The answer that I found is the incorrect answer

A) 101* (x+12) <br> B)63* (7x-14) <br> C)(2x+3) (x+41)* <br> Please help :)

IceJOKER [234]

Answer:

A. x+12 = 101

-12 -12

x=89

B. 7x-14 = 63

+14 +14

7x=77

/7 /7

x=11

C. 2x+3 = x+ 41

-x -3 -x -3

x=38

im a god *0*

4 0

2 years ago

1. PART A: Determine the missing lengths:

Otrada [13]

#1

Divide into two rectangles

$\\ \tt\hookrightarrow "1"=14-12=2$

$\\ \tt\hookrightarrow "2"=3+1=4$

#2

$\\ \tt\hookrightarrow P=14+1+12+3+2+4=36$

#3

$\\ \tt\hookrightarrow A=14(1)+3(2)=14+6=20units^2$

3 0

2 years ago

Read 2 more answers

Write a polynomial in standard form that meets the following conditions. Assume a=1 and your function is f(x), The zeros are 5 a

jeka57 [31]

Polynomials are equations that uses variables and several terms

The polynomial in standard form is f(x) = x^2 - x -20

<h3>How to determine the polynomial</h3>

The polynomial has 2 zeros.

So, the form of the polynomial is:

f(x) = a(x - x1)(x - x2)

The zeros of the polynomial are 5 and -4.

So, the equation becomes

f(x) = a(x - 5)(x + 4)

The value of a = 1.

So, we have;

f(x) = 1(x - 5)(x + 4)

This gives

f(x) = (x - 5)(x + 4)

Expand

f(x) = x^2 - x -20

Hence, the polynomial in standard form is f(x) = x^2 - x -20

Read more about polynomials at:

brainly.com/question/2833285

7 0

2 years ago

You are designing a rectangular sign for your school’s football team. The sign will be 6 yd wide and 9 ft high. How many square

loris [4]

You need to find the area of the rectangle which is 9 times 6. You will need 54 square yards of material

5 0

3 years ago

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