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1 year ago

Find the dimensions of a rectangle

Mathematics

1 answer:

gayaneshka [121]1 year ago

4 0

The length of the shorter side is 7 longer side is 12 7+7+12+12 is 38 and 7x12 is 84

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Use the IVT to prove that one solution to <img src="https://tex.z-dn.net/?f=x%5E%7B3%7D%20%2Bx-5%3D0" id="TexFormula1" title="x^

Dafna11 [192]

Let f(x) = x³ + x - 5. f(x) is a polynomial so it's continuous everywhere on its domain (all real numbers). Since

f (1) = 1³ + 1 - 5 = -3 < 0

and

f (2) = 2³ + 2 - 5 = 5 > 0

it follows by the intermediate value theorem that there at least one number x = c between 1 and 2 for which f(c) = 0.

7 0

2 years ago

I need help on this for brainly

Gre4nikov [31]

Answer:

Step-by-step explanation:

(7,1)

7 0

2 years ago

Help there are five people at the carnival they are 15 people at the other carnival how many people are there

faltersainse [42]

I think there are 20 people

7 0

2 years ago

What are the domain and range of the function x^2-2x^2-4?

jonny [76]

${x}^{2} - 2 {x}^{2} - 4 \\ - {x}^{2} - 4$

The Quadratic Function has the domain as the set of all real numbers.

For the range, start from minimum value to maximum value.

But because the parabola is downward as a < 0. Thus, there are no minimum value but the maximum value instead.

Therefore the range is y <= -4

7 0

2 years ago

What are the ordered pairs of the

9966 [12]

Start by writing the system down, I will use $y$ to represent $f(x)$

$y=x^2-5x+3\wedge y=-3$

Substitute the fact that $y=-3$ into the first equation to get,

$-3=x^2-5x+3$

Simplify into a quadratic form ( $ax^2+bx+c=0$ ),

$x^2-5x+6=0$

Now you can use Vieta's rule which states that any quadratic equation can be written in the following form,

$x^2+x(m+n)+mn=0$

which then must factor into

$(x+m)(x+n)=0$

And the solutions will be $m,n$ .

Clearly for small coefficients like ours $5,6$ , this is very easy to figure out. To get 5 and 6 we simply say that $m=3, n=2$ .

This fits the definition as $5=3+2$ and $6=2\cdot3$ .

So as mentioned, solutions will equal to $m=3, n=2$ but these are just x-values in the solution pairs of a form $(x,y)$ .

To get y-values we must substitute 3 for x in the original equation and then also 2 for x in the original equation. Luckily we already know that substituting either of the two numbers yields a zero.

So the solution pairs are $(2,0)$ and $(3,0)$ .

Hope this helps :)

5 0

2 years ago