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2 years ago

What is one fifth of 15

Mathematics

2 answers:

Andre45 [30]2 years ago

6 0

$\frac{1}{5}\cdot15= \frac{1}{5}\cdot \frac{15}{1}= \frac{1}{5}\cdot \frac{3\cdot5}{1}=3$

Finger [1]2 years ago

4 0

1/5 of 15=1/5 ×15=3
whenever there is of it means u need 2 multiply first no. with second no.

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A cone has a diameter of 18 inches. and a height of 14 inches. What is the volume of the cone?

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3 years ago

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finlep [7]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

The square of a number decreased by 3 times the number 28 find all possible values for the number

stealth61 [152]

Question:

The square of a number decreased by 3 times the number is 28 find all possible values for the number

Answer:

The possible values of number are 7 and -4

Solution:

Given that the square of a number decreased by 3 times the number is 28

To find: all possible values of number

Let "a" be the unknown number

From given information,

square of a number decreased by 3 times the number = 28

$a^2 - 3a = 28$

$a^2 - 3a - 28 = 0$

Let us solve the above quadratic equation

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0$

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Using the above formula,

$\text { For } a^{2}-3 a-28=0 \text { we have } a=1, b=-3, c=-28$

$\begin{aligned}&a=\frac{-(-3) \pm \sqrt{(-3)^{2}-4(1)(-28)}}{2 \times 1}\\\\&a=\frac{3 \pm \sqrt{9+112}}{2}\\\\&a=\frac{3 \pm \sqrt{121}}{2}=\frac{3 \pm 11}{2}\\\\&a=\frac{3+11}{2} \text { or } a=\frac{3-11}{2}\\\\&a=7 \text { or } a=-4\end{aligned}$

Thus the possible values of number are 7 and -4

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3 years ago

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valentinak56 [21]

Answer:

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Step-by-step explanation:

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3 0

2 years ago

Provide reasons for the proof. Given line h is parallel to line k. line j is perpendicular to line h. Prove line j is perpendicu

dusya [7]

Answer: Provided.

Step-by-step explanation: We are given two lines 'h' and 'k' which are parallel to each other. Also, there is another line 'j' that is perpendicular to line 'h'.

We are to prove that line 'j' is perpendicular to line 'k'.

Let, m, n and p be the slopes of lines 'h', 'k' and 'j' respectively.

Now, since line 'h' and 'k' are parallel, so their slopes will be equal. i.e., m = n.

Also, lines 'h' and 'j' are perpendicular, so the product of their slopes is -1. i.e.,

m×p = -1.

Hence, we can write from the above two relations

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Thus, the line 'j' is perpendicular to line 'k'.

Proved.

8 0

3 years ago