Explain why a square can never be a trapezoid

5.Karen is thinking of a number. If she multiplies her number by −3 and then adds 6, she has −4 times the original number. What

alekssr [168]

-3x+6 = -4x

6 = -x

x = -6

check:

-6*-3 = 18 +6 =24

-6*-4 = 24

number = -6

5 0

2 years ago

What is 59m • tan 46

kotykmax [81]

59m • tan 46 is -123.110198 meters.

5 0

3 years ago

Solve the inequality 47. 75 x Less-than-or-equal-to 50 to determine how much more weight can be added to Li’s suitcase without g

vlada-n [284]

The value of x is less than or equal to 2.25.

<h2>Linear system</h2>

It is a system of an equation in which the highest power of the variable is always 1. A one-dimension figure that has no width. It is a combination of infinite points side by side.

<h3>Simplification</h3>

Simplification is to make something easier to do or understand and to make something less complicated.

Given

The expression is 47.75 + x ≤ 50.

The value of x.

<h3>How to find the value of x?</h3>

The expression is 47.75 + x ≤ 50.

On simplifying,

47.75 + x ≤ 50

x ≤ 50 - 47.75

x ≤ 2.25

Thus, the value of x is less than or equal to 2.25.

More about the linear system link is given below.

brainly.com/question/20379472

3 0

1 year 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

1. Which Transformation would result in an image NOT congruent to its pre-image?

irga5000 [103]

A is the correct answer

pls mark me as the brainliest

4 0

3 years ago

Read 2 more answers