All categories

2 years ago

Olivia ate 1/6 of the pepperoni pizza and 1/3 of the vegetarian pizza and 1/6 of the cheese pizza how much of the pizza she eat

Mathematics

2 answers:

Romashka [77]2 years ago

8 0

She ate 4/6 or 2/3 of the pizza

tester [92]2 years ago

5 0

Answer: She ate 4/6 of the pizza.

You might be interested in

N/n^2-1 over 3n/n^2+8n+7

lisabon 2012 [21]

Answer:

Step-by-step explanation:

i did it

6 0

2 years ago

Susan threw a softball 42 years on her first try and 51 1/3 yard on her second try. How much farther did she throw the softball

Andru [333]

Answer:

28 feet farther than 1st ball.

Step-by-step explanation:

We have been given that Susan threw a softball 42 yards on her first try and $51\frac{1}{3}$ yard on her second try.

To find second ball is how much farther from the 1st ball, we will subtract 42 yards from $51\frac{1}{3}$ yards.

$\text{The second ball is farther from 1st ball}=51\frac{1}{3}\text{ yards}-42\text{ yards}$

$\text{The second ball is farther from 1st ball}=\frac{154}{3}\text{ yards}-42\text{ yards}$

Let us have a common denominator.

$\text{The second ball is farther from 1st ball}=\frac{154}{3}\text{ yards}-\frac{42*3}{3}\text{ yards}$

$\text{The second ball is farther from 1st ball}=\frac{154}{3}\text{ yards}-\frac{126}{3}\text{ yards}$

$\text{The second ball is farther from 1st ball}=\frac{154-126}{3}\text{ yards}$

$\text{The second ball is farther from 1st ball}=\frac{28}{3}\text{ yards}$

$\text{1 yard}=3\text{ feet}$

$\frac{28}{3}\text{ yards}=\frac{28}{3}\times 3\text{ feet}$

$\frac{28}{3}\text{ yards}=28\text{ feet}$

Therefore, Susan thrown the second ball 28 feet farther from the 1st ball.

7 0

3 years ago

Joseph claims that a scatterplot in which the y-values increase as the x-values increase must have a linear association. Amy cla

My name is Ann [436]

Answer: Choice C

Amy is correct because a nonlinear association could increase along the whole data set, while being steeper in some parts than others. The scatterplot could be linear or nonlinear.

======================================================

Explanation:

Just because the data points trend upward (as you go from left to right), it does not mean the data is linearly associated.

Consider a parabola that goes uphill, or an exponential curve that does the same. Both are nonlinear. If we have points close to or on these nonlinear curves, then we consider the scatterplot to have nonlinear association.

Also, you could have points randomly scattered about that don't fit either of those two functions, or any elementary math function your teacher has discussed so far, and yet the points could trend upward. If the points are not close to the same straight line, then we don't have linear association.

-----------------

In short, if the points all fall on the same line or close to it, then we have linear association. Otherwise, we have nonlinear association of some kind.

Joseph's claim that an increasing trend is not enough evidence to conclude the scatterplot is linear or not.

8 0

2 years ago

What is the answer 5i/(2+3i)^2

Nataly_w [17]

Simplify the following:5 i/(3 i + 2)^2
(3 i + 2)^2 = 4 + 6 i + 6 i - 9 = -5 + 12 i:(5 i)/12 i - 5
Multiply numerator and denominator of (5 i)/(12 i - 5) by 5 + 12 i:(5 i (12 i + 5))/((12 i - 5) (12 i + 5))
(12 i - 5) (12 i + 5) = -5×5 - 5×12 i + 12 i×5 + 12 i×12 i = -25 - 60 i + 60 i - 144 = -169:(5 i (12 i + 5))/-169
i (12 i + 5) = -12 + 5 i:(5 5 i - 12)/(-169)
Multiply numerator and denominator of (5 (5 i - 12))/(-169) by -1:Answer: (-5 (5 i - 12))/169

3 0

2 years ago

Draw the line through the point (4,5) that is parallel to the y-axis. Write the equation of this line and state its slope

ollegr [7]

Parallel to the y axis...that would mean that it is a vertical line.
The equation would be : x = 4 and it would have an undefined slope

7 0

2 years ago