Let
x---------> the length of the first piece
y---------> the length of the second piece
z---------> the length of the third piece
we know that
x+y+z=72 -------> equation 1
y=x+6 -----> solve for x
x=y-6 -------> equation 2
z=6+y -------> equation 3
substitute equation 2 and equation 3 in equation 1
[y-6]+y+[6+y]=72
3y=72
y=(72/3)=24 in
<u>Find the value of x</u>
x=y-6 -------> equation 2
x=24-6=18 in
<u>Find the value of z</u>
z=6+y -------> equation 3
z=6+24=30 in
therefore
<u>the answer is</u>
the length of the first piece is 18 in
the length of the second piece 24 in
the length of the third piece is 30 in
Answer:
A or D
Step-by-step explanation:
You know that triangles are Similar if they have like same angles. If too triangles are the same the 3rd one has to be too!
Hope this helped and I hope you do well on your test!! Good Luck!
*Sends good vibes* :)) <33
"You were born to be real, Not perfect.."
-SUGA-
1. Rational numbers can be written as a ratio (fraction)
Whole numbers are rational. 5 = 5/1, for example.
Square roots are NOT rational. Example: √3
However, square roots of square numbers can be simplified, and are therefore rational. <span>√4 = 2, rational.</span>
√4 + <span>√16 = 2 + 4 = 6. rational
</span>√5 + √36...<span> irrational
</span>√9 + <span>√24... irrational
</span>2 × <span>√4 = 2 × 2 = 4. rational
</span>√49 × <span>√81 = 7 × 9 = 63. rational
</span>3√12... irrational
2.
![n^\frac12=\sqrtn](https://tex.z-dn.net/?f=n%5E%5Cfrac12%3D%5Csqrtn)
![9^\frac32=9^3\times\frac12=\sqrt{9^3}=\sqrt{729}=29](https://tex.z-dn.net/?f=9%5E%5Cfrac32%3D9%5E3%5Ctimes%5Cfrac12%3D%5Csqrt%7B9%5E3%7D%3D%5Csqrt%7B729%7D%3D29)
3.
![\frac{n^a}{n^b}=n^{a-b}](https://tex.z-dn.net/?f=%5Cfrac%7Bn%5Ea%7D%7Bn%5Eb%7D%3Dn%5E%7Ba-b%7D)
![\frac{a^\frac13}{a^\frac14}=a^{\frac13-\frac14}=a^{\frac{1}{12}}](https://tex.z-dn.net/?f=%20%5Cfrac%7Ba%5E%5Cfrac13%7D%7Ba%5E%5Cfrac14%7D%3Da%5E%7B%5Cfrac13-%5Cfrac14%7D%3Da%5E%7B%5Cfrac%7B1%7D%7B12%7D%7D)
4.
![n^\frac1x=\sqrt[x]n](https://tex.z-dn.net/?f=n%5E%5Cfrac1x%3D%5Csqrt%5Bx%5Dn)
![\sqrt[3]{m^2n^5}=m^{\frac23}n^{\frac53}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bm%5E2n%5E5%7D%3Dm%5E%7B%5Cfrac23%7Dn%5E%7B%5Cfrac53%7D)
5.
![\sqrt{a}\times\sqrt{b}=\sqrt{ab}](https://tex.z-dn.net/?f=%5Csqrt%7Ba%7D%5Ctimes%5Csqrt%7Bb%7D%3D%5Csqrt%7Bab%7D)
![\sqrt{3}\times\sqrt{12}=\sqrt{3\times12}=\sqrt{36}=6](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%5Ctimes%5Csqrt%7B12%7D%3D%5Csqrt%7B3%5Ctimes12%7D%3D%5Csqrt%7B36%7D%3D6)
A, since neither 3 nor 12 is a square but we end up with 6.
Answer:
5 months
Step-by-step explanation:
We assume that y represents production capacity, rather than <em>increase</em> in production capacity. Then we want to solve the 6th-degree equation ...
x^6 -25x^4 +199x^2 -4975 = 0
This can be factored in groups as ...
x^4(x^2 -25) + 199(x^2 -25) = 0
(x^4 +199)(x^2 -25) = 0
This has 4 complex solutions and 2 real solutions.
x^2 = 25
x = ±5
The duration required for capacity to reach 4975 units is 5 months.
Answer:
if those are variables they can equal anything you want like r=4 and p=2 r/p = 2
Step-by-step explanation: