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Basile [38]
2 years ago
15

Simply the expresstion: 5+8 (3 + x )

Mathematics
2 answers:
DIA [1.3K]2 years ago
7 0
The answer is going to be 8x+29
Sidana [21]2 years ago
5 0

Answer:

29+8x

Step-by-step explanation:

5+8(3+x)

Expand 8(3+x)

= 5+24+8x (add the numbers)

= 29+8x

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The scale on a map of Wisconsin states that 1 cm: 25 km, if the distance on the map between Madison and Milwaukee measures 5cm,
damaskus [11]
I think the answer is c 175
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3 years ago
Identify the factors of the terms of the expression. <br>3). 5+6a+11b<br>4). 13m-2n​
xeze [42]

Answer:

factors are 5, 6, a, 11, b

Step-by-step explanation:

3)   5 + 6a + 11b

<u>Algebraic expressions: 5 + 6a + 11b </u>

Are combinations of numbers, variables, and at least one arithmetic.

<u>Term are: 5, 6a, 11b</u>

Each expression is made up of terms. A term can be a signed number, a variable, or a constant multiplied by a variable(s).

<u>Factors are: 5, 6, a, 11, b </u>

Something which is multiplied by something else. A factor can be a variable, number,  a term, or a longer expression.

<u>Coefficient: 6, 11 </u>

The numerical factor of a multiplication expression that contains a variable.

 

<u>Constant: 5 </u>

A number that cannot change its value.

4).  13m-2n​ (note: it applies as above)

7 0
3 years ago
How can I express logbx-(logby+logbz) as a single log?
GuDViN [60]
\bf log_{{  a}}(xy)\implies log_{{  a}}(x)+log_{{  a}}(y)&#10;\\ \quad \\&#10;% Logarithm of rationals&#10;log_{{  a}}\left(  \frac{x}{y}\right)\implies log_{{  a}}(x)-log_{{  a}}(y)\\\\&#10;-----------------------------\\\\&#10;log_b(x)-[log_b(y)+log_b(z)]\implies log_b(x)-[log_b(yz)]&#10;\\\\\\&#10;log_b\left( \cfrac{x}{yz} \right)
3 0
4 years ago
Need help with number 11 ASAP!!!!
Korolek [52]

Answer:

78%, no he didn't score high enough

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
How to 2 ft 5 in plus 9 in
Elina [12.6K]
1 ft equals 12 in. So 2ft equals 24 in. 24 in plus 5 in equals 29 in, and 29 in plus 9 in equals 38 in. 38/12 equals 3 ft 2 in.
4 0
3 years ago
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