All categories

3 years ago

Combine the like terms to create an equivalent expression. \large{7n+4n}7n+4n

Mathematics

1 answer:

julsineya [31]3 years ago

6 0

Answer:

11n

Step-by-step explanation:

The expression is 7n + 4n. Since 7n and 4n are like terms (they both are variables with n), we can combine them so the expression becomes 7n + 4n = 11n.

You might be interested in

Tim is playing a game on his computer. He missed the first 12 questions, and each question is worth 6 points. If his starting sc

Aleksandr-060686 [28]

Answer:

Total Tim's score = 0 points

Step-by-step explanation:

Given:

Total number of question missed = 12

Each correct question mark = 6 points

Find:

Total Tim's score

Computation:

Total Tim's score = Total correct question x Each correct question mark

Total Tim's score = 0 x 6 points

Total Tim's score = 0 points

5 0

3 years ago

Can someone round this to the nearest tenth

Dominik [7]

Answer:

I think its A

Step-by-step explanation:

I did the math and thats what i got even rounded up

6 0

2 years ago

Frankie descends 56 ft in 4 minutes. what is frankies average change in position per minute from which he started.

ArbitrLikvidat [17]

Answer:

14?

Step-by-step explanation:

6 0

3 years ago

In ΔABC, the measure of ∠C=90°, the measure of ∠B=64°, and BC= 6.6 Find the length of CA the nearest tenth of a foot

Marrrta [24]

Answer:

CA = 13.5

Step-by-step explanation:

tan 64 = CA/6.6

2.05 = CA/6.6

CA = 13.5

4 0

2 years ago

Please answer correctly !!!!!!!!! Will<br> Mark brainliest !!!!!!!!!!!!!

il63 [147K]

Answer:

$x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

Step-by-step explanation:

$w=-5\left(x-8\right)\left(x+4\right)\\\mathrm{Expand\:}-5\left(x-8\right)\left(x+4\right):\quad -5x^2+20x+160\\w=-5x^2+20x+160\\Switch\:sides\\-5x^2+20x+160=w\\\mathrm{Subtract\:}w\mathrm{\:from\:both\:sides}\\-5x^2+20x+160-w=w-w\\Simplify\\-5x^2+20x+160-w=0\\Solve\:with\:the\:quadratic\:formula\\\mathrm{Quadratic\:Equation\:Formula:}\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

6 0

3 years ago