Answer:
The table:
x | y
2 5
3 7.5
6 15
Explanation:
Yes, Y is proportional to X and Y because it follows the equation which is
y = 5/2x
Let's solve this problem step-by-step.
STEP-BY-STEP EXPLANATION:
Let's first establish that triangle BCD is a right-angle triangle.
Therefore, we can use Pythagoras theorem to find BC and solve this problem. Pythagoras theorem is displayed below:
a^2 + b^2 = c^2
Where c = hypotenus of right-angle triangle
Where a and c = other two sides of triangle
Now we can solve the problem by substituting the values from the problem into the Pythagoras theorem as displayed below:
Let a = BC
b = DC = 24
c = DB = 26
a^2 + b^2 = c^2
a^2 + 24^2 = 26^2
a^2 = 26^2 - 24^2
a = square root of ( 26^2 - 24^2 )
a = square root of ( 676 - 576 )
a = square root of ( 100 )
a = 10
Therefore, as a = BC, BC = 10.
If we want to check our answer, we can substitute the value of ( a ) from our answer in conjunction with the values given in the problem into the Pythagoras theorem. If the left-hand side is equivalent to the right-hand side, then the answer must be correct as displayed below:
a = BC = 10
b = DC = 24
c = DB = 26
a^2 + b^2 = c^2
10^2 + 24^2 = 26^2
100 + 576 = 676
676 = 676
FINAL ANSWER:
Therefore, BC is equivalent to 10.
Please mark as brainliest if you found this helpful! :)
Thank you and have a lovely day! <3
Answer:
Step-by-step explanation:
Ratio of cakes baked by Calvin and Susie = 3 : 7
Cakes baked by Calvin = 3x
Cakes baked by Susie = 7x
Cakes baked by Calvin = Cakes baked by Susie - 60
3x = 7x - 60
7x - 60 = 3x
7x = 3x + 60
7x - 3x = 60
4x = 60
x = 60/4
x = 15
Cakes baked by Calvin = 3*15 = 45
Cakes baked by Susie = 7 *15 = 105
Total cakes baked = 45 + 105 = 150
They both baked 150 cakes altogether.
![\bf \cfrac{\sqrt[4]{63}}{4\sqrt[4]{6}}\qquad \begin{cases} 63=3\cdot 3\cdot 7\\ 6=2\cdot 3 \end{cases}\implies \cfrac{\sqrt[4]{3\cdot 3\cdot 7}}{4\sqrt[4]{2\cdot 3}}\implies \cfrac{\underline{\sqrt[4]{3}}\cdot \sqrt[4]{3}\cdot \sqrt[4]{7}}{4\sqrt[4]{2}\cdot \underline{\sqrt[4]{3}}} \\\\\\ \cfrac{\sqrt[4]{3}\cdot \sqrt[4]{7}}{4\sqrt[4]{2}}\implies \cfrac{\sqrt[4]{3\cdot 7}}{4\sqrt[4]{2}}\implies \cfrac{\sqrt[4]{21}}{4\sqrt[4]{2}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B63%7D%7D%7B4%5Csqrt%5B4%5D%7B6%7D%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0A63%3D3%5Ccdot%203%5Ccdot%207%5C%5C%0A6%3D2%5Ccdot%203%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B3%5Ccdot%203%5Ccdot%207%7D%7D%7B4%5Csqrt%5B4%5D%7B2%5Ccdot%203%7D%7D%5Cimplies%20%5Ccfrac%7B%5Cunderline%7B%5Csqrt%5B4%5D%7B3%7D%7D%5Ccdot%20%5Csqrt%5B4%5D%7B3%7D%5Ccdot%20%5Csqrt%5B4%5D%7B7%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%5Ccdot%20%5Cunderline%7B%5Csqrt%5B4%5D%7B3%7D%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%5Csqrt%5B4%5D%7B3%7D%5Ccdot%20%5Csqrt%5B4%5D%7B7%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B3%5Ccdot%207%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D)
![\bf \textit{now, rationalizing the denominator}\\\\ \cfrac{\sqrt[4]{21}}{4\sqrt[4]{2}}\cdot \cfrac{\sqrt[4]{2^3}}{\sqrt[4]{2^3}}\implies \cfrac{\sqrt[4]{21}\cdot \sqrt[4]{8}}{4\sqrt[4]{2}\cdot \sqrt[4]{2^3}}\implies \cfrac{\sqrt[4]{21\cdot 8}}{4\sqrt[4]{2\cdot 2^3}}\implies \cfrac{\sqrt[4]{168}}{4\sqrt[4]{2^4}} \\\\\\ \cfrac{\sqrt[4]{168}}{4\cdot 2}\implies \cfrac{\sqrt[4]{168}}{8}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bnow%2C%20rationalizing%20the%20denominator%7D%5C%5C%5C%5C%0A%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D%5Ccdot%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B2%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B2%5E3%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%7D%5Ccdot%20%5Csqrt%5B4%5D%7B8%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%5Ccdot%20%5Csqrt%5B4%5D%7B2%5E3%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%5Ccdot%208%7D%7D%7B4%5Csqrt%5B4%5D%7B2%5Ccdot%202%5E3%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B168%7D%7D%7B4%5Csqrt%5B4%5D%7B2%5E4%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%5Csqrt%5B4%5D%7B168%7D%7D%7B4%5Ccdot%202%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B168%7D%7D%7B8%7D)
and is all you can simplify from it.
so... all we did, was rationaliize it, namely, "getting rid of the pesky radical at the bottom", we do so by simply multiplying it by something that will raise the radicand, to the same degree as the root, thus the radicand comes out.
