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2 years ago

Evaluate y^5 for y = 1.

Mathematics

1 answer:

kkurt [141]2 years ago

8 0

Answer:

Step-by-step explanation:

note that $1^{n}$ = 1 for any value of n, thus

$1^{5}$ = 1

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Can somebody help me on these two

Zarrin [17]

to find the slope for the first one, use the formula $\frac{y2-y1}{x2-x1}$

(It's the slope formula)

so... $3=y2, 3=y1, 2=x2, 10=x1$

then plug in... $\frac{3-3}{2-10} =\frac{0}{-8} = 0$

this line has a slope of 0

As for the miles problem, to find mph, the miles have to be at 1.

so... $4x=242\\x=60.5$

therefore, he drove 60.5 mph

6 0

2 years ago

How do you do this question?

erik [133]

Step-by-step explanation:

Use the first fundamental theorem of calculus.

∫₆¹⁰ f'(x) dx = f(10) − f(6) = 8 − 8 = 0

5 0

2 years ago

In a recent NBA basketball game, the Hornets were trying to earn a playoff spot with a win

bonufazy [111]

Answer:

they are tied at 45

Step-by-step explanation:

no they would still be tied

4 0

3 years ago

The first figure is dilated to form the second figure.

Alja [10]

1) The scale factor is 0.25 is true.

Step-by-step explanation:

Step 1:

The first figure is dilated to form the second figure. The shape is a rhombus.

It is given that the side length of the first figure is 5.8 and the side length of the second figure is 1.45.

Step 2:

To calculate the scale factor, we divide the measurement after scaling by the same measurement before scaling. In this case, it is the side length of the rhombus.

The scale factor $= \frac{1.45}{5.8} = \frac{1}{4} = 0.25.$

So the scale factor is 0.25. This is the first option.

5 0

3 years ago

Janet deposited $9,600 into an account that pays 4.4 interest, compounded daily. At the end of nine months, how much interest ha

neonofarm [45]

bearing in mind that 9 months is not even a year, but since there are 12 months in a year, then 9 months is really 9/12 years.

$\bf ~~~~~~ \stackrel{\textit{daily}}{\textit{Continuously}} \textit{Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$9600\\ r=rate\to 4.4\%\to \frac{4.4}{100}\dotfill &0.044\\ t=years\to \frac{9}{12}\dotfill &\frac{3}{4} \end{cases} \\\\\\ A=9600e^{0.044\cdot \frac{3}{4}}\implies A=9600e^{0.033}\implies A\approx 9922.09 \\\\\\$

$\bf \stackrel{\textit{interest earned}}{9922.09-9600\implies 322.09}$

5 0

3 years ago