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3 years ago

Find the consecutive positive odd integers whose product is 99

Mathematics

1 answer:

dezoksy [38]3 years ago

4 0

Answer:

9 and 11

Step-by-step explanation:

So let's say that the first integer is x.

That means that the second integer is x+2, since it us the next odd number.

This problem can be easily solved with our mind the answer is 9 and 11, but I'll show you the steps to slove this.

We make an equation using these two numbers:

$(x)*(x+2)=99\\$

Now all we gotta do is solve for x:

$x^2 +2x = 99\\x^2+2x-99=0$

Now we use either splitting the middle term or quadratic formula:

$x^2+11x-9x+99=0\\x(x+11)-9(x+11)=0\\(x+11)(x-9)=0$

Now we split each terma and solve for x:

$(x-9)=0\\x=9\\(x+11)=0\\x=-11$

Now the question states <em>positive </em>so we can rule out x = -11.

Now we have the first integer x = 9,

the second integer is x+2 = 9+2 =11

So the two consecutive positive odd integers are 9 and 11

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What is the area of the figure?

Effectus [21]

Answer:

0.77 M

Step-by-step explanation:

I'm going to assume this is a half-circle

Formula:

A = (πr^2)/ 2

D = 1.4

r = 0.7

A = (π 0.7^2)/2

A = (0.49π)/2

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8 0

3 years ago

The radius of a right circular cylinder is increasing at the rate of 7 in./sec, while the height is decreasing at the rate of 6

Arlecino [84]

Answer:

$\approx \bold{6544\ in^3/sec}$

Step-by-step explanation:

Given:

Rate of change of radius of cylinder:

$\dfrac{dr}{dt} = +7\ in/sec$

(This is increasing rate so positive)

Rate of change of height of cylinder:

$\dfrac{dh}{dt} = -6\ in/sec$

(This is decreasing rate so negative)

To find:

Rate of change of volume when r = 20 inches and h = 16 inches.

Solution:

First of all, let us have a look at the formula for Volume:

$V = \pi r^2h$

Differentiating it w.r.to 't':

$\dfrac{dV}{dt} = \dfrac{d}{dt}(\pi r^2h)$

Let us have a look at the formula:

$1.\ \dfrac{d}{dx} (C.f(x)) = C\dfrac{d(f(x))}{dx} \ \ \ (\text{C is a constant})\\2.\ \dfrac{d}{dx} (f(x).g(x)) = f(x)\dfrac{d}{dx} (g(x))+g(x)\dfrac{d}{dx} (f(x))$

$3.\ \dfrac{dx^n}{dx} = nx^{n-1}$

Applying the two formula for the above differentiation:

$\Rightarrow \dfrac{dV}{dt} = \pi\dfrac{d}{dt}( r^2h)\\\Rightarrow \dfrac{dV}{dt} = \pi h\dfrac{d }{dt}( r^2)+\pi r^2\dfrac{dh }{dt}\\\Rightarrow \dfrac{dV}{dt} = \pi h\times 2r \dfrac{dr }{dt}+\pi r^2\dfrac{dh }{dt}$

Now, putting the values:

$\Rightarrow \dfrac{dV}{dt} = \pi \times 16\times 2\times 20 \times 7+\pi\times 20^2\times (-6)\\\Rightarrow \dfrac{dV}{dt} = 22 \times 16\times 2\times 20 +3.14\times 400\times (-6)\\\Rightarrow \dfrac{dV}{dt} = 14080 -7536\\\Rightarrow \dfrac{dV}{dt} \approx \bold{6544\ in^3/sec}$