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2 years ago

7

Anybody know the answer?

1 answer:

stepan [7]2 years ago

8 0

Put 53/x = k/2hours I think the equation is something like that

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5/9 -7/15 divided by 1 5/9 ×7/15

Greeley [361]

Hi there,

the first part equals 4/45 and the second part equals 7/9

4/45 divided by 7/9 = 0.00141093474
Hope this helps :)<span>
</span>

3 0

3 years ago

Read 2 more answers

Find the solution of the system of equations.<br> 10x+10y=40 <br> 5x+3y=8

ddd [48]

Answer:

x = -2 & y = 6

Step-by-step explanation:

$10x+10y=40$

$10x+10y+ -10y=40+-10y$

$10x=-10y+40$

$\frac{10x}{10} =\frac{-10y+40}{10}$

$x=-y+4$

$5x+3y=8$

$5(-y+4)+3y=8$

$-2y+20=8$

$-2y+20+-20=8+-20$

$-2y=-12$

$\div2$

$y=6$

$x=-y+4$

$x=-6+4$

$x=-2$

Hope this helps.

7 0

2 years ago

Find each quotient. Write in simplest form. keep answer as fraction - not decimal

kirill [66]

So we are given the expression:

$\frac{mn^{2}}{4}$ ÷ $\frac{m^{2}n}{8}$

When we divide fractions, we must flip the second term and change the sign to multiplication:

$\frac{mn^{2}}{4} *\frac{8}{m^{2}n}$

And then we multiply across:

$\frac{mn^{2}}{4} *\frac{8}{m^{2}n}= \frac{8mn^{2}}{4m^{2}n}$

Then we can break apart all of the like variables for simplification:

$\frac{8mn^{2}}{4m^{2}n}=\frac{8}{4} * \frac{m}{m^{2}} *\frac{n^{2}}{n}$

When we simplify variables through division, we subtract the exponent of the numerator from the exponent of the denominator. So we then have:

$\frac{8}{4} =2$

$\frac{m}{m^{2}}=m^{-1} =\frac{1}{m}$

$\frac{n^{2}}{n}=n^{1}=\frac{n}{1}$

So then we multiply all of these simplified parts together:

$\frac{2}{1}*\frac{1}{m}*\frac{n}{1} =\frac{2n}{m}$

So now we know that the simplified form of the initial expression is: $\frac{2n}{m}$ .

3 0

3 years ago

Tell whether the given value is a solution of the inequality

topjm [15]

Answer:

the given value <u>is</u> a solution of the inequality

Step-by-step explanation:

12 + m ≥ -4 if m = -3

12 + (-3) ≥ -4

9 ≥ -4

9 is greater than or equal to -4

5 0

2 years ago

Given that 'n' is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction

Oliga [24]

The base case is the claim that

$\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}$

which reduces to

$\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86$

which is true.

Assume that the inequality holds for <em>n</em> = <em>k </em>; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}$

We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}$

By the induction hypothesis,

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}$

Now compare this to the upper bound we seek:

$\dfrac{2k+1}{k+1} > \dfrac{2k+2}{k+2}$

because

$(2k+1)(k+2) > (2k+2)(k+1)$

in turn because

$2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0$

6 0

2 years ago

Read 2 more answers