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2 years ago

5

Molly is baking for the Moms and Muffins event at her school. She will bake 4 batches of banana muffins. She needs 1 3/4 cups of

bananas for each batch of muffins.
Part A

Molly completed the multiplication below and said she needed 8 cups of bananas for 4 batches of muffins. What is Molly's error?

4x1 3/4 =4x 8/4 =32/4=8

Part B

What is the correct number of cups Molly needs for 4 batches of muffins? Explain how you found your answer

2 answers:

jasenka [17]2 years ago

7 0

Part A: it should be 7/4 not 8/4

Part B: it is 7 because 1 3/4 times 4 is 7

Grace [21]2 years ago

3 0

The correct answer is 1 and 8 because 4 times 1 will be 1 and 8 times 4 equals 32 and will be 8

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You buy 2 shirts and 1 pair of pants for $34. Your friend buys 3 shirts and 3 pairs of pants for $69. What is the price of 1 shi

V125BC [204]

You buy 3 shirts and 1 pair of pants for $32. Your friend buys 2 shirts and 3 pairs of pants for $40. How much did each item cost?

First let’s define a shirt as X and pants as Y!

In the first scenario, the equation will be 3X + Y =$32

In the second scenario, the equation will be 2X + 3Y = $40

Now because we have two different variables (X & Y) let’s solve for one of them (Ex: Y) to simplify the equations!

3X + Y= $32—> subtract 3X on both sides of the equation

-3X. - 3X

Net result: Y= 32- 3X

Now we can insert this into the second equation 2X + 3Y = $40 in order to solve for X variable

2X + 3(32-3X) = $40

2X + 96 - 9X = $40 —> simplify the equation

-7X + 96 = $40 —> substrate 96 from both sides

-7X = -56 —> divide both sides by -7

X= 8

Now reinsert 8= X into 3X + Y = $32

3(8) + Y = $32

24+ Y = $32 —> substract 24 from both sides

y = 8

ANSWER: X= 8, Y= 8

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2 years ago

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3 years ago

Find the exact value of the expression.<br> tan( sin−1 (2/3)− cos−1(1/7))

Sonja [21]

Answer:

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

Step-by-step explanation:

I'm going to use the following identity to help with the difference inside the tangent function there:

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

Let $a=\sin^{-1}(\frac{2}{3})$ .

With some restriction on $a$ this means:

$\sin(a)=\frac{2}{3}$

We need to find $\tan(a)$ .

$\sin^2(a)+\cos^2(a)=1$ is a Pythagorean Identity I will use to find the cosine value and then I will use that the tangent function is the ratio of sine to cosine.

$(\frac{2}{3})^2+\cos^2(a)=1$

$\frac{4}{9}+\cos^2(a)=1$

Subtract 4/9 on both sides:

$\cos^2(a)=\frac{5}{9}$

Take the square root of both sides:

$\cos(a)=\pm \sqrt{\frac{5}{9}}$

$\cos(a)=\pm \frac{\sqrt{5}}{3}$

The cosine value is positive because $a$ is a number between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ because that is the restriction on sine inverse.

So we have $\cos(a)=\frac{\sqrt{5}}{3}$ .

This means that $\tan(a)=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}$ .

Multiplying numerator and denominator by 3 gives us:

$\tan(a)=\frac{2}{\sqrt{5}}$

Rationalizing the denominator by multiplying top and bottom by square root of 5 gives us:

$\tan(a)=\frac{2\sqrt{5}}{5}$

Let's continue on to letting $b=\cos^{-1}(\frac{1}{7})$ .

Let's go ahead and say what the restrictions on $b$ are.

$b$ is a number in between 0 and $\pi$ .

So anyways $b=\cos^{-1}(\frac{1}{7})$ implies $\cos(b)=\frac{1}{7}$ .

Let's use the Pythagorean Identity again I mentioned from before to find the sine value of $b$ .

$\cos^2(b)+\sin^2(b)=1$

$(\frac{1}{7})^2+\sin^2(b)=1$

$\frac{1}{49}+\sin^2(b)=1$

Subtract 1/49 on both sides:

$\sin^2(b)=\frac{48}{49}$

Take the square root of both sides:

$\sin(b)=\pm \sqrt{\frac{48}{49}$

$\sin(b)=\pm \frac{\sqrt{48}}{7}$

$\sin(b)=\pm \frac{\sqrt{16}\sqrt{3}}{7}$

$\sin(b)=\pm \frac{4\sqrt{3}}{7}$

So since $b$ is a number between $0$ and $\pi$ , then sine of this value is positive.

This implies:

$\sin(b)=\frac{4\sqrt{3}}{7}$

So $\tan(b)=\frac{\sin(b)}{\cos(b)}=\frac{\frac{4\sqrt{3}}{7}}{\frac{1}{7}}$ .

Multiplying both top and bottom by 7 gives:

$\frac{4\sqrt{3}}{1}= 4\sqrt{3}$ .

Let's put everything back into the first mentioned identity.

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

$\tan(a-b)=\frac{\frac{2\sqrt{5}}{5}-4\sqrt{3}}{1+\frac{2\sqrt{5}}{5}\cdot 4\sqrt{3}}$

Let's clear the mini-fractions by multiply top and bottom by the least common multiple of the denominators of these mini-fractions. That is, we are multiplying top and bottom by 5:

$\tan(a-b)=\frac{2 \sqrt{5}-20\sqrt{3}}{5+2\sqrt{5}\cdot 4\sqrt{3}}$

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

4 0

3 years ago

The Lopez family and the Russell family each used their sprinklers last summer. The water output rate for the Lopez family's spr

grin007 [14]

L = hours used by the Lopez's sprinkler

R = hours used by the Russell's sprinkler

so, we know the Lopez's sprinkler uses 15 Liters per hour, so say after 1 hour it has used 15(1), after 2 hours it has used 15(2), after 3 hours it has used 15(3) liters and after L hours it has used then 15(L) or 15L.

likewise, the Russell's sprinkler, after R hours it has used 40R, since it uses 40 Liters per hour.

we know that both sprinklers combined went on and on for 45 hours, therefore whatever L and R are, L + R = 45.

we also know that the output on those 45 hours was 1050 Liters, therefore, we know that 15L + 40R = 1050.

$\bf \begin{cases}
L+R=45\implies \boxed{L}=45-R\\
15L+40R=1050\\
----------\\
15\left(\boxed{45-R} \right)+40R=1050
\end{cases}
\\\\\\
675-15R+40R=1050\implies 25R=375\implies R=\cfrac{375}{25}\\\\\\ R=15$

how long was the Lopez's on for? well, L = 45 - R.

7 0

3 years ago

What is 0.506 rounded in the hundredths

Aleksandr [31]

0.51 The hundredths place increases when it is higher than or equal to 5 in the thousandths place.

4 0

3 years ago

Read 2 more answers