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3 years ago

Need help with this....

Mathematics

1 answer:

Yakvenalex [24]3 years ago

6 0

Answer:

$\boxed{\dfrac{f(a+h)-f(a)}{h}=-3a^2-3ah-h^2}$

Step-by-step explanation:

$f(x)=-x^3+2\\\\\dfrac{f(a+h)-f(a)}{h}\\\\f(a+h)\to\text{substitute x = a + h}\\f(a)\to\text{substitute x = a}\\\\\dfrac{f(a+h)-f(a)}{h}=\dfrac{-(a+h)^3+2-(-a^3+2)}{h}\\\\\text{use}\ (x+y)^3=x^3+3x^2y+3xy^2+y^3\\\\=\dfrac{-(a^3+3a^2h+3ah^2+h^3)+2-(-a^3)-2}{h}\\\\=\dfrac{-a^3-3a^2h-3ah^2-h^3+2+a^3-2}{h}\\\\\text{combine like terms}$

$=\dfrac{(-a^3+a^3)-3a^2h-3ah^2-h^3+(2-2)}{h}\\\\=\dfrac{-3a^2h-3ah^2-h^3}{h}\\\\=\dfrac{h(-3a^2-3ah-h^2)}{h}\\\\=-3a^2-3ah-h^2$

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What is the answer to 2 3/4 ÷ 1 5/8

alex41 [277]

Exact form:
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3 years ago

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Answer: G and F are mutually exclusive because they cannot occur together

Step-by-step explanation:

According to the definition of mutually exclusive events,

The events which can not occur together and probability of them occurring together is 0 are known as mutually exclusive events.

The first statement gives an implication that if one happens then other happens meaning they could both still happen so it is not true.

The second statement contradict the question about being mutually exclusive events.

The third statement also is a implication that if one event occurs then other does or does not occur.

The last statement is correct one that conforms with the question and obeys the definition of mutually exclusive events.

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3 years ago

I need help 6x-2y=10 x-2y=-5 solve by elimination

dem82 [27]

<h3><u>Explanation</u></h3>

Given the system of equations.

$\begin{cases} 6x - 2y = 10 \\ x - 2y = - 5 \end{cases}$

Solve the system of equations by eliminating either x-term or y-term. We will eliminate the y-term as it is faster to solve the equation.

To eliminate the y-term, we have to multiply the negative in either the first or second equation so we can get rid of the y-term. I will multiply negative in the second equation.

$\begin{cases} 6x - 2y = 10 \\ - x + 2y = 5 \end{cases}$

There as we can get rid of the y-term by adding both equations.

$(6x - x) + ( - 2y + 2y) = 10 + 5 \\ 5x + 0 = 15 \\ 5x = 15 \\ x = \frac{15}{5} \longrightarrow \frac{ \cancel{15}}{ \cancel{5}} = \frac{3}{1} \\ x = 3$

Hence, the value of x is 3. But we are not finished yet because we need to find the value of y as well. Therefore, we substitute the value of x in any given equations. I will substitute the value of x in the second equation.

$x - 2y = - 5 \\ 3 - 2y = - 5 \\ 3 + 5 = 2y \\ 8 = 2y \\ \frac{8}{2} = y \\ y = \frac{8}{2} \longrightarrow \frac{ \cancel{8}}{ \cancel{2}} = \frac{4}{1} \\ y = 4$

Hence, the value of y is 4. Therefore, we can say that when x = 3, y = 4.

Answer Check by substituting both x and y values in both equations.

<u>First</u><u> </u><u>Equation</u>

$6x - 2y = 10 \\ 6(3) - 2(4) = 10 \\ 18 - 8 = 10 \\ 10 = 10 \longrightarrow \sf{true} \: \green{ \checkmark}$

<u>Second</u><u> </u><u>Equation</u>

$x - 2y = - 5 \\ 3 - 2(4) = - 5 \\ 3 - 8 = - 5 \\ - 5 = - 5 \longrightarrow \sf{true} \: \green{ \checkmark}$

Hence, both equations are true for x = 3 and y = 4. Therefore, the solution is (3,4)

<h3><u>Answer</u></h3>

$\begin{cases} x = 3 \\ y = 4 \end{cases} \\ \sf \underline{Coordinate \: \: Form} \\ (3,4)$

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emmainna [20.7K]

First, disregard the sign for absolute value and solve for x.

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Step-by-step explanation:

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