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11111nata11111 [884]

3 years ago

13

A school club sold 300 shirts. 31% were sold to fifth graders, 52% were sold to sixth graders, and the rest were sold to teacher

s. How many shirts were sold to each group-fifth graders, 6th graders, and teachers? Explain or show your reasoning. Please help I'm late on turning in homework!

2 answers:

Tamiku [17]3 years ago

4 0

300x0.31 = 93 shirts sold to 5th graders
300x0.52 = 156 shirts sold to 6th graders
156+93 = 249 shirts sold to students
300-249 = 51 shirts sold to teachers

Aleks [24]3 years ago

3 0

Answer:

5th graders = 93 people 6th graders = 156 teachers = 51

Step-by-step explanation:

93 5th graders bought shirts, 156 6th graders bought shirts, 51 teachers bought shirts

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NNADVOKAT [17]

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3 years ago

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timurjin [86]

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2 years ago

Two numbers that multiply to -100 and add to 15

Whitepunk [10]

Answer:

-5 and 20

Step-by-step explanation:

-5 + 20 = 15

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3 0

3 years ago

Evaluate the interval (Calculus 2)

Darya [45]

Answer:

$2 \tan (6x)+2 \sec (6x)+\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}$

If the terms are multiplied by constants, take them outside the integral:

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x$

Multiply by the conjugate of 1 - sin(6x) :

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x$

$\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:$

$\implies \sin^2 (6x) + \cos^2 (6x)=1$

$\implies \cos^2 (6x)=1- \sin^2 (6x)$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

Expand:

$\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

$\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:$

$\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x$

$\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$

$\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}$

$\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}$

Simplify:

$\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}$

$\implies 2 \tan (6x)+2 \sec (6x)+\text{C}$

Learn more about indefinite integration here:

brainly.com/question/27805589

brainly.com/question/28155016

3 0

2 years ago

Need help....... thx

Blababa [14]

I got 75582.

Explanation:

First, group the 40 identical candies into 20 pairs. It doesn't matter how since the candies are identical. This grouping will ensure that any assigment will contain at least two candies.

Then think of the 20 groups a 20 beads on a string. We are looking to place 11 separators between them to obtain 12 segments, each with a varying number of beads between them. How many ways are there to place 11 separators to 19 potential spaces between beads? The asnwer is $n = {19\choose{11}}=75582$

5 0

3 years ago