All categories

3 years ago

Which phrases can be represented by the algebraic expression x + 14? Select three options.

Mathematics

2 answers:

Lady_Fox [76]3 years ago

7 0

Answer:

number plus 14

fourteen more than a number

a number increased by 1 :D

Step-by-step explanation:

Snezhnost [94]3 years ago

4 0

Answer:

a number plus 14

fourteen more than a number

a number increased by 14

Step-by-step explanation:

The phrases which can be represented by the algebraic expression x + 14 are

Let x = the unknown number

a number plus 14

= x + 14

fourteen more than a number

= x + 14

a number increased by 14

x + 14

You might be interested in

Vanessa and zack are playing a game where a player with the lower score wins at the end of the game Vanessa has a score of -3 5/

Novosadov [1.4K]

Complete question :

Vanessa and Zack are playing a game where the player with the lower score wins. At the end of the game, Vanessa has a score of -3 5/8 and Zack has a score of -3 2/3. Which statement explains who won?

Vanessa won. If you compare the decimal equivalents of their scores, -3.625 < -3.666.

Vanessa won. If you compare the decimal equivalents of their scores, -3.666 < -3.625.

Zack won. If you compare the decimal equivalents of their scores, -3.625 < -3.666.

Zack won. If you compare the decimal equivalents of their scores, -3.666 < -3.625

Answer: Zack won. If you compare the decimal equivalents of their scores, -3.666 < -3.625

Step-by-step explanation:

Given that :

Player with the lowest score wins ;

Vanessa's score = - 3 5/8

Zack's score = - 3 2/3

Converting to decimal:

Vanessa :

-3 5/8 = - 29 /8 = - 3.625

Zack:

-3 2/3 = - 11/3 = - 3.666

Comparing the scores :

From the negaryce side of a number line :

-3.666 comes before - 3.625, hence, - 3.666 is lower or lesser than - 3.625

-3.666 < - 3.625, hence Zack won

3 0

3 years ago

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$