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Lapatulllka [165]

3 years ago

11

Agrocery store sells brand a olive oil in a 33-fluid-ounce bottle for $16.50. the store also sells brand b olive oil in a large

dispenser that customers can buy for $4.80 per cup.

1 answer:

Ivan3 years ago

4 0

Brand A olive oil is cheaper, if that’s the question

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8 0

3 years ago

A lid on a circular container has a diameter of 9 cm. What is the approximate area of the lid? Use 3.14 to approximate pi and ro

kkurt [141]

Area=pir^2

diameter/2=radius
diameter=9
d/2=9/2=4.5=r

area=pi4.5^2
area=pi20.25
aproxpi=3.14
area=63.585
round hundreth
area=63.59

answer is 63.59cm^2

7 0

3 years ago

Read 2 more answers

This question is from the tangent ratio section in our chapter and im not sure how to solve for this

VladimirAG [237]

The large triangle is an isosceles since both angles at the base each equal 42°.
In an isosceles triangle the altitude z is at the same time median , then it bisects the opposite side in the middle . So w = 120/2 = 60

Now let's calculate z:
tan 42° = (opposite side) / (adjacent side) = z/60
tan 42° = 0.9,
0.9 = z/60 and z = 54

6 0

3 years ago

I need help with 8-11....

MaRussiya [10]

8. Quadrants
9. Proportion
10. Proportionality
11. Dimensional analysis

8 0

3 years ago

I need help! Will mark Brainliest for full answer!!!

Softa [21]

<u><em>Answer:</em></u>

Part a .............> x = 11

Part b .............> k = 57.2

Part c .............> y = 9.2

<u><em>Explanation:</em></u>

The three problems deal with inverse variation between two variables

An inverse variation relation between two variables means that when one of the variables increases, the other will decrease (and vice versa)

<u>Mathematically, an inverse variation relation is represented as follows:</u>

$y = \frac{k}{x}$

where x and y are the two variables and k is the constant of variation

<u><em>Now, let's check the givens:</em></u>

<u>Part a:</u>

We are given that y = 3 and k = 33

<u>Substitute in the original relation and solve for x as follows:</u>

$y = \frac{k}{x}\\ \\3 = \frac{33}{x}\\ \\x=\frac{33}{3}=11$

<u>Part b:</u>

We are given that y = 11 and x = 5.2

<u>Substitute in the original relation and solve for k as follows:</u>

$y=\frac{k}{x}\\ \\11=\frac{k}{5.2}\\ \\k=11*5.2=57.2$

<u>Part c:</u>

We are given that x=7.8 and k=72

<u>Substitute in the original relation and solve for y as follows:</u>

$y=\frac{k}{x}=\frac{72}{7.8}=9.2$ to the nearest tenth

Hope this helps :)

6 0

3 years ago