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Iteru [2.4K]
3 years ago
15

Find the missing number of each unit rate. 28/7 = ?/1. and 60/12 = ?/1​

Mathematics
2 answers:
Lerok [7]3 years ago
6 0

Answer:

28/7=4 and 60/12=5

Step-by-step explanation:

Zanzabum3 years ago
6 0

Answer:

\frac{28}{7} = \frac{4}{1} \\\\\frac{60}{12} = \frac{5}{1}

Explanation: 28 is divisible by 7, when multiplied by 4.

                    60 is divisible by 12, when multiplied by 5.

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when 6 is subtracted from the square of a number, the result is 5 times the number. Find the negative solution.
sveta [45]

When 6 is subtracted from the square of a number, the result is 5 times the number, then the negative solution is -1

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

Given that when 6 is subtracted from the square of a number, the result is 5 times the number

To find: negative solution

Let "a" be the unknown number

Let us analyse the given sentence

square of a number = a^2

6 is subtracted from the square of a number = a^2 - 6

5 times the number = 5 \times a

<em><u>So we can frame a equation as:</u></em>

6 is subtracted from the square of a number = 5 times the number

a^2 - 6 = 5 \times a\\\\a^2 -6 -5a = 0\\\\a^2 -5a -6 = 0

<em><u>Let us solve the above quadratic equation</u></em>

For a quadratic equation ax^2 + bx + c = 0 where a \neq 0

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

Here in this problem,

a^2-5 a-6=0 \text { we have } a=1 \text { and } b=-5 \text { and } c=-6

Substituting the values in above quadratic formula, we get

\begin{array}{l}{a=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(-6)}}{2 \times 1}} \\\\ {a=\frac{5 \pm \sqrt{25+16}}{2}=\frac{5 \pm \sqrt{49}}{2}} \\\\ {a=\frac{5 \pm 7}{2}}\end{array}

We have two solutions for "a"

\begin{array}{l}{a=\frac{5+7}{2} \text { and } a=\frac{5-7}{2}} \\\\ {a=\frac{12}{2} \text { and } a=\frac{-2}{2}}\end{array}

<h3>a = 6 or a = -1</h3>

We have asked negative solution. So a = -1

Thus the negative solution is -1

6 0
3 years ago
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What is the equation of a parabola with (−2, 4) as its focus and y = 6 as its directrix? Enter the equation in the box.
antiseptic1488 [7]

Check the picture below.  So the parabola looks more or less like so.

let's recall that the vertex is half-way between the focus point and the directrix, at "p" units away from both.

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\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \stackrel{\textit{we'll use this one}}{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-2\\ k=5\\ p=-1 \end{cases}\implies 4(-1)(y-5)=[x-(-2)]^2\implies -4(y-5)=(x+2)^2 \\\\\\ y-5=-\cfrac{1}{4}(x+2)^2\implies y=-\cfrac{1}{4}(x+2)^2+5

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