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Serggg [28]
4 years ago
11

(-ab+5a-8)-(-4ab-5) find the difference

Mathematics
1 answer:
sergij07 [2.7K]4 years ago
8 0
(-ab+5a-8) +4ab + 5 Distribute the negative

3ab+5a-3
is the answer after you combine like terms.
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PLEASE HELP!!
inn [45]
You can use the Pythagorean Theorem to find the length of the third side AB (Identified as "x" in the figure attached in the problem), which says that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the legs:
a² = b²+c²
As we can see the figure, the triangle does not have an angle of 90°, but it can be divided into two equal parts, leaving two triangles with a right angle. We already have the values of the hypotenuse and a leg in triangle "A" , so we can find the value of the other leg:
b = √(a²-c²) b = √(10²-4²) b = 9.16
With these values, we can find the hypotenuse in the triangle "B": x = √b²+c² x = √(9.16)²+(4)² x = 10
4 0
3 years ago
I accidentally clicked that it isn’t the answer so help? What’s the correct answer?
katrin2010 [14]

The answer is A

-2 with a multi. of 2

4 with a multi. of 1

-1 with a multi. of 3

:)))

6 0
3 years ago
Read 2 more answers
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.
6 0
4 years ago
A car that travels 308 miles on 11 gallons of gasoline
DanielleElmas [232]
28 miles per gallon of gasoline
6 0
3 years ago
Read 2 more answers
What is q + (-5)<br> _____. =3<br> 3.
maria [59]
You can switch the addition and subtraction signs to get q-(+5) =? So you want to find what 5 can be subtracted from to get 3, which is 8. 8+(-5) =3
4 0
3 years ago
Read 2 more answers
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