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2 years ago

What is the equivalent expression for 2( x + 4)

Mathematics

2 answers:

lidiya [134]2 years ago

6 0

The answer is 2x + 8

weeeeeb [17]2 years ago

5 0

Answer:

2x + 8

Step-by-step explanation:

What we have here is a case of distributive property.

Multiply 2 by x and 4 to get the the equivalent expression.

2(x + 4)

2x + 8 is the equivalent expression.

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A recipe for individual chocolate hazelnut tarts calls for ½ cup of hazelnuts per tart. If 1 cup of hazelnuts weighs 4 ounces, w

DerKrebs [107]

Answer:

The amount is sufficient to make 75 tarts.

Step-by-step explanation:

We have been given that a recipe for individual chocolate hazelnut tarts calls for ½ cup of hazelnuts per tart and 1 cup of hazelnuts weighs 4 ounces.

The half cup of hazelnuts will weigh 2 ounces $(\frac{4}{2}=2)$ .

1 kg equals 35.274 ounces.

$\text{5 kg}=5\times 35.274\text{ ounces}$

$\text{5 kg}=176.37\text{ ounces}$

Since each tart needs ½ cup of hazelnuts and half cup of hazelnuts will weigh 2 ounces, so we will divide 176.37 ounces by 2 to find number of tarts.

$\frac{176.37}{2}=88.185\approx 88$

Since we can make 88 tarts from 5 kg hazelnuts, therefore, the 5-kilogram bag of hazelnuts be sufficient to make 75 tarts.

7 0

3 years ago

Which expression represents the greatest common factor (GCF) of 70 and 112? A. 2 x 2 x 2 x 7 B. 2 x 2 x 2 x 5 C. 2 x 7 D. 2 x 2

Mekhanik [1.2K]

Answer:

$\huge\boxed{C. \ \ GCF = 2 * 7}$

Step-by-step explanation:

<u>Firstly, we'll do prime factorization of these numbers.</u>

70 = 2 × 5 × 7

112 = 2 × 2 × 2 × 2 × 7

<u>So, The Greatest Common Factor can be written as:</u>

GCF = 2 × 7 <u>[Since Only one 2 and one 7 is common]</u>

<u></u>

Hope this helped!

<h2>~AnonymousHelper1807</h2>

3 0

3 years ago

Which statement describes the solutions of the equation? 4x/3x+1 = x/2x+10

stiks02 [169]

Answer:

<h3> The equation has one valid solution and no extraneous of solutions.</h3>

Step-by-step explanation:

Given the expression;

4x/3x+1 = x/2x+10

We are to get the nature of the value of x

Cross multiply;

x(3x+1) = 4x(2x+10)

3x²+x = 8x²+40x

Collect like terms;

3x²-8x² + x - 40x = 0

-5x²+x -40x = 0

-5x²-39x = 0

-5x² = 39x

-5x = 39

x = -39/5

<em>Since we have just one value of x hence, the equation has one valid solution and no extraneous of solutions.</em>

4 0

2 years ago

Read 2 more answers

Suppose <img src="https://tex.z-dn.net/?f=m" id="TexFormula1" title="m" alt="m" align="absmiddle" class="latex-formula"> men and

ollegr [7]

Firstly, we'll fix the postions where the $n$ women will be. We have $n!$ forms to do that. So, we'll obtain a row like:

$\underbrace{\underline{~~~}}_{x_2}W_2 \underbrace{\underline{~~~}}_{x_3}W_3 \underbrace{\underline{~~~}}_{x_4}... \underbrace{\underline{~~~}}_{x_n}W_n \underbrace{\underline{~~~}}_{x_{n+1}}$

The n+1 spaces represented by the underline positions will receive the men of the row. Then,

$x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m~~~(i)$

Since there is no women sitting together, we must write that $x_2,x_3,...,x_{n-1},x_n\ge1$ . It guarantees that there is at least one man between two consecutive women. We'll do some substitutions:

$\begin{cases}x_2=x_2'+1\\x_3=x_3'+1\\...\\x_{n-1}=x_{n-1}'+1\\x_n=x_n'+1\end{cases}$

The equation (i) can be rewritten as:

$x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m\\\\
x_1+(x_2'+1)+(x_3'+1)+...+(x_{n-1}'+1)+x_n+x_{n+1}=m\\\\
x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-(n-1)\\\\
x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-n+1~~~(ii)$

We obtained a linear problem of non-negative integer solutions in (ii). The number of solutions to this type of problem are known: $\dfrac{[(n)+(m-n+1)]!}{(n)!(m-n+1)!}=\dfrac{(m+1)!}{n!(m-n+1)!}$