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2 years ago

$72 with a 40% markup

Mathematics

1 answer:

alex41 [277]2 years ago

8 0

Answer:

$100.80

Step-by-step explanation:

40%/100%×72=$28.80

$72+$28.80=$100.80

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Which sentence would be a good counter example to this statement ?

ArbitrLikvidat [17]

Answer:

The answer is "Option D".

Step-by-step explanation:

A line is a horizontal 1D-dimensional representation without any thickness and extends in every way. It sometimes is also known as the straight line. The line, which connects two planes lies simultaneously on both planes, that's why in this question only "option D" is correct.

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3 years ago

8 and 3/15 - 4 and 3/15

Iteru [2.4K]

Answer:

Step-by-step explanation:

8 3/15 - 4 3/15 the 3/15 cancel each other out so then there is only 8-4 which is 4

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3 years ago

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How do you know what to split 26 into ?

madam [21]

Answer:

see explanation

Step-by-step explanation:

Assuming you are factoring the expression

Given

4y² + 26y + 30 ← factor out 2 from each term

= 2(2y² + 13y + 15) ← factor the quadratic

Consider the factors of the product of the coefficient of the y² term and the constant term which sum to give the coefficient of the y- term.

product = 2 × 15 = 30 and sum = 13

the factors are 10 and 3

Use these factors to split the y- term

2y² + 10y + 3y + 15 ( factor the first/second and third/fourth terms )

= 2y(y + 5) + 3(y + 5) ← factor out (y + 5) from each term

= (y + 5)(2y + 3)

Thus

4y² + 26y + 30

= 2(y + 5)(2y + 3)

5 0

3 years ago

Round 85714.285714285714

djverab [1.8K]

Answer:

to the nearest what

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

2 years ago