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4 years ago

The difference between two number is 9 and the product of the number is 162 find the 2 number

Mathematics

1 answer:

erik [133]4 years ago

5 0

Answer:

18 and 9

-18 and -9

Step-by-step explanation:

Let:

first number = x

second number = y

According to given conditions:

x - y = 9 ---------- eq1

x * y = 162 ------- eq2

From eq1:

x = 9 + y

Put it in eq2:

(9+ y)*y = 162

9y + y^2 - 162 = 0

y^2 + 9y -162 = 0

By factorizing we get:

y^2 -9y + 18y - 162 = 0

Taking common:

y(y - 9) + 18(y - 9) = 0

(y - 9)(y + 18) = 0

So:

y = -18 and y = 9

ignoring negative integer

Putting values of y in eq1:

x - 9 = 9

x = 9 + 9

x = 18

So the two numbers are:

18 and 9

-18 and -9

i hope it will help you!

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M(x)=1/x f(x)=2x-10 find the domain restriction of m(R(x))

AlexFokin [52]

Answer:

x ≠ 5

Step-by-step explanation:

Given

m(x) = 1/x
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Composite function

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3 years ago

Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your

Veronika [31]

The expression of integral as a limit of Riemann sums of given integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ is 4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$ from i=1 to i=n.

Given an integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ .

We are required to express the integral as a limit of Riemann sums.

An integral basically assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinite data.

A Riemann sum is basically a certain kind of approximation of an integral by a finite sum.

Using Riemann sums, we have :

$\int\limits^b_a {f(x)} \, dx$ = $\lim_{n \to \infty}$ ∑f(a+iΔx)Δx ,here Δx=(b-a)/n

$\int\limits^5_1 {x/(2+x^{3}) } \, dx$ =f(x)= $x/2+x^{3}$

⇒Δx=(5-1)/n=4/n

f(a+iΔx)=f(1+4i/n)

f(1+4i/n)= $[n^{2}(n+4i)]/2n^{3}+(n+4i)^{3}$

$\lim_{n \to \infty}$ ∑f(a+iΔx)Δx=

$\lim_{n \to \infty}$ ∑ $n^{2}(n+4i)/2n^{3}+(n+4i)^{3}4/n$

=4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$

Hence the expression of integral as a limit of Riemann sums of given integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ is 4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$ from i=1 to i=n.

Learn more about integral at brainly.com/question/27419605

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