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3 years ago

Sanjay has 2 1/3 cups of pretzels, Raul has 1 5/6 cups of chocolate pieces, and Sonny has 3 7/9 cups of granola. If they

1 answer:

4 0

Sanjay and craig best friends you know that tv show from cartoon network yeah anyways they can make about 1,000 bags of trail mix

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John gets a new job and receive a 500 signing bonus after he makes 200 a day

FromTheMoon [43]

Okay, I will start talking you through it, it'll be quick but I need your response. Just leave a comment! Loves,
-Brainiac224

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3 years ago

in order to be invited to tutoring or extension activities a student's performance must be at least 20 points higher or lower th

Ad libitum [116K]

Answer:

p ≤ 85 or p ≥ 125

Step-by-step explanation:

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3 years ago

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How to show these 2 problems are inverses

nataly862011 [7]

$\bf \begin{cases}
f(x)=\sqrt[3]{7x-2}\\\\
g(x)=\cfrac{x^3+2}{7}
\end{cases}\\\\
-----------------------------\\\\
now
\\\\
f[\ g(x)\ ]\implies f\left[ \frac{x^3+2}{7} \right]\implies \sqrt[3]{7\left[ \frac{x^3+2}{7} \right]-2}\implies \sqrt[3]{x^3+2-2}
\\\\\\
\sqrt[3]{x^3}\implies x\\\\
-----------------------------\\\\
or
\\\\
g[\ f(x)\ ]\implies g\left[\sqrt[3]{7x-2}\right]\implies \cfrac{\left[\sqrt[3]{7x-2}\right]^3+2}{7}
\\\\\\
\cfrac{7x-2+2}{7}\implies \cfrac{7x}{7}\implies x$

thus f[ g(x) ] = x indeed, or g[ f(x) ] =x, thus they're indeed inverse of each other

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3 years ago

If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w

MatroZZZ [7]

Let $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}$ be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

$\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.$

2. If the mean weight of the 7 other players is 202 lb, then

$\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.$

3. From the previous statements you have that

$x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.$

Add these two equalities and then divide by 11:

$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.$

Answer: the mean weight of the 11-person team is $208\dfrac{10}{11}\ lb.$

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3 years ago

A number, a power of a variable, or a product of the two is a monomial, while a _________ is the sum of monomials

andrew-mc [135]

A polynomial is the sum of monomials

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3 years ago

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