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3 years ago

Three mangos and two apples sell for $1.25. Five mangos and one apple sell for $1.50. What is the cost for each mango

Mathematics

1 answer:

Pepsi [2]3 years ago

6 0

Answer:

mango is .25 and apple is .25 i took a picture of my work- dont mind the purple paper it was my notes packet xdd

Step-by-step explanation:

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A. 1.5 <br> B. 2<br> C. 2.5 <br> D. 3

Crank

Answer:

a because you just adding

Step-by-step explanation:

3 0

3 years ago

The movie theater received a shipment of candy boxes in a carton that has a volume of about 64,000 cubic inches. Each candy box

vova2212 [387]

Volumen carton= 64000 inches³
Volume of a candy box= 6 inx 5 inx 1/2
Volume of a candy box=15 inches³

Then, you have that the total candy boxes in the carton is:

x=640000 inches³/15 inches³
x=4267 candy boxes

About how many candy boxes are in the carton? There are 4267 candy boxes

4 0

3 years ago

Read 2 more answers

Please help!! what is the value of x and please do step by step if possible !

alexira [117]

Answer:

x=71 degrees

Step-by-step explanation:

it is a isosceles triangle so the blank angle is also x

angles of a triangle add up to 180

180=38+2x

2x=142

x=71

8 0

2 years ago

8/15 + 2/5 pls solve this for me someone pls

photoshop1234 [79]

Answer: 8/15 + 2/5 = 0.93etc or 14/15

Step-by-step explanation:

7 0

2 years ago

18 POINTS! At the AoPS office, mice vary inversely with cats, that is, mice=k/cats, for some value of k. When there are 3r-19 ca

Blizzard [7]

Answer:

k=30

Step-by-step explanation:

The number of mice vary inversely with cats. This is written as:

$Mice \propto \dfrac{1}{Cat} \\Mice = \dfrac{k}{Cat}$

When there are 3r-19 cats, there are 2r+1 mice

$2r+1 = \dfrac{k}{3r-19}\\$Cross multiply$\\k=(3r-19)(2r+1)$

When there are 6r-27 mice, there are r-5 cats.

$6r-27 = \dfrac{k}{r-5}\\$Cross multiply$\\k=(6r-27)(r-5)$

Taking the values of k above, we have:

$k=(3r-19)(2r+1) =(6r-27)(r-5)\\(3r-19)(2r+1) =(6r-27)(r-5)\\6r^2+3r-38r-19=6r^2-30r-27r+135\\$Collect like terms\\6r^2-6r^2+3r+30r-38r+27r-19-135=0\\22r-154=0\\22r=154\\$Divide both sides by 22\\r=7$

Therefore:

$k=(3r-19)(2r+1) \\=(3*7-19)(2*7+1)\\=(21-19)(14+1)\\=2*15\\k=30$

8 0

3 years ago