You see how these 2 angles marked are both inside the 2 parallel lines?
And they are on opposite side of the transversal, the line crossing the 2 parallel lines?
These 2 angles are alternate interior angles and they are equal, I think you can do the last part by yourself.
Answer:
Step-by-step explanation:
1. Find two numbers that add to make the coefficient of x (in this case, -5) and that multiply to make the constant term multiplied by the coefficient of x^2 (in this case, -2 x 3 = -6)
Two numbers that work are -6 and +1
-6 x +1 = -6
-6 + -1 = -5
2. Split the middle term into the two numbers that you found.
3x^2 -6x +x -2 = 0
I've put the -6 on the left side because in our next step, when we factorise, it will be easier than having the numbers the other way around.
3. Factorise the left side by taking out common factors from each pair. The pairs I'm talking about here are '3x^2 and -6x', and 'x and -2'
3x (x-2) +1 (x-2) = 0
4. You now have two numbers both being multiplied by the term x-2. We can rearrange this equation to give us two brackets being multiplied by each other.
(3x + 1) (x-2) = 0
5. According to the Null Factor Law, if two terms are multiplied together and the result is 0, then one of those terms must be 0. Make both terms equal to 0 and solve each for x.
3x + 1 = 0 x-2 = 0
3x = -1 x = 2
x = -1/3
6. The solutions to this equation are x = 2 and x = -1/3
Answer:
17rx2−23rx−71x+75
Step-by-step explanation:
(17x−23)(xr−4)−(3x+17)
=(17x−23)(xr−4)+−1(3x+17)
=(17x−23)(xr−4)+−1(3x)+(−1)(17)
=(17x−23)(xr−4)+−3x+−17
=(17x)(xr)+(17x)(−4)+(−23)(xr)+(−23)(−4)+−3x+−17
=17rx2+−68x+−23rx+92+−3x+−17
=17rx2+−68x+−23rx+92+−3x+−17
=(17rx2)+(−23rx)+(−68x+−3x)+(92+−17)
=17rx2+−23rx+−71x+75
Answer:
602.6
Step-by-step explanation:
The six is in the hundreds place. Since no number is specified for the tens, a 0 goes there as a placeholder. The two, since it has no tag, is assumed to go in the ones. Another six goes in the tenths place after the decimal.
9514 1404 393
Answer:
not a right triangle
Step-by-step explanation:
The perimeter is 22, so we have ...
AB +BC +CA = 22
8 + 5 + CA = 22
CA = 22 -13 = 9
If this is a right triangle, the sum of the squares of the short sides will be equal to the square of the long side.
AB² +BC² = 8² +5² = 64 +25 = 89
This is not the same as ...
CA² = 9² = 81
The triangle is not a right triangle.
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<em>Additional comment</em>
If you're familiar with a few Pythagorean triples, you know that a couple involving a side length 5 are (3, 4, 5) and (5, 12, 13). That is, (5, 8, 9) is NOT a Pythagorean triple, so you know immediately that those side lengths do not form a right triangle. (No calculation is necessary.)
