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Harlamova29_29 [7]

2 years ago

12

Katie can buy four cans of soup for $10. She can buy 18 cans of soup for $30. Kate wants to buy soup for the lowest possible pri

ce at which store should she shop explain .

1 answer:

amid [387]2 years ago

5 0

She should buy at the second store due to the fact that you are getting 6 mora for the same price (if you were to buy 30$ worth from the first one)

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Assuming the probability that a student is accepted to UCF is 0.6. If 5 students from the same school apply, what is the probabi

igor_vitrenko [27]

Answer:

0.006

Step-by-step explanation:

0.6x0.005x0.02

7 0

1 year ago

the sum of three numbers is 128.The third number is 10 more than the first .The second number 4 times the third number. what are

Ray Of Light [21]

128 = a + + 4(a + 10) + (a + 10)
128 = a + 4a + 40 + a + 10
128 = 6a + 50
128-50 = 6a
78 = 6a
13 = a

1st = a = 13
2nd = 4(a + 10) = 4(23) = 92
3rd = a + 10 = 23

3 0

2 years ago

How many meters in 1 kilometer

mixas84 [53]

1 kilometer = 10000 meters
1 kilometer = 100000 centimeters
I can so more and explain more is you want Hope it helps
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7 0

2 years ago

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In the diagram, the length of segment VS is 39 units.

Darina [25.2K]

Answer:

$TV=38\ units$

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

we know that

The figure shows a kite

The kite has two pairs of consecutive and congruent sides and the diagonals are perpendicular

That means

TS=VS

TQ=VQ

TR=RV

we have

$VS=39\ units$

$TS=6x-3$

so

$6x-3=39$

solve for x

$6x=39+3\\6x=42\\x=7$

<em>Find the value of RV</em>

$RV=2x+5$

substitute the value of x

$RV=2(7)+5=19\ units$

Remember that

$TV=TR+RV$ ---> by segment addition postulate

we have

$TR=RV=19\ units$

so

$TV=19+19=38\ units$

5 0

2 years ago

PLSSSSSS!!!! HELP ME I WILL MARK YOU!!!!!!!!!!

joja [24]

Answer:

2.25

Step-by-step explanation: Replace the z with the value given (-18) and then multiply that by 1.1 you can then add it to 21.3 and then do the exponent

5 0

2 years ago

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