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3 years ago

6

Andrea paid $60.75 for 9 sandwiches. Each sandwich costs the same amount. Use the drop down menus to write an equation for the p

rice, p, of one sandwich.

1 answer:

satela [25.4K]3 years ago

8 0

The answer is p=60.75/9

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kobusy [5.1K]

$\text{Given that,}~\\\\\cot \theta = \dfrac 1{\sqrt 3} \\\\\\\text{So,}~~ \tan \theta = \dfrac 1{\cot \theta} = \sqrt 3\\\\\\\sec^2 \theta + \csc^2 \theta \\\\=(1+\tan^2 \theta) + (1 + \cot^2 \theta)\\\\=1 +\left (\sqrt 3 \right)^2 + 1 + \left(\dfrac 1{\sqrt 3} \right)^2\\\\=1 +3 + 1 + \dfrac 13 \\\\= 5 + \dfrac 13 \\\\= 5 \dfrac 13$

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2 years ago

What is the answer to 2(12-18x)=x-11x

Ivan

Answer:

x=0

Step-by-step explanation:

move all terms to the left. Then add all the numbers together. Then multiply parenthesis to get rid of them finally add all numbers together

4 0

3 years ago

There is a 12-ft fence on one side of a rectangular garden. The gardener has 44 ft of fencing to the enclose the other three sid

Morgarella [4.7K]

Answer:

16 ft

Step-by-step explanation:

One of the other three sides is 12 ft. After that fence is subtracted from the total, there are 32 ft of fence left for the other two sides of the rectangle. That allows those sides to be 32 ft/2 = 16 ft long.

The rectangle can be 12 ft by 16 ft. The longer dimension is 16 ft.

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3 years ago

What is half of a teaspoon of vanilla

Ksivusya [100]

Answer:

I believe the answer would be: Half of a teaspoon of vanilla.

Step-by-step explanation:

6 0

3 years ago

On one day a pet store sells 2 birds 6 gerbils 3 fish and 3 hamsters whats the data and relative frequency as a percent?

bonufazy [111]

Answer:

Birds $F_b=14.3\%$

Gerbils $F_g=42.9\%$

Fish $F_f=21.4\%$

Hamsters $F_h=21.4\%$

Step-by-step explanation:

From the question we are told that:

Sales

Number of birds $n_b=2$

Number of gerbils $n_g=6$

Number of fish $n_f=3$

Number of hamsters $n_h=3$

Generally the frequency for birds $F_b$ is mathematically given b

$F_b=\frac{n_b}{n}*100$

$F_b=\frac{2}{14}*100$

$F_b=14.3\%$

Generally the frequency for Gerbils $F_g$ is mathematically given b

$F_g=\frac{n_g}{n}*100$

$F_g=\frac{6}{14}*100$

$F_g=42.9\%$

Generally the frequency for fish $F_f$ is mathematically given b

$F_f=\frac{n_f}{n}*100$

$F_f=\frac{3}{14}*100$

$F_f=21.4\%$

Generally the frequency for fish $F_h$ is mathematically given b

$F_h=\frac{n_h}{n}*100$

$F_h=\frac{3}{14}*100$

$F_h=21.4\%$

5 0

3 years ago