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Klio2033 [76]
2 years ago
6

For 6 consecutive days, Alejandro studied for 5 minutes more each day than he did the previous day. Which best represents the ch

ange in the amount of time that Alejandro studied on the first day to the amount of time he studied on the last day?
–30 minutes - This the answer guys if rong sorry
–11 minutes
11 minutes
30 minutes
Also mark me brainly please!
Mathematics
2 answers:
kap26 [50]2 years ago
8 0

Answer:

I believe its 30

Step-by-step explanation:

because he studied for five minutes the first day and then added five minutes per day for 6 days so i think its 30

weqwewe [10]2 years ago
7 0
Answer : the change in the amount of time is 30 minutes Step - by - step explanation : The computation of the change in the amount of time is shown below = Number of consecutive ways * number of minutes more each day = 6 days ~ 5 minutes = 30 minutes Hence , the change in the amount of time is 30 minutes Therefore the correct option is D. We simply applied the above formula so that the correct value could come And , the same is to be considered( answer is 30 )
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35. For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.
Montano1993 [528]

Answer:

The linear equation for the line which passes through the points given as (-2,8) and $(4,6), is written in the point-slope form as $y=-\frac{1}{3} x-\frac{26}{3}$.

Step-by-step explanation:

A condition is given that a line passes through the points whose coordinates are (-2,8) and (4,6).

It is asked to find the linear equation which satisfies the given condition.

Step 1 of 3

Determine the slope of the line.

The points through which the line passes are given as (-2,8) and (4,6). Next, the formula for the slope is given as,

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Substitute $6 \& 8$ for $y_{2}$ and $y_{1}$ respectively, and 4&-2 for $x_{2}$ and $x_{1}$ respectively in the above formula. Then simplify to get the slope as follows, $m=\frac{6-8}{4-(-2)}$

$$\begin{aligned}&m=\frac{-2}{6} \\&m=-\frac{1}{3}\end{aligned}$$

Step 2 of 3

Write the linear equation in point-slope form.

A linear equation in point slope form is given as,

$y-y_{1}=m\left(x-x_{1}\right)$

Substitute $-\frac{1}{3}$ for m,-2 for $x_{1}$, and 8 for $y_{1}$ in the above equation and simplify using the distributive property as follows, y-8=-\frac{1}{3}(x-(-2))$\\ $y-8=-\frac{1}{3}(x+2)$\\ $y-8=-\frac{1}{3} x-\frac{2}{3}$

Step 3 of 3

Simplify the equation further.

Add 8 on each side of the equation $y-8=-\frac{1}{3} x-\frac{2}{3}$, and simplify as follows, $y-8+8=-\frac{1}{3} x-\frac{2}{3}+8$

$y=-\frac{1}{3} x-\frac{2+24}{3}$$\\ $$y=-\frac{1}{3} x-\frac{26}{3}$$

This is the required linear equation.

6 0
1 year ago
**Spam answers will not be tolerated**
Morgarella [4.7K]

Answer:

f'(x)=-\frac{2}{x^\frac{3}{2}}

Step-by-step explanation:

So we have the function:

f(x)=\frac{4}{\sqrt x}

And we want to find the derivative using the limit process.

The definition of a derivative as a limit is:

\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

Therefore, our derivative would be:

\lim_{h \to 0}\frac{\frac{4}{\sqrt{x+h}}-\frac{4}{\sqrt x}}{h}

First of all, let's factor out a 4 from the numerator and place it in front of our limit:

=\lim_{h \to 0}\frac{4(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x})}{h}

Place the 4 in front:

=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}

Now, let's multiply everything by (√(x+h)(√(x))) to get rid of the fractions in the denominator. Therefore:

=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}(\frac{\sqrt{x+h}\sqrt x}{\sqrt{x+h}\sqrt x})

Distribute:

=4\lim_{h \to 0}\frac{({\sqrt{x+h}\sqrt x})\frac{1}{\sqrt{x+h}}-(\sqrt{x+h}\sqrt x)\frac{1}{\sqrt x}}{h({\sqrt{x+h}\sqrt x})}

Simplify: For the first term on the left, the √(x+h) cancels. For the term on the right, the (√(x)) cancel. Thus:

=4 \lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }

Now, multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x+h)). Thus:

= 4\lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }(\frac{\sqrt x +\sqrt{x+h})}{\sqrt x +\sqrt{x+h})}

The numerator will use the difference of two squares. Thus:

=4 \lim_{h \to 0} \frac{x-(x+h)}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Simplify the numerator:

=4 \lim_{h \to 0} \frac{x-x-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}\\=4 \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Both the numerator and denominator have a h. Cancel them:

=4 \lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}

Now, substitute 0 for h. So:

=4 ( \frac{-1}{(\sqrt{x+0}\sqrt x)(\sqrt x+\sqrt{x+0})})

Simplify:

=4( \frac{-1}{(\sqrt{x}\sqrt x)(\sqrt x+\sqrt{x})})

(√x)(√x) is just x. (√x)+(√x) is just 2(√x). Therefore:

=4( \frac{-1}{(x)(2\sqrt{x})})

Multiply across:

= \frac{-4}{(2x\sqrt{x})}

Reduce. Change √x to x^(1/2). So:

=-\frac{2}{x(x^{\frac{1}{2}})}

Add the exponents:

=-\frac{2}{x^\frac{3}{2}}

And we're done!

f(x)=\frac{4}{\sqrt x}\\f'(x)=-\frac{2}{x^\frac{3}{2}}

5 0
2 years ago
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