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Scorpion4ik [409]
4 years ago
15

How do you get the answer to a math problem -k+h+kh k=-5 h=-2

Mathematics
2 answers:
Aloiza [94]4 years ago
4 0

Answer:

Final answer is 13.

Step-by-step explanation:

Given expression is

-k+h+kh

We also have been given values of h and k as:

k=-5 h=-2

to simplify the given expression, we just need to plug the given values of k=-5 and h=-2 into -k+h+kh

-k+h+kh

=-(-5)+(-2)+(-5)(-2)

=5+(-2)+(-5)(-2)

=5-2+(-5)(-2)

=5-2+(10)

=5-2+10

=3+10

= 13

Hence final answer is 13.

dalvyx [7]4 years ago
3 0

Answer:

7

Step-by-step explanation:

you have to put the numbers in for the variables

-5+2+5(2)

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List the first six common multiples of 6 and 9.
Zepler [3.9K]

6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96,102,108

9,18,27,36,45,54,63,72,81,90,99,108

So the first 6 common multiples of 6 and 9 are:

Answer: 18, 36, 54, 72, 90, 108

Hope this Helps!!

8 0
4 years ago
A number increased by 19 is the same as 9 less than five times a number
IgorLugansk [536]

Answer:

the number should be 7

Step-by-step explanation:

7 times 5 eqauls 35. 35 minus 9 equals 26. if you add 7 and 19 you get 26.

4 0
3 years ago
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What are the orders of 3,7,9,11,13,17 and 19(mod20)?does 20 have primitive roots?
bezimeni [28]
3\equiv3\mod{20}
3^2\equiv9\mod{20}
3^3\equiv27\equiv7\mod{20}
3^4\equiv3\cdot3^3\equiv3\cdot7\equiv21\equiv1\mod{20}

7\equiv7\mod{20}
7^2\equiv49\equiv9\mod{20}
7^3\equiv7\cdot7^2\equiv63\equiv3\mod{20}
7^4\equiv7\cdot7^3\equiv21\equiv1\mod{20}

9\equiv9\mod{20}
9^2\equiv3^4\equiv1\mod{20}

11\equiv11\mod{20}
11^2\equiv121\equiv1\mod{20}

13\equiv-7\equiv13\mod{20}
13^2\equiv169\equiv9\mod{20}
13^3\equiv13\cdot13^2\equiv(-7)9\equiv-63\equiv-3\mod{20}
13^4\equiv13\cdot13^3\equiv(-7)(-3)\equiv21\equiv1\mod{20}

17\equiv-3\equiv17\mod{20}
17^2\equiv(-3)^2\equiv9\mod{20}
17^3\equiv(-3)^3\equiv-27\equiv3\mod{20}
17^4\equiv(-3)^4\equiv81\equiv1\mod{20}

19\equiv-1\equiv19\mod{20}
19^2\equiv19(-1)\equiv-19\equiv1\mod{20}

Generally speaking, a number x coprime to n will be a primitive root of n if we have x^n\equiv x\mod{n}, or x^{n-1}\equiv1\mod{n}. In other words, if x is of order n-1 modulo n, then x is a primitive root of n.

Since none of these numbers has order 19, it follows that 20 does not have any primitive roots.
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Which of the following phrases are equations?
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Annette [7]
-12n - 19 = 77
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