All categories

3 years ago

Miranda buys 6 pounds of nuts . 3/4 pounds in her bag how many bags

Mathematics

2 answers:

PIT_PIT [208]3 years ago

6 0

Answer:

8 bags

Step-by-step explanation:

Take the number of nuts and divide by the number of pounds in a bag

6 lbs ÷ 3/4 lbs

Copy dot flip

6 * 4/3

8 bags

Andrews [41]3 years ago

5 0

Answer:

$\Huge \boxed{\mathrm{8 \ bags}}$

Step-by-step explanation:

We can solve the problem with algebra.

$\displaystyle \frac{3}{4} x=6$

Let x be the number of bags.

Multiply both sides by $\frac{4}{3}$

$\displaystyle \frac{3}{4} x \cdot \frac{4}{3} =6 \cdot \frac{4}{3}$

$x=8$

The number of bags is 8.

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3 years ago

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1 year ago

Verify that:

Lelu [443]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to verify that:

$\displaystyle \left(\cos(x)\right)\left(\cot(x)\right)=\csc(x)-\sin(x)$

Note that cot(x) = cos(x) / sin(x). Hence:

$\displaystyle \left(\cos(x)\right)\left(\frac{\cos(x)}{\sin(x)}\right)=\csc(x)-\sin(x)$

Multiply:

$\displaystyle \frac{\cos^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Recall that Pythagorean Identity: sin²(x) + cos²(x) = 1 or cos²(x) = 1 - sin²(x). Substitute:

$\displaystyle \frac{1-\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Split:

$\displaystyle \frac{1}{\sin(x)}-\frac{\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Simplify:

$\csc(x)-\sin(x)=\csc(x)-\sin(x)$

Problem 2)

We want to verify that:

$\displaystyle (\csc(x)-\cot(x))^2=\frac{1-\cos(x)}{1+\cos(x)}$

Square:

$\displaystyle \csc^2(x)-2\csc(x)\cot(x)+\cot^2(x)=\frac{1-\cos(x)}{1+\cos(x)}$

Convert csc(x) to 1 / sin(x) and cot(x) to cos(x) / sin(x). Thus:

$\displaystyle \frac{1}{\sin^2(x)}-\frac{2\cos(x)}{\sin^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor out the sin²(x) from the denominator:

$\displaystyle \frac{1}{\sin^2(x)}\left(1-2\cos(x)+\cos^2(x)\right)=\frac{1-\cos(x)}{1+\cos(x)}$

Factor (perfect square trinomial):

$\displaystyle \frac{1}{\sin^2(x)}\left((\cos(x)-1)^2\right)=\frac{1-\cos(x)}{1+\cos(x)}$

Using the Pythagorean Identity, we know that sin²(x) = 1 - cos²(x). Hence:

$\displaystyle \frac{(\cos(x)-1)^2}{1-\cos^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor (difference of two squares):

$\displaystyle \frac{(\cos(x)-1)^2}{(1-\cos(x))(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor out a negative from the first factor in the denominator:

$\displaystyle \frac{(\cos(x)-1)^2}{-(\cos(x)-1)(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Cancel:

$\displaystyle \frac{\cos(x)-1}{-(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Distribute the negative into the numerator. Therefore:

$\displaystyle \frac{1-\cos(x)}{1+\cos(x)}=\displaystyle \frac{1-\cos(x)}{1+\cos(x)}$

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3 years ago