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2 years ago

NEED HELP !! please word problems

Mathematics

1 answer:

Anika [276]2 years ago

5 0

Answer:

216.108280255π

(maybe round it to 216.1π or 216π)

Step-by-step explanation:

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A baker makes apple tarts and apple pies each day. Each tart, t, requires 1 apple, and each pie, p, requires 8 apples. The baker

Lynna [10]

For this case, the first thing we must do is define variables.
We have then:
t: number of tarts
p: number of pies
We now write the system of equations:
Each tart, t, requires 1 apple, and each pie, p, requires 8 apples. The baker receives a shipment of 184 apples every day:
8p + t ≤ 184
the baker makes no more than 40 tarts per day:
t ≤ 40
Answer:
A system of inequalities that can be used to find the possible number of pies and tarts the baker can make is:
D. t ≤ 40
8p + t ≤ 184

5 0

3 years ago

Read 2 more answers

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Marina86 [1]

Answer:

B I think..................................

4 0

1 year ago

Pls help it is 7th grade math

JulsSmile [24]

$39+\dfrac 15 w \\\\=39+\dfrac 15 \left(-\dfrac 12 \right)\\\\\\=39 - \dfrac 1{10}\\\\\\=\dfrac{389}{10}\\\\\\=38\dfrac{9}{10}$

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2 years ago

A box at a yard sale contains 3 different china dinner sets, each consisting of 5 plates. A customer will randomly select 2 plat

marshall27 [118]

Answer:

$\dfrac{2}{7}$

Step-by-step explanation:

3 different china dinner sets, each consisting of 5 plates consist of 15 plates.

A customer can select 2 plates in

$C^{15}_2=\dfrac{15!}{2!(15-2)!}=\dfrac{15!}{13!\cdot 2!}=\dfrac{13!\cdot 14\cdot 15}{2\cdot 13!}=7\cdot 15=105$

different ways.

2 plates can be selected from the same dinner set in

$3\cdot C^5_2=3\cdot \dfrac{5!}{2!(5-2)!}=3\cdot \dfrac{3!\cdot 4\cdot 5}{2\cdot 3!}=3\cdot 2\cdot 5=30$

different ways.

Thus, the probability that the 2 plates selected will be from the same dinner set is

$Pr=\dfrac{30}{105}=\dfrac{6}{21}=\dfrac{2}{7}$

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3 years ago

Why is the derivative of a constant zero

andreev551 [17]

A constant function is exactly that - constant - meaning it exhibits no change whatsoever with respect to any change in its input. If $f(x) = 1$ , then it doesn't matter what value of $x$ I give you, the value of $f(x)$ will always be nothing other than 1.

That the derivative of such a function is zero follows immediately from the definition of the derivative. If $c\in\Bbb R$ and $f(x) = c$ , then

$f'(x) = \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \lim_{h\to0} \frac{1 - 1}h = \lim_{h\to0}\frac0h = 0$

8 0

1 year ago