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3 years ago

Help me solve this gcf

Mathematics

2 answers:

Sedaia [141]3 years ago

6 0

Answer:

4(7 + 6)

Step-by-step explanation:

28 + 24

both can be divide by 4

28 / 4 = 7

24 / 4 = 6

28 + 24 = 4(7 + 6)

finlep [7]3 years ago

5 0

Answer:

Can we say

1×(28+24)

...

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Harry is trying to solve the equation 0 = 2x2 − x − 6 using the quadratic formula. He has made an error in one of the steps belo

bija089 [108]

<h3>Given - </h3>

➙ a quadratic equation in which Harry lagged due to an error made by him, 2x² - x - 6= 0

<h3>To solve - </h3>

➙ the given quadratic equation.

<h3>Concept applied - </h3>

➙ We will apply the quadratic formula as given in the question. So, let's study about quadratic equation first because we are supposed to apply the formula in equation.

What is quadratic equation?

➙ A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, c are real numbers, a ≠ 0.

Now, what is quadratic formula?

➙The roots of a quadratic equation ax + bx + c = 0 are given by $\sf{\:\frac{-b \pm\: \sqrt {b ^ 2 - 4ac}}{2a}}$ provided b - 4ac ≥ 0.

<h3>Solution - </h3>

here as per the given quadratic equation,

a = 2, b = -1 and c = -6

putting in the formula,

$\implies\sf{x=\frac{-(-1) \pm\: \sqrt {(-1)^2 - 4(2)(-6)}}{2(2)}}$

$\implies\sf{x=\frac{1 \pm\: \sqrt {1+48}}{4}}$

$\implies\sf{x=\frac{1 \pm\: \sqrt {49}}{4}}$

$\implies\sf{x=\frac{1 \pm\: 7}{4}}$

Solving one by one,

$\implies\sf{x=\frac{1 + \: 7}{4}}$

$\implies\sf{x=\frac{8}{4}}$

$\implies{\boxed{\bf{x=2}}}$

________________

$\implies\sf{x=\frac{1 - \: 7}{4}}$

$\implies\sf{x=\frac{-6}{4}}$

$\implies{\boxed{\bf{x=\frac{-3}{2}}}}$

________________________________

Note - Hey dear user!! You haven't provided the solution which was solved by Harry (A.T.Q). Please go through the solution as it will help you to find the error done by Harry.

________________________________

Hope it helps!! (:

4 0

2 years ago

If a and b are positive numbers, find the maximum value of f(x) = x^a(2 − x)^b on the interval 0 ≤ x ≤ 2.

Ad libitum [116K]

Answer:

The maximum value of f(x) occurs at:

$\displaystyle x = \frac{2a}{a+b}$

And is given by:

$\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

Step-by-step explanation:

Answer:

Step-by-step explanation:

We are given the function:

$\displaystyle f(x) = x^a (2-x)^b \text{ where } a, b >0$

And we want to find the maximum value of f(x) on the interval [0, 2].

First, let's evaluate the endpoints of the interval:

$\displaystyle f(0) = (0)^a(2-(0))^b = 0$

And:

$\displaystyle f(2) = (2)^a(2-(2))^b = 0$

Recall that extrema occurs at a function's critical points. The critical points of a function at the points where its derivative is either zero or undefined. Thus, find the derivative of the function:

$\displaystyle f'(x) = \frac{d}{dx} \left[ x^a\left(2-x\right)^b\right]$

By the Product Rule:

$\displaystyle \begin{aligned} f'(x) &= \frac{d}{dx}\left[x^a\right] (2-x)^b + x^a\frac{d}{dx}\left[(2-x)^b\right]\\ \\ &=\left(ax^{a-1}\right)\left(2-x\right)^b + x^a\left(b(2-x)^{b-1}\cdot -1\right) \\ \\ &= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right] \end{aligned}$

Set the derivative equal to zero and solve for x:

$\displaystyle 0= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right]$

By the Zero Product Property:

$\displaystyle x^a (2-x)^b = 0\text{ or } \frac{a}{x} - \frac{b}{2-x} = 0$

The solutions to the first equation are x = 0 and x = 2.

First, for the second equation, note that it is undefined when x = 0 and x = 2.

To solve for x, we can multiply both sides by the denominators.

$\displaystyle\left( \frac{a}{x} - \frac{b}{2-x} \right)\left((x(2-x)\right) = 0(x(2-x))$

Simplify:

$\displaystyle a(2-x) - b(x) = 0$

And solve for x:

$\displaystyle \begin{aligned} 2a-ax-bx &= 0 \\ 2a &= ax+bx \\ 2a&= x(a+b) \\ \frac{2a}{a+b} &= x \end{aligned}$

So, our critical points are:

$\displaystyle x = 0 , 2 , \text{ and } \frac{2a}{a+b}$

We already know that f(0) = f(2) = 0.

For the third point, we can see that:

$\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(2- \frac{2a}{a+b}\right)^b$

This can be simplified to:

$\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

Since a and b > 0, both factors must be positive. Thus, f(2a / (a + b)) > 0. So, this must be the maximum value.

To confirm that this is indeed a maximum, we can select values to test. Let a = 2 and b = 3. Then:

$\displaystyle f'(x) = x^2(2-x)^3\left(\frac{2}{x} - \frac{3}{2-x}\right)$

The critical point will be at:

$\displaystyle x= \frac{2(2)}{(2)+(3)} = \frac{4}{5}=0.8$

Testing x = 0.5 and x = 1 yields that:

$\displaystyle f'(0.5) >0\text{ and } f'(1)$

Since the derivative is positive and then negative, we can conclude that the point is indeed a maximum.

Therefore, the maximum value of f(x) occurs at:

$\displaystyle x = \frac{2a}{a+b}$

And is given by:

$\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

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Volume=Length x Width x Height
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