Answer:
![y\ =\ 2\sec\left(\frac{\pi}{2}x\right)\ +1](https://tex.z-dn.net/?f=y%5C%20%3D%5C%202%5Csec%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B2%7Dx%5Cright%29%5C%20%2B1)
Step-by-step explanation:
i) ![y\ =\ 2\sec\left(\frac{\pi}{2}x\right)\ +1](https://tex.z-dn.net/?f=y%5C%20%3D%5C%202%5Csec%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B2%7Dx%5Cright%29%5C%20%2B1)
Answer: Second Option
(Point in Quadrant I)
Step-by-step explanation:
The solution to a system of linear equations is the point where the two lines intersect.
Note that in this case we have two lines with different slope . By definition, if two lines have different slopes and are contained in the same plane, then there will always be an intersection between them at some point in the plane.
Looking at the image, you can see that the lines get closer as x and y increase. Then they will intercept in the first quadrant.
Answer:
16. D
17. 5^-1 · 5^-1 and 25^-1
8^0 · 8^3 and 8^3
Step-by-step explanation:
When you multiply two same numbers with exponents, just think of it as adding up exponents.
When you divide two same numbers with exponents, just think of it as subtracting exponents.
If a parenthesis has an exponent outside, then every exponent inside the parenthesis will be multiplied by that outside parenthesis.
(5^-1)^-2 · 5^3 ÷ 5^7
5^2 · 5^3 ÷ 5^7
5^5 ÷ 5^7
5^-2
When exponents are negative, you have to flip them or put them on the other side
5^-2 = 1/5^2 = 1/25
Let's try to find the equivalent pairs
6^7 ÷ 6^-2 = 6^5
6^9 = 6^5 Not equivalent
(6^3)^0 = 6
6^0 = 1 (Any number with a zero exponent is always 1)
1 = 6 Not equivalent
5^-1 · 5^-1 = 25^-1
5^-2 = 25^-1
Flip em
1/5^2 = 1/25
1/25 = 1/25 Equivalent
8^0 · 8^3 = 8^3
8^3 = 8^3 Equivalent
4^-3 ÷ 4^2 = 4^5
4^-5 = 4^5 Not equivalent
Answer:
Step-by-step explanation:
P = 1800
R = 10%
T = 2 years
I = PRT
= 1800 * 10/100 * 2
= 1800 * 0.1 * 2
I = £360
Answer:
yes the relation between the sides and angles of a right triangles bases off math