The picture will now be 7.5 inches high by 18 inches wide.
Answer:
0.5
Step-by-step explanation:
you just need see at one point. i see at point (1, 0.5)
Answer:
Final answer is ![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3Dx)
.
Step-by-step explanation:
Given problem is ![\sqrt[3]{x}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
.
Now we need to simplify this problem.
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
Apply formula
![\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Ep%7D%5Ccdot%5Csqrt%5Bn%5D%7Bx%5Eq%7D%3D%5Csqrt%5Bn%5D%7Bx%5E%7Bp%2Bq%7D%7D)
so we get:
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B1%2B2%7D%7D)
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B3%7D%7D)
Hence final answer is .
Multiply 2x-5y= -21 by 3 to make it 6x-15y= -63
Multiply 3x-3y= -18 by -5 to make it -15x+15y=90
This cancels the y’s out which leaves us with
6x=-63
&
-15x=90
x for 6x=-63 equals - 10.5 so x is - 10.5 and for -15x=90, x= -6
Then you plug in x into any equation you’d like to find y.
Let’s plug in - 10.5 into 6x... equation.
6(- 10.5)-15y=-63
63-15y= -63
-63 -63
-15y=0
y=0 and x= - 10.5. When you plug in this values it makes the equation true!
But the correct answer is the first one north. Sorry if I’m doing too much hahah
If I’m confusing here’s the right answer...
6x-15y= -63
-15x+15y=90
Sin(D) = 24/25
tan(D) = 24/7
sin(E) = 7/25
