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VashaNatasha [74]

2 years ago

13

What does use a multiplicative comparison to explain its meaning mean?

1 answer:

Montano1993 [528]2 years ago

5 0

<span>Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison.

I hope this helps you.
</span>

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Round each number to the nearest hundred and estimate the sum.

murzikaleks [220]

1841 ≈ 1800 (nearest hundred)
964 ≈ 1000 (nearest hundred}

<span>1,841 + 964 </span>≈ 1800 + 1000 = 2800 (Answer B)

8 0

2 years ago

Read 2 more answers

a fireplace contains 46 bricks along its bottom row. if each row above decreases by 4 bricks, what is the function that shows th

m_a_m_a [10]

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
a_1=46\\
d=-4\\
n=12
\end{cases}
\\\\\\
a_{12}=46+(12-1)(-4)$

4 0

2 years ago

Read 2 more answers

For the funtion f(x)= (x+7)^5, Find f^-1 (x)

quester [9]

Answer:

The inverse of function $f(x)= (x+7)^5$ is $\mathbf{f^{-1} (x)=\sqrt[5]{x}+7}$

Option A is correct option.

Step-by-step explanation:

For the function $f(x)= (x+7)^5$ , Find $f^{-1} (x)$

For finding inverse of x,

First let:

$y=(x+7)^5$

Now replace x with y and y with x

$x=(y+7)^5$

Now, solve for y

Taking 5th square root on both sides

$\sqrt[5]{x}=\sqrt[5]{(y+7)^5}\\\sqrt[5]{x}=y+7\\=> y+7=\sqrt[5]{x}\\y=\sqrt[5]{x}-7$

Now, replace y with $f^{-1} (x)$

$f^{-1} (x)=\sqrt[5]{x}+7$

So, the inverse of function $f(x)= (x+7)^5$ is $\mathbf{f^{-1} (x)=\sqrt[5]{x}+7}$

Option A is correct option.

6 0

2 years ago

James is participating in a 5-mile walk to raise money for a charity. he has received $200 in fixed pledgesand raises $20 extra

xxMikexx [17]

5x20=100 100+200=300 or atlas what i think if this is wrong i am sry but if isn't I'm glad I'm able to help ^u^

7 0

2 years ago

Just need the equation

Anna007 [38]

$a. $ x = -2\\b. $ x = \frac{3}{2} \\c. $ x = 0\\d. $ x = -\frac{1}{5}$

Each of the equations can be solved as shown below:

a. $3x + 7 = - x - 1$

$3x + 7 + x = - x - 1 + x$ (<em>addition property of equality</em>)

$4x + 7 = - 1\\4x + 7 - 7 = -1-7$ (<em>Subtraction property of equality</em>)

$4x = -8\\\frac{4x}{4} = \frac{-8}{4}$ <em> (Division property of equality)</em>

<em /> $x = -2$

b. $1 - 2x + 5 = 4x - 3\\$

<em>Add like terms </em>

<em /> $6 - 2x = 4x - 3\\\\6 - 2x - 6 = 4x - 3 - 6$ (<em>Subtraction property of equality</em>)

$- 2x = 4x - 9\\-2x - 4x = 4x -9 - 4x$ <em> (Subtraction property of equality)</em>

<em /> $-6x = -9 \\\frac{-6x}{-6} = \frac{-9}{-6}$ <em> (Division property of equality)</em>

<em /> $x = \frac{3}{2}$

c. $4x - 2 + x =-2 + 2x\\$

$5x - 2 = -2 + 2x\\5x - 2x = -2 + 2\\3x = 0\\ x = \frac{0}{3} \\x = 0$

d. $3x - 4 + 1 = - 2x -5 + 5x\\$

$3x - 3 = - 7x -5\\3x + 7x = 3 - 5\\10x = -2\\x = \frac{-2}{10} \\x = -\frac{1}{5}$

<em>The value of x in each </em><em>equation</em><em> are:</em>

<em /> $a. $ x = -2\\b. $ x = \frac{3}{2} \\c. $ x = 0\\d. $ x = -\frac{1}{5}$

Learn more here:

brainly.com/question/1527981

5 0

2 years ago