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antoniya [11.8K]

2 years ago

14

MAY YALL PLZ HELP MEH????????????

2 answers:

pentagon [3]2 years ago

7 0

Answer:

small = $2.80

large = $21.60

Step-by-step explanation:

Small = We know for 1, it's equal to 2.80. So, if they only need 1 it would be 2.80.

Large = We know for 1, it's equal to 3.60. So, if they need 6, we would do 3.60 times 6 to get the answer of 21.60.

Andreas93 [3]2 years ago

5 0

Answer:

$24.40

Step-by-step explanation:

3.60×6=21.6

21.60+2.80=$24.40

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The students at Walden Middle School are selling tins of popcorn to raise money for new uniforms. They sold 42 tins in the first

sdas [7]

Answer:

$388.50

Step-by-step explanation:

42 tins

$9.25/tin

$42*9.25=388.5$

So, they raised $388.50

3 0

2 years ago

Read 2 more answers

For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instant

nirvana33 [79]

Answer:

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

Step-by-step explanation:

a) Geometrically speaking, the average rate of change of $y$ with respect to $x$ over the interval by definition of secant line:

$r = \frac{y(b) -y(a)}{b-a}$ (1)

Where:

$a$ , $b$ - Lower and upper bounds of the interval.

$y(a)$ , $y(b)$ - Function exaluated at lower and upper bounds of the interval.

If we know that $y = 3\cdot x^{2}$ , $a = 3$ and $b = 6$ , then the average rate of change of $y$ with respect to $x$ over the interval is:

$r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}$

$r = 27$

The average rate of change of $y$ with respect to $x$ over the interval $[3,6]$ is 27.

b) The instantaneous rate of change can be determined by the following definition:

$y' = \lim_{h \to 0}\frac{y(x+h)-y(x)}{h}$ (2)

Where:

$h$ - Change rate.

$y(x)$ , $y(x+h)$ - Function evaluated at $x$ and $x+h$ .

If we know that $x = 3$ and $y = 3\cdot x^{2}$ , then the instantaneous rate of change of $y$ with respect to $x$ is:

$y' = \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-3\cdot x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{(x+h)^{2}-x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}$

$y' = 6\cdot \lim_{h \to 0} x +3\cdot \lim_{h \to 0} h$

$y' = 6\cdot x$

$y' = 6\cdot (3)$

$y' = 18$

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

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2 years ago

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Yuki888 [10]

The equations (-75)(-49)=3,675 the answer

4 0

2 years ago

Three less than the product of 18

ololo11 [35]

Answer:

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Step-by-step explanation:

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3 years ago

F(x) = x^2 + 4x. G(x) = 1 + e^2x. As a single logarithm find fg(x) = 21

vfiekz [6]

Answer:

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Step-by-step explanation:

I hope you understand

8 0

2 years ago