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2 years ago

John bought 8 watches for $56 and his friend victor bought 7 watches for $63. whose watch is expensive ?

Mathematics

2 answers:

Hunter-Best [27]2 years ago

8 0

John=$7 each watch
Victor =$9 each watch

bogdanovich [222]2 years ago

5 0

Victors was more because if I were to buy the same amount of each (56).
63*8= 504
56*7=382.

John bought 56 for 382
Victor bought 56 for 504.

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Jim's soccer team is making fruit baskets for a fundraiser. They have 88 peaches, 60 bananas, and 54 kiwis to use. If each baske

cluponka [151]

The greatest number of fruit baskets they can make is 2 with each one having 44 peaches, 30 bananas and 27 kiwis

<h3><u>Solution:</u></h3>

Jim’s fruit basket has 88 peaches, 60 bananas and 54 Kiwis to use.

If each basket have same number of each type, we have to determine the greatest number of fruit baskets they can make

We need to find greatest common factor of 88, 60 and 54

When we find all the factors of two or more numbers, and some factors are the same ("common"), then the largest of those common factors is the Greatest Common Factor.

<em><u>Greatest common factor of 88, 60 and 54:</u></em>

The factors of 54 are: 1, 2, 3, 6, 9, 18, 27, 54

The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

The factors of 88 are: 1, 2, 4, 8, 11, 22, 44, 88

Then the greatest common factor is 2

On dividing the number of fruits by 2 we get

$\begin{array}{l}{\text { Apples in each basket }=\frac{88}{2}=44} \\\\ {\text { Bananas in each basket }=\frac{60}{2}=30} \\\\ {\text { Kiwis in each basket }=\frac{54}{2}=27}\end{array}$

Hence, there can be 2 basket with each one having 44 peaches, 30 bananas and 27 kiwis

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4 years ago

Evaluate the expression a2+6ab for a=−3 and b=4.

xxTIMURxx [149]

(-3)2+6(-3)(4) is -63

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John is a runner what is the speed if you Ran<br> hundred metter sprint in 10 seconds

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The hypotenuse of a right triangle is 15 inches, and one of its legs is 11 inches. Find the length of the other leg.

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Step-by-step explanation:

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3 years ago

Let $a$ and $b$ be nonzero real numbers such that

algol [13]

<h3>Answer: -7/2</h3>

=========================================================

Explanation:

Let's expand out (2 - 7i)(a + bi) using the FOIL rule

(2 - 7i)(a + bi) = 2a + 2bi - 7ai - 7bi^2

(2 - 7i)(a + bi) = 2a + 2bi - 7ai - 7b(-1)

(2 - 7i)(a + bi) = 2a + 2bi - 7ai + 7b

(2 - 7i)(a + bi) = (2a+7b) + (2bi-7ai)

(2 - 7i)(a + bi) = (2a+7b) + (2b-7a)i

We're told the result is purely imaginary. What this means is that the real part (2a+7b) is zero, while the imaginary part (2b-7a) is nonzero. If both are zero, then we have 0+0i = 0 which is purely real.

For example, the complex numbers 0-7i and 0+2i are purely imaginary.

Let's use the fact that 2a+7b must be zero to do the following steps:

2a+7b = 0

2a = -7b

a = -7b/2

a/b = -7/2 which is the final answer

We must check to see if 2b-7a is nonzero

2b - 7a = 2b - 7(-7b/2)

2b - 7a = 2b + 24.5b

2b - 7a = 26.5b

The result is nonzero if and only if b is nonzero. Luckily we're told b is nonzero at the top of the problem. So we don't have any worries that (2b-7a) is zero.

Therefore, (2a+7b) + (2b-7a)i will be purely imaginary with a/b = -7/2

------------------

A concrete example:

Let a = -14 and b = 4

a/b = -14/4 = -7/2

(2-7i)(a+bi) = (2-7i)(-14+4i) = 0 + 106i which is purely imaginary.

I'll let you do the steps in expanding that out using the FOIL rule.

4 0

1 year ago