Answer:
24
Step-by-step explanation:
I use the Pythagoras theorem.
16 x 36= 576
Answer: option B. it has the highest y-intercept.
Explanation:
1) point -slope equation of the line
y - y₁ = m (x - x₁)
2) Replace (x₁, y₁) with the point (5,3):
y - 5 = m (x - 3)
3) Expand using distributive property and simplify:
y - 5 = mx - 3m ⇒ y = mx + 5 - 3m
4) Compare with the slope-intercept equation of the line: y = mx + b, where m is the slope and b is the y-intercept
⇒ slope = m
⇒ b = 5 - 3m = y - intercept.
Therefore, for the same point (5,3), the greater m (the slope of the line) the less b (the y-intercept); and the smaller m (the slope) the greater the y - intercept.
Then, the conclusion is: the linear function with the smallest slope has the highest y-intercept (option B).
Answer:
<h2>2^7</h2>
Multiply
2^3 by 2^4 by adding the exponents.
Use the power rule
a^m a^n=a^m+n
to combine exponents.
2^3+4
Add 3 and 4.
2^7
Raise 2 to the power of 7.
128
Step-by-step explanation:
Hope it is helpful....
Hello there! I can help you! The formula for compound interest is P(1 + r)^t, where P= principal (initial amount), r = interest rate (in decimal form), and t = time (in years). Let's do this step by step. First off, we add the rate into 1. 4% is the interest rate (0.04 in decimal form). 1 + 0.04 is 1.04. Now, what we will do is raise that number to the 2nd power, because the time that elapses is 2 years. 1.04² is 1.0816. That's that. Now, multiply 7,500 to find the total amount of money. 1.0816 * 7,500 is 8,112. There. Toby's savings account balance in 2 years is £8,112.
Note: To solve for compound interest questions like it, add 1 to the percentage rate in decimal form, raise that number to a power based on the number of years (for example, raise the number to the 7th power if we are looking for the balance after 7 years), and then multiply that number by the starting amount. After you raise the number by a power, there may be a lot of numbers behind it. Whatever you do, DO NOT delete the number. Keep it there and multiply it by the principal.