Explain how to use grid paper to model 0.2 and 0.02. describe any diffrences

Bob need to mix 2 cups of orange juice concentrate with 3.5 cups of water to make orange juice. bob has 6 cups of concentrate. h

USPshnik [31]

3 x 2 = 6 cups of concentrate

3.5 x 3 = 10.5 cups of water

Bob can make 6 cups of orange juice.

8 0

3 years ago

Read 2 more answers

GEO, TRIGONOMETRY! 45 POINTS AND BRAINLIEST FOR CORRECT ANSWER. NEED HELP ASAP PLEASE :) Isabell is standing at the top of a wat

IRINA_888 [86]

Answer:

h ≈ 39 ft

Step-by-step explanation:

When we draw out a picture, we should see that we have to use cos∅ to solve this:

cos31.66° = x/46

46cos31.66° = x

x = 39.1542

5 0

3 years ago

Plz help, I'm taking an Advanced class (Algebra) Will give brainliest ✔✔✔✔✔✔✔✔✔

bija089 [108]

Part A:

The average rate of change refers to a function's slope. Thus, we are going to need to use the slope formula, which is:

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

$(x_1, y_1)$ and $(x_2, y_2)$ are points on the function

You can see that we are given the x-values for our interval, but we are not given the y-values, which means that we will need to find them ourselves. Remember that the y-values of functions refers to the outputs of the function, so to find the y-values simply use your given x-value in the function and observe the result:

$h(0) = 3(5)^0 = 3 \cdot 1 = 3$

$h(1) = 3(5)^1 = 3 \cdot 5 = 15$

$h(2) = 3(5)^2 = 3 \cdot 25 = 75$

$h(3) = 3(5)^3 = 3 \cdot 125 = 375$

Now, let's find the slopes for each of the sections of the function:

Section A

$m = \dfrac{15 - 3}{1 - 0} = \boxed{12}$

Section B

$m = \dfrac{375 - 75}{3 - 2} = \boxed{300}$

Part B:

In this case, we can find how many times greater the rate of change in Section B is by dividing the slopes together.

$\dfrac{m_B}{m_A} = \dfrac{300}{12} = 25$

It is 25 times greater. This is because $3(5)^x$ is an exponential growth function, which grows faster and faster as the x-values get higher and higher. This is unlike a linear function which grows or declines at a constant rate.