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2 years ago

Anybody know how to do this ?

Mathematics

1 answer:

s344n2d4d5 [400]2 years ago

7 0

Answer:

The answer is though, 8

Step-by-step explanation:

10/ x Is equal to 15/12

so it’s 8

hope this helps :))

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Brianna has x nickels and y pennies. She has no less than 20 coins worth no more

Bond [772]

Answer:

$y \ge 20 - x$

$y \le 80 -5x$

$(x,y) = (15,5)$

Step-by-step explanation:

Given

$Nickels = x$

$Pennies = y$

$Amount = \$0.80$ maximum

$Coins = 20$ minimum

Required

Solve graphically

First, we need to determine the inequalities of the system.

For number of coins, we have:

$x\ +\ y\ge 20$ because the number of coins is not less than 20

For the worth of coins, we have:

$0.05x\ +\ 0.01y\ \le0.80$ because the worth of coins is not more than 0.80

So, we have the following equations:

$x\ +\ y\ge 20$

$0.05x\ +\ 0.01y\ \le0.80$

Make y the subject in both cases:

$y \ge 20 - x$

$0.01y \le 0.80 - 0.05x$

Divide through by 0.01

$\frac{0.01y}{0.01} \le \frac{0.80}{0.01} -\frac{ 0.05x}{0.01}$

$y \le \frac{0.80}{0.01} -\frac{ 0.05x}{0.01}$

$y \le 80 -5x$

The resulting inequalities are:

$y \ge 20 - x$

$y \le 80 -5x$

The two inequalities are plotted on the graph as shown in the attachment.

$y \ge 20 - x$ --- Blue

$y \ge 80 -5x$ --- Green

Point A on the attachment are possible solutions

At A:

$(x,y) = (15,5)$

4 0

2 years ago

Look at the pic try to show ur work for those equations the best u can I’ll give you the brainly thing who ever does the best

-Dominant- [34]

Alrighty! KFC! Keep, change, flip! When dividing fractions, you keep the first fraction, flip the second, and change the divide sign to multiply. Here is work as requested.

6 0

2 years ago

Determine whether each integral is convergent or divergent. If it is convergent evaluate it. (a) integral from 1^(infinity) e^(-

AVprozaik [17]

Answer:

a) So, this integral is convergent.

b) So, this integral is divergent.

c) So, this integral is divergent.

Step-by-step explanation:

We calculate the next integrals:

$\int_1^{\infty} e^{-2x} dx=\left[-\frac{e^{-2x}}{2}\right]_1^{\infty}\\\\\int_1^{\infty} e^{-2x} dx=-\frac{e^{-\infty}}{2}+\frac{e^{-2}}{2}\\\\\int_1^{\infty} e^{-2x} dx=\frac{e^{-2}}{2}\\$

So, this integral is convergent.

$\int_1^{2}\frac{dz}{(z-1)^2}=\left[-\frac{1}{z-1}\right]_1^2\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\frac{1}{1-1}+\frac{1}{2-1}\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\infty\\$

So, this integral is divergent.

$\int_1^{\infty} \frac{dx}{\sqrt{x}}=\left[2\sqrt{x}\right]_1^{\infty}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=2\sqrt{\infty}-2\sqrt{1}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=\infty\\$