4. x^10
5. x^3
I hope this helped! Mark me Brainliest! :) -Raven❤️
Answer:
a = 9
Step-by-step explanation:
you can simplify the proportion to:
use cross- multiplication
8a = 72
a = 9
It will cost about $0.72 for three pair of shoes assuming shoe box is rectangular shape having 12 square inches of packing paper.
Step-by-step explanation:
- Assuming a shoe box rectangle in shape and the same size is required.
- Length of rectangle box assuming to be 4 inches.
- Width of rectangle box assuming to be 3 inches.
- as we know area of rectangle is Length * Breath(Width) .
- 12 square inches is the total packing paper required for each box.
- To cover cost of three pairs of shoes $0.02 per square inch.
- Each box would cost about 12 square inches * 0.02 pr square inches.
- 0.24 square inches of packing paper will be required.
- let three pairs of shoes be x
- Total pairs of shoes would be= (12 * 0.02)*x
- Answer would be for 3 pairs of shoes = 0.24*3 = 0.72 square inches.
Answer:
the answer is B
f(x)=-2x+3
Step-by-step explanation:
to get the -2x I did --->
pick two points I used (0,3) and (1,1) and used the equation 
where y2 would be 1 and y1 would be 3
and where x2 would be 1 and x1 would be 0
so you would put the numbers in the equation which would look like
and equal -2 which would be -2x in the answer
And the +3 is the y-intercept
The first solution is quadratic, so its derivative y' on the left side is linear. But the right side would be a polynomial of degree greater than 1, so this is not the correct choice.
The third solution has a similar issue. The derivative of √(x² + 1) will be another expression involving √(x² + 1) on the left side, yet on the right we have y² = x² + 1, so that the entire right side is a polynomial. But polynomials are free of rational powers, so this solution can't work.
This leaves us with the second choice. Recall that
1 + tan²(x) = sec²(x)
and the derivative of tangent,
(tan(x))' = sec²(x)
Also notice that the ODE contains 1 + y². Now, if y = tan(x³/3 + 2), then
y' = sec²(x³/3 + 2) • x²
and substituting y and y' into the ODE gives
sec²(x³/3 + 2) • x² = x² (1 + tan²(x³/3 + 2))
x² sec²(x³/3 + 2) = x² sec²(x³/3 + 2)
which is an identity.
So the solution is y = tan(x³/3 + 2).
