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ra1l [238]
2 years ago
5

7 - (2x + 1) + 4x please show solving steps as well

Mathematics
1 answer:
xeze [42]2 years ago
7 0

Answer:

2x + 6

Step-by-step explanation:

7 - (2x + 1) + 4x

Distribute the negative inside 2x+1

7 - 2x - 1 + 4x

Remind that <u>negative</u><u> </u><u>multiply</u><u> </u><u>negative</u><u> </u><u>must</u><u> </u><u>equal</u><u> </u><u>positive</u><u>.</u>

variables cannot sum/add/subtract with constants.

They must be the same terms.

4x - 2x - 1 + 7 \\ 2x + 6

Therefore, our answer is 2x+6

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Answer:

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Step-by-step explanation:

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HELP HELP MEEEEEEE Determine whether the triangles are similar. If they are write similarity statement. Explain your reasoning.
Alex_Xolod [135]

Answer:

no

Step-by-step explanation:

every triangles degrees equal 180.

78+21=99

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2 years ago
Monica walks 2 miles each day Monday through Friday. She walks 3 miles on Saturday. She does not walk on Sunday. Monica calculat
tekilochka [14]
Lets calculate how much she walks each weeks and then we multiply that by the number of weeks.
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To make 2 batches of nut bars, jayda needs to use 4 eggs. How many eggs are used in each nut bar?
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3 years ago
Show that if a and b are positive integers, then ab = gcd (a, b). lcm (a, b) . [Hint: Use prime factorizations of a and b and th
AleksandrR [38]

Answer:

ab=\gcd(a,b)\cdot \text{lcm}(a,b)

Step-by-step explanation:

Using the hint, write a and b in the following prime factorization:

a=p_1^{x_1}p_2^{x_2}\cdots p_t^{x_t}\cdot q,

b=p_1^{y_1}p_2^{y_2}\cdots p_t^{y_t}\cdot r,

where  \gcd(p_i,r)=1,\ \gcd(p_i,q)=1,\ \gcd(r,q)=1,\ \gcd(p_i,p_j)=1, for i ≠ j.

Then by the formulae for gcd(a,b) and lcm(a,b) we know that:

\gcd(a,b)=p_1^{\min(x_1,y_1)}p_2^{\min (x_2,y_2)} \cdots p_t^{\min(x_t,y_t)}

\text{lcm}(a,b)= q\cdot r\cdot p_1^{\max(x_1,y_1)}p_2^{\max(x_2,y_2)}\cdots p_t^{\max(x_t,y_t)}

Note that the expression \min(x_i,y_i)+\max(x_i,y_i)=x_i+y_i for all i, since if the minimum is, <em>without loss of generality</em>, x_i, then the maximum must be y_i, and viceversa. Then, it is straightforward to verify that when we multiply gcd(a, b) and lcm(a, b) its prime factorization matches the prime factorization of ab, and so we can see the equaility holds:

\gcd(a,b)\cdot \text{lcm}(a,b)=ab.

7 0
3 years ago
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