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2 years ago

13

Cameron brought a set of 7 classic paperbacks to a secondhand store to sell. He was paid cash for the entire set. Before he left

, Cameron used $4 of his earnings to purchase a used book light. He had $6.50 remaining . What was the value of each paperback in the set?

1 answer:

BigorU [14]2 years ago

4 0

Answer:

6.44

Step-by-step explanation:

subtract 6.50 by 4 which will give you 6.44

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Evaluate 2/3+3/4 <br> 1and5/12

chubhunter [2.5K]

2/3 + 3/4 = 8/12 + 9/12 = 17/12 = 1 5/12

1 + 5/12 = 12/12 + 5/12 = 17/12 = 1 5/12

2/3 + 3/4 = 1 + 5/12

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2 years ago

HELP ASAP!! 10 points!

masha68 [24]

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2 years ago

Classify ABC by its sides. Then determine whether it is a right triangle.

m_a_m_a [10]

Answer:

∴Given Δ ABC is not a right-angle triangle

a= AB = √45 = 3√5

b = BC = 12

c = AC = √45 = 3√5

Step-by-step explanation:

Given vertices are A(3,3) and B(6,9)

$AB = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

AB = $\sqrt{(9-3)^{2}+(6-3)^{2} } = \sqrt{6^{2}+3^{2} } =\sqrt{45}$

Given vertices are B(6,9) and C( 6,-3)

$B C = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

= $\sqrt{(-3-9)^{2}+(6-6)^{2} } =\sqrt{12^{2} } = 12$

BC = 12

Given vertices are A(3,3) and C( 6,-3)

$AC = \sqrt{(6-3)^{2}+(-3-3)^{2} } = \sqrt{9+36} = \sqrt{45}$

AC² = AB²+BC²

45 = 45+144

45 ≠ 189

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2 years ago

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murzikaleks [220]

Answer: Option a) 3

Step-by-step explanation:

The formula for calculating combinations is as follows

$nCr = \frac{n!}{r!(n-r)!}$

Where "n" is the amount of items in a set and you can choose "r" from them

3C1 reads as: The combination of 1 in 3. You have a set of 3 elements and choose 1 of them.

$n = 3\\\\r = 1$

So :

$3C1=\frac{3!}{1!(3-1)!}\\\\3C1=\frac{3!}{1!*2!}\\\\3C1=\frac{3*2*1}{1*2*1}\\\\3C1=3$

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2 years ago

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