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3 years ago

If Gerald bought a 12-case of Dr. Pepper for $4.89, what would be the unit rate for each can?

Mathematics

2 answers:

iren2701 [21]3 years ago

4 0

It 1 at dollar tree

OverLord2011 [107]3 years ago

4 0

$0.4075 per can would be your answer

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A study found that 9% of dog owners brush their dog’s teeth. Of 578 dog owners, about how many would be expected to brush their

WINSTONCH [101]

It’s about 52. 578•.09=52.02, which rounds to 52.

6 0

3 years ago

Help me and i’ll mark you brainliest plus you get 20 points !!!!

Anton [14]

Answer:

Angle bisector

Step-by-step explanation:

Find the measure of the angle COA. By angle addition postulate,

$m\angle COX=m\angle AOX+m\angle COA$

From the diagram,

$m\angle COX=80^{\circ}\\ \\m\angle AOX=40^{\circ},$

then

$80^{\circ}=m\angle COA+40^{\circ}\\ \\m\angle COA=80^{\circ}-40^{\circ}=40^{\circ}$

Find the measure of the angle BOA. By angle addition postulate,

$m\angle BOX=m\angle AOX+m\angle BOA$

From the diagram,

$m\angle BOX=60^{\circ}\\ \\m\angle AOX=40^{\circ},$

then

$60^{\circ}=m\angle BOA+40^{\circ}\\ \\m\angle BOA=60^{\circ}-40^{\circ}=20^{\circ}$

Find the measure of the angle COB. By angle addition postulate,

$m\angle COX=m\angle BOX+m\angle COB$

From the diagram,

$m\angle BOX=60^{\circ}\\ \\m\angle COX=80^{\circ},$

then

$80^{\circ}=m\angle COB+60^{\circ}\\ \\m\angle COB=80^{\circ}-60^{\circ}=20^{\circ}$

This means, the measures of angles COB and BOA are the same and are equal half the measure of angle COA, so angles COB and BOA are congruent. This means, the ray OB is the angle bisector of angle COA

6 0

3 years ago

What is 2/3 times 1 whole

VMariaS [17]

Answer:

2/3

Step-by-step explanation:

2/3 times 1 is 2/3 of 1

5 0

3 years ago

Read 2 more answers

Let v be the vector from initial point upper p 1p1 to terminal point upper p 2p2. write v in terms of i and j.

Umnica [9.8K]

I reallly dont know but that long is going to end up long

6 0

3 years ago

How do you solve this equation: -2m+3=|15+m|

Elden [556K]

Any number inside the modulus sign becomes positive. This means $|15+m|=|-(15+m)|=|-15-m|$ and so we have,

$-2m+3=|15+m| \Rightarrow \left \Big\{ {{-2m+3=15+m} \atop{-2m+3=-15-m}} \right$

Solving these gives us

$-2m+3=15+m \Rightarrow m=-4$

$-2m+3=-15-m \Rightarrow = m=18$

However if we check the second solution in the original equation we obtain $-2(18)+3=|15+18| \Rightarrow -33=33$ . This is false and so $m=18$ can't be a solution.

Therefore the only solution is $m=-4$ .

(Note: I'm not sure why the second solution didn't work but when there's a modulus sign involved it always pays to check your final answers to be sure. I'll have a think about it but in case you find out before I do, I'd be interested to know in the comments.)

8 0

3 years ago