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Korvikt [17]
2 years ago
13

20 years from now I will be 4 times as old as my present age what is my present age ?​

Mathematics
2 answers:
Nina [5.8K]2 years ago
8 0

Answer:

Your present is age 5

your present age = 20/4 = 5

Shkiper50 [21]2 years ago
4 0

Step-by-step explanation:

your present age = 20/4 =5

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Evaluate the expression. P(9, 3) · P(5, 4)
boyakko [2]

Answer:

P(9,3) *P(5,4)

And if we use the permutation formula given by:

nPx = \frac{n!}{(n-x)!}

And replacing we got:

\frac{9!}{6!} \frac{5!}{4!}= 504*5 = 2520

Step-by-step explanation:

For this problem we want to find the following expressionÑ

P(9,3) *P(5,4)

And if we use the permutation formula given by:

nPx = \frac{n!}{(n-x)!}

And replacing we got:

\frac{9!}{6!} \frac{5!}{4!}= 504*5 = 2520

6 0
3 years ago
For which nonnegative value of x is the expression 5+x/25-x^2
erica [24]

Answer:

Step-by-step explanation:

\frac{5+x}{25-x^2} >0\\=\frac{5+x}{(5+x)(5-x)} \\=\frac{1}{5-x} >0\\if~5-x>0\\or~5>x\\or x

8 0
3 years ago
Give an example of a polynomial that has the asymptotes of x=5 and y=0.5
BaLLatris [955]
(x²+4x+3)/2(x²-10x+25)

the horizontal asymptote when the numerator and the denominator have the same degree (in this case, both of a degree of 2) is ration of the coefficients of the numerator and denominator. In this case, the coefficient for numerator x² is 1, and the coefficient for the denominator 2x² is 2, so the horizontal asymptote is y=1/2=0.5

the vertical asymptote is the x value. the denominator cannot be zero, if x²-10x+25=0, x would be 5, so the vertical asymptote is x=5 

this is just one example. There can be others:
(2x²+5x+2)/[(4x-7)(x-5)]  for another example, but this example has a second vertical asymptote 4x-7=0 =>x=7/4
5 0
2 years ago
Solve the inequality <br><br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B2x%7D%7Bx%20%2B%203%7D%20%20%5Cleqslant%203" id="Tex
Fofino [41]

Domain:x\neq-3\\\\\dfrac{2x}{x+3}\leq3\qquad\text{subtract 3 from both sides}\\\\\dfrac{2x}{x+3}-3\leq0\\\\\dfrac{2x}{x+3}-\dfrac{3(x+3)}{x+3}\leq0\\\\\dfrac{2x}{x+3}-\dfrac{3x+9}{x+3}\leq0\\\\\dfrac{2x-(3x+9)}{x+3}\leq0\\\\\dfrac{2x-3x-9}{x+3}\leq0\\\\\dfrac{-x-9}{x+3}\leq0\Rightarrow(x+3)(-x-9)\leq0\\\\x=-3,\ x=-9\\\\x\in(-\infty,\ -9]\ \cup\ (-3,\ \infty)

3 0
3 years ago
R
Andreas93 [3]

TWINKLE TWINKLE LITTLE STAR

HOW I WONDER WHAT YOU ARE

UP ABOVE THE WORLD SO HIGH

LIKE A DIAMOND IN THE SKY

lol sorry-

your question ain't properly readable-

3 0
3 years ago
Read 2 more answers
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