and

$m = \min \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/16} = 7$

Then the integral is bounded by

$\displaystyle 7\left(\frac\pi{12} - \frac\pi{16}\right) \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le 7\sqrt3 \left(\frac\pi{12} - \frac\pi{16}\right)$

$\implies \displaystyle \boxed{\frac{7\pi}{48}} \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le \boxed{\frac{7\sqrt3\,\pi}{48}}$

7 0

2 years ago