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Zepler [3.9K]
3 years ago
11

Help with math please

Mathematics
1 answer:
JulijaS [17]3 years ago
5 0
You can first multiply your coefficients (7x9), which gives you 63. You can then simplify the exponents by using the rule that when two terms with the same variable are being multiplied, you can simplify by adding exponents. Since 7+3= 10, your simplified answer would be 63x^10 :) hope it helped!
You might be interested in
Simplify the expression completely. 7b+3(4b−2)+6÷3
Karo-lina-s [1.5K]

Answer:

19b-4

Step-by-step explanation:

7b+3(4b-2)+6÷3

Bracket comes first so we distribute 3 over the terms in the bracket

= 7b + 12b -6 +6 ÷ 3

First we solve division

= 7b + 12b -6 +6 ÷ 3

= 7b + 12b -6 +2

= (7b+12b)+(-6+2)

= 19b -4

This can't be simplified further as it can't be solved anymore

4 0
3 years ago
Read 2 more answers
the ratio of men and women in a certain factory is 3 to 4. there are 198 men. How many workers are there?
LUCKY_DIMON [66]

3 men

4 women

7 total


3 men/ 7 total = 198 men/ x total

using cross products

3*x = 7 * 198

divide each side by 3

x = 7*198/3

x = 462

There are 462 workers


If you mean 3 men and 1 women for a total of 4 workers when you state a  ratio of men and women in a certain factory is 3 to 4.

3/4 =198/x

Using cross products

3x = 4* 198

Divide each side by 3

3x/3 = 4*198/3

x =264

264 workers


It all depends on how you define ratio of men and women in a certain factory is 3 to 4.   This is incorrect phrasing and I took it to be men to women.   You cannot have a ratio of men and women.




5 0
3 years ago
Is the answer 200? Or 4.5??? Or is none ? I don't know it's confusing :/ can I get help :/
skelet666 [1.2K]
Let x = the number of penguins we are finding

x ÷ 100 × 30 = 15
x = 15 ÷ 30 × 100
x = 50

hope this helps and have a great day :)


6 0
3 years ago
Read 2 more answers
Determine the range of the following graph:
BaLLatris [955]

Answer:

The range of a function is the set of outputs the function can give

The y-axis on the graph shows as the output of the function

From the graph, we can see that the outputs of this specific function range from 0 to 5

Therefore, the range of this function is:   [0 , 5]

3 0
3 years ago
write a function g whose graph represents a translation 2 units to the right followed by a horizontal stretch by a factor or 2 o
Alenkinab [10]
\bf f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}\\\\
f(x)=&{{  A}} \left|{{ B }}x+{{  C}}  \right|+{{  D}}
\\\\
--------------------\\\\

\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{  D}}\\
\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\
\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind, let's see,
 
\bf f(x)=|x| \implies \begin{array}{lllccll}
f(x)=&1|&1x&+0|&+0\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}

so, to shift it to the right by 2 units, simply set C = 2.
to stretch it by 2, set A = 1/2.

the smaller A is, the wider it opens, the larger it is, the more it shrinks.
5 0
3 years ago
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