It’d be 2=-2 ... but then subtract two from both side and the answer is a null set. Also known as “no solution”..
Best of luck in your studies :)
-Nicky
Answer:
B.
Step-by-step explanation:
The answer is B. The problem says that the slope is 2 and it as the points (3,10) on its line. When looking at the graph, you can see that the line crosses at four on the y-intercept which is why four will be your constant. So, your equation in slope intercept form will become y=2x+4. With this, you can start eliminating the given answers.
You can immediately eliminate c and d because the 2 is negative when it isn't in its slope form. It leaves you with a and b.
When looking at both a and b you now have to look at what your y will become when you sinplify both of them. In choice a, the 2 multiplies with the 3 and gives you 6. Since you have to leave the y by itself you have to subtract 10 from both sides which will leave you with -4. Since your y-intercept isn't negative you know that a isnt the ansewr.
When checking b, you multiply the 2 and -3 to get -6. Since you have to leave the y by itself, you add 10 to each side and end up with 4 which is the same number that crosses the y-axis. and that is how you know it's the right answer.
Answer:
5
Step-by-step explanation:
.05 ×100=5 per cent
a nickel is 5 cent
![\bf \begin{cases} f(x)=\sqrt[3]{7x-2}\\\\ g(x)=\cfrac{x^3+2}{7} \end{cases}\\\\ -----------------------------\\\\ now \\\\ f[\ g(x)\ ]\implies f\left[ \frac{x^3+2}{7} \right]\implies \sqrt[3]{7\left[ \frac{x^3+2}{7} \right]-2}\implies \sqrt[3]{x^3+2-2} \\\\\\ \sqrt[3]{x^3}\implies x\\\\ -----------------------------\\\\ or \\\\ g[\ f(x)\ ]\implies g\left[\sqrt[3]{7x-2}\right]\implies \cfrac{\left[\sqrt[3]{7x-2}\right]^3+2}{7} \\\\\\ \cfrac{7x-2+2}{7}\implies \cfrac{7x}{7}\implies x](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Af%28x%29%3D%5Csqrt%5B3%5D%7B7x-2%7D%5C%5C%5C%5C%0Ag%28x%29%3D%5Ccfrac%7Bx%5E3%2B2%7D%7B7%7D%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Anow%0A%5C%5C%5C%5C%0Af%5B%5C%20g%28x%29%5C%20%5D%5Cimplies%20f%5Cleft%5B%20%5Cfrac%7Bx%5E3%2B2%7D%7B7%7D%20%5Cright%5D%5Cimplies%20%5Csqrt%5B3%5D%7B7%5Cleft%5B%20%5Cfrac%7Bx%5E3%2B2%7D%7B7%7D%20%5Cright%5D-2%7D%5Cimplies%20%5Csqrt%5B3%5D%7Bx%5E3%2B2-2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7Bx%5E3%7D%5Cimplies%20x%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Aor%0A%5C%5C%5C%5C%0Ag%5B%5C%20f%28x%29%5C%20%5D%5Cimplies%20g%5Cleft%5B%5Csqrt%5B3%5D%7B7x-2%7D%5Cright%5D%5Cimplies%20%5Ccfrac%7B%5Cleft%5B%5Csqrt%5B3%5D%7B7x-2%7D%5Cright%5D%5E3%2B2%7D%7B7%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B7x-2%2B2%7D%7B7%7D%5Cimplies%20%5Ccfrac%7B7x%7D%7B7%7D%5Cimplies%20x)
thus f[ g(x) ] = x indeed, or g[ f(x) ] =x, thus they're indeed inverse of each other
