PLEASE HELPPPP!!!

Can any one help with this one ?

loris [4]

Answer:

48,000

Step-by-step explanation:

320,000÷100=3200

1/100=3200

×15=48,000

therefore 15/100=48,000

7 0

3 years ago

Read 2 more answers

Whoever gets this answered will get 15 points. Pls answer quickly

mihalych1998 [28]

Answer:

76%

Step-by-step explanation:

First parts right

75.7% i think maybe

8 0

3 years ago

Which equations for the measures of the unknown

MAXImum [283]

Answer: Option B, Option C, Option E

Step-by-step explanation:

The options written correctly, are:

$A)x = cos^{-1}(\frac{a}{c})\\\\B) x = sin^{-1}(\frac{c}{b})\\\\C)x = tan^{-1}(\frac{c}{a})\\\\D) y = sin^{-1}(\frac{a}{c})\\\\ E)y = cos^{-1}(\frac{c}{b})$

For this exercise you need to use the following Inverse Trigonometric Functions:

$1)\ sin^{-1}(x)\\\\2)\ cos^{-1}(x)\\\\3)\ tan^{-1}(x)\\\\$

When you have a Right triangle (a triangle that has an angle that measures 90 degrees) and you know that lenght of two sides, you can use the Inverse Trigonometric Functions to find the measure of an angle $\alpha$ :

$1)\alpha = sin^{-1}(\frac{opposite}{hypotenuse}) \\\\2)\ \alpha =cos^{-1}(\frac{adjacent}{hypotenuse})\\\\3)\ \alpha=tan^{-1}(\frac{opposite}{adjacent})$

Therefore, the conclusion is that the angles "x" and "y" can be found with these equations:

$x=sin^{-1}(\frac{c}{b})\\\\x= cos^{-1}(\frac{a}{b})\\\\x=tan^{-1}(\frac{c}{a})\\\\\\ y=sin^{-1}(\frac{a}{b})\\\\y=cos^{-1}(\frac{c}{b})\\\\y=tan^{-1}(\frac{a}{c})$

5 0

3 years ago

Read 2 more answers

I want to get as many bags of chips as possible to each cost $2 I also told my mom I would get her a carton of eggs eggs cost $5

Sav [38]

Answer:

y(total)=2x+5

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

You need to invest $1000 in a bank account and are give two options. The first option is to earn $50 every month you leave the m

garri49 [273]

Answer:

<h3><u>Option 1</u></h3>

Earn $50 every month.

Let x = number of months the money is left in the account
Let y = the amount in the account
Initial amount = $1,000

$\implies y = 50x + 1000$

This is a <u>linear function</u>.

<h3><u>Option 2</u></h3>

Earn 3% interest each month.

(Assuming the interest earned each month is <u>compounding interest</u>.)

Let x = number of months the money is left in the account
Let y = the amount in the account
Initial amount = $1,000

$\implies y = 1000(1.03)^x$

This is an <u>exponential function</u>.

<h3><u>Table of values</u></h3>

<u />

$\large \begin{array}{| c | l | l |}\cline{1-3} & \multicolumn{2}{|c|}{\sf Account\:Balance} \\ \cline{1-3} & \sf Option\:1 & \sf Option\:2 \\\sf Month & \sf \$50\:per\:mth & \sf 3\%\:per\:mth \\\cline{1-3} 0 & \$1000 & \$1000 \\\cline{1-3} 1 & \$1050 & \$1030 \\\cline{1-3} 2 & \$1100 & \$1060.90 \\\cline{1-3} 3 & \$1150 & \$1092.73 \\\cline{1-3} 4 & \$1200 & \$1125.51 \\\cline{1-3} 5 & \$1250 & \$1159.27 \\\cline{1-3} 6 & \$1300 & \$1194.05 \\\cline{1-3} 7 & \$1350 & \$1229.87 \\\cline{1-3}\end{array}$

From the table of values, it appears that <u>Account Option 1</u> is the best choice, as the accumulative growth of this account is higher than the other account option.

However, there will be a point in time when Account Option 2 starts accruing more than Account Option 2 each month. To find this, graph the two functions and find the <u>point of intersection</u>.

From the attached graph, Account Option 1 accrues more until month 32. From month 33, Account Option 2 accrues more in the account.

<h3><u>Conclusion</u></h3>

If the money is going to be invested for less than 33 months then Account Option 1 is the better choice. However, if the money is going to be invested for 33 months or more, then Account Option 2 is the better choice.

3 0

1 year ago