Step-by-step explanation:
Hey there!
While factorising you remember to make it take common in most of the expression.
Here;
=mx+cx+my+cy
Take common 'x' in "mx+cx" and 'y' in my + cy.
= x(m+c) + y(m+c)
Now, "(m+c)" common again.
= (m+c) (x+y)
Therefore the factorized form of the expression in (m+c)(x+y).
<u>Hope it helps</u><u>.</u><u>.</u><u>.</u>
Let the three consecutive integers be x, x+1 and x+2.
According to the question, x+x+1+x+2=51
=>3x+3=51
=>3x+3-3=51-3
=>3x=48
=>x=16
Hence, first integer = 16,
second integer = 16 + 1 = 17 and
third integer = 16 + 2 = 18.
5.2x10^6 is the right answer because the decimal is moved 6 times to the right which makes the 5 the first number because it can’t be more than 10
Answer:
![y=\sqrt{55}](https://tex.z-dn.net/?f=y%3D%5Csqrt%7B55%7D)
which agrees with answer A
Step-by-step explanation:
Notice there are three right angle triangles for which we can apply the Pythagorean theorem:
In the small triangle at the bottom we have the Pythagorean theorem rendering:
(a)
![5^2+y^2=x^2\\x^2=25+y^2](https://tex.z-dn.net/?f=5%5E2%2By%5E2%3Dx%5E2%5C%5Cx%5E2%3D25%2By%5E2)
in the second right angle triangle on top of the previous one, if we call the vertical side on the right side "z", we have:
(b)
![11^2+y^2=z^2\\z^2=121+y^2](https://tex.z-dn.net/?f=11%5E2%2By%5E2%3Dz%5E2%5C%5Cz%5E2%3D121%2By%5E2)
and finally in the large right angle triangle:
(c)
![z^2+x^2=16^2\\z^2=256-x^2](https://tex.z-dn.net/?f=z%5E2%2Bx%5E2%3D16%5E2%5C%5Cz%5E2%3D256-x%5E2)
We can combine equations b and c to obtain:
![121+y^2=256-x^2\\x^2+y^2=256-121=135\\x^2=135-y^2](https://tex.z-dn.net/?f=121%2By%5E2%3D256-x%5E2%5C%5Cx%5E2%2By%5E2%3D256-121%3D135%5C%5Cx%5E2%3D135-y%5E2)
and then combine this and (a) to get:
![25+y^2=135-y^2\\2\,y^2=135-25\\2y^2=110\\y^2=55\\y=\sqrt{55}](https://tex.z-dn.net/?f=25%2By%5E2%3D135-y%5E2%5C%5C2%5C%2Cy%5E2%3D135-25%5C%5C2y%5E2%3D110%5C%5Cy%5E2%3D55%5C%5Cy%3D%5Csqrt%7B55%7D)