How to solve a quadratic equation that does not factor?

What is the area of the paper used to wrap a can of soup if the can is 11 cm tall and has a diameter of 7.5 cm. Show all formula

Dimas [21]

Notice that a can is a cylinder, which is basically two circles and one rectangle. Now, the question wants area of this rectangle.

You know the width of the rectangle (or the paper), which is 11cm. How do you find the length then?

Notice that the length of the rectangle is also the circumference of the circles. From here, use C = πd where C is circumference and d is diameter. The d is given, which is 7.5 cm. Depending on the question, π can be 3.14, 22/7 or the actual pi.

multiply π and 7.5cm and you get the circumference. Multiply this to 11cm and you get area of the rectangle.

6 0

2 years ago

Which ratio has a unit rate of 4 choose all that apply

Murrr4er [49]

Answer:

Where are the answer choices my dude

Step-by-step explanation:

8 0

2 years ago

E^0.5ln9=? no calculator

Neko [114]

$e^{0.5\ln 9}=e^\ln{9^{0.5}}=9^{0.5}= \sqrt{9} =3$

Use the property of logs: a ln b = ln (b^a)
Also e and log are inverses, so the cancel each other

5 0

3 years ago

Choose the correct answer below. A. Increasing the level of confidence has no effect on the interval. B. Increasing the level of

murzikaleks [220]

Answer:

$ME = t_{\alpha/2} \frac{s}{\sqrt{n}}$

And the width for the confidence interval is given by:

$Width =2ME=2t_{\alpha/2} \frac{s}{\sqrt{n}}$

And we want to see the effect if we increase the confidence level for a interval. On this case if we increase the confidence level then the critical value for the confidence interval $t_{\alpha/2}$ would be higher and then the width of the interval would increase. So then the best answer for this case would be:

B. Increasing the level of confidence widens the interval.

Step-by-step explanation:

Let's assume that we have a parameter of interest $\mu$ who represent for example the true mean for a population. And we can construct a confidence interval in order to estimate this parameter if we know the distribution for the statistic let's say $\bar X$ and for this particular example the confidence interval is given by:

$\bar X \pm ME$

Where ME represent the margin of error for the estimation and this margin of error is given by:

$ME = t_{\alpha/2} \frac{s}{\sqrt{n}}$

And the width for the confidence interval is given by:

$Width =2ME=2t_{\alpha/2} \frac{s}{\sqrt{n}}$

And we want to see the effect if we increase the confidence level for a interval. On this case if we increase the confidence level then the critical value for the confidence interval $t_{\alpha/2}$ would be higher and then the width of the interval would increase. So then the best answer for this case would be:

B. Increasing the level of confidence widens the interval.

3 0

2 years ago

2

Gemiola [76]

9514 1404 393

Answer:

14 nickels
6 dimes

Step-by-step explanation:

The system of equations can be written ...

n + d = 20 . . . . . . . . coins in the jar

5n +10d = 130 . . . . . value in cents

_____

Using the first equation, we can write an expression for n:

n = 20 -d

We can substitute this into the second equation:

5(20 -d) +10d = 130

100 +5d = 130 . . . . . . simplify

5d = 30 . . . . . . . . . . . . subtract 100

d = 6 . . . . . . . . . . . . divide by 5

n = 20-d = 14

The jar contains 14 nickels and 6 dimes.

7 0

2 years ago