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3 years ago

4+5(6p+2=24 ........

Mathematics

1 answer:

guajiro [1.7K]3 years ago

7 0

Answer:

$4+5\left(6p+2\right)=24$

$4+5\left(6p+2\right)-4=24-4$

$5\left(6p+2\right)=20$

$\frac{5\left(6p+2\right)}{5}=\frac{20}{5}$

$6p+2=4$

$6p+2-2=4-2$

$6p=2$

$p=\frac{1}{3}$

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3 years ago

78 : 120 :: ? : 66 please give answer with explanation

yawa3891 [41]

Step-by-step explanation:

Method 1:

Given that:

78 : 120 :: ? : 66

Here,

Product of extremes = 78 and 66

Product of means = 120 and ?

so,

(78*66) = (120 * ?)

⇛5148 = 120 * ?

⇛5148 = 120?

⇛120? = 5148

Shift the number 120 from LHS to RHS, changing it's sign.

⇛? = 5148/120

Therefore, (?) = 42.9

Method 2:

78 : 120 :: ? : 66

Convert each ratio into fraction.

⇛78/120 = ?/66

On applying cross multiplication then

since, (a/b) = (c/d) ⇛cd = ad

Where, a = 78, b = 120, c = ? and d = 66

Now,

⇛(120)? = (78)66

⇛120? = 5148

⇛(?) = 5148/120

Therefore, (?) = 42.9

Answer: Hence the value of (?) for the given problem is 42.9.

Verification:

Check whether the value of (?) is true or false.

78 : 120 :: ? : 66

Substitute the value of (?).

⇛78:120 = 42.9:66

Product of extremes should be equal to product of means.

Product of extremes:- 78*66 = 5148

Product of means:- 120*42.9 = 5148

⇛5148 = 5148

LHS = RHS

Hence, verified.

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2 years ago

En 6 - 2x = 6x - 10x + 8 .

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Step-by-step explanation:

6-2x=6x-10x+8

6-2x=-4x+8

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3 years ago

A company has 10 male and 11 female employees, and needs to nominate 2 men and 2 women for the company bowling team. How many di

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

Since we need to nominate 2 men from a company of 10 males, the number of ways this can be done = $C(10,2)=\frac{10!}{(10-2)!(2)!} =45\ ways$

Also, we need to nominate 2 women from a company of 11 females, the number of ways this can be done = $C(10,2)=\frac{11!}{(11-2)!(2)!} =55\ ways$