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3 years ago

Ryan makes 65 ounces of apple jelly. He puts the same amount of jelly into 6jars. About how much jelly is in each jar?

Mathematics

1 answer:

vfiekz [6]3 years ago

7 0

There would be 10.8 ounces in each jar

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HELP HELP I MIGHT DIE - Can a 24 volt controller handle a 36V battery

Flura [38]

Nooo im pretty sure it cannot

3 0

3 years ago

What are the four points of intersection between 4x^2 + y^2 - 4y - 32 = 0 and x^2 - y - 7 = 0 ? Solve algebraically

tensa zangetsu [6.8K]

Remember (a²-b²)=(a-b)(a+b)

solve for a single variable
solve for y in 2nd

add y to both sides
x²-7=y
sub (x²-7) for y in other equaiton

4x²+(x²-7)²-4(x²-7)-32=0
expand
4x²+x⁴-14x²+49-4x²+28-32=0
x⁴-14x²+45=0
factor
(x²-9)(x²-5)=0
(x-3)(x+3)(x-√5)(x+√5)=0
set each to zero

x-3=0
x=3

x+3=0
x=-3

x-√5=0
x=√5

x+√5=0
x=-√5

sub back to find y

(x²-7)=y

for x=3
9-7=2
(3,2)

for x=-3
9-7=2
(-3,2)

for √5
5-7=-2
(√5,-2)

for -√5
5-7=-2
(-√5,-2)

the intersection points are

(3,2)
(-3,2)
(√5,-2)
(-√5,-2)

8 0

4 years ago

Jaxon has a bag of marbles. He draws one marble from the bag. Before drawing another marble from the bag,

icang [17]

A because if you I want the outcome to be independent then you have to take it out

5 0

3 years ago

Read 2 more answers

Pls answer this quick

AfilCa [17]

Answer:

I belive its $\sqrt{7}$

Mark me as brainliest pls

5 0

3 years ago

A train is leaving in 11 minutes and you are one mile from the station. Assuming you can walk at 4mph and run at 8mph, how much

ddd [48]

7 minutes is the time can you afford to walk

Solution:

Given that, train is leaving in 11 minutes and you are one mile from the station

Let "x" represent how much time can you afford to walk

Then, (11 - x) is the time you run

You can walk at 4mph and run at 8mph

Convert to miles per minute

1 hour = 60 minutes

$4\ mph = 4 \times \frac{1}{60} \text{ miles per minute } = \frac{1}{15} \text{ miles per minute }\\\\8\ mph = 8 \times \frac{1}{60}\ \text{ miles per minute } = \frac{2}{15} \text{ miles per minute }$

We know that,

$Distance = speed \times time$

Given that, you are one mile from the station

Therefore,

1 mile = distance walk + distance run

$1 = x \times \frac{1}{15} + (11-x) \times \frac{2}{15}\\\\1 = \frac{1}{15}(x + 2(11-x))\\\\1 = \frac{1}{15}(x + 22 - 2x)\\\\x + 22 - 2x = 15\\\\x = 22 - 15\\\\x = 7$

Thus, 7 minutes is the time can you afford to walk

6 0

4 years ago