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2 years ago

11

guys honestly my day hasnt been that good but anyway here your daily free pints, hope you guys are having a good day keep your h

eads held up high kings and queens

2 answers:

Ksju [112]2 years ago

8 0

Answer:

You have a nice day

Step-by-step explanation:

Don’t let anyone keep you down queen UwU

Rina8888 [55]2 years ago

6 0

Well everyone have bad and good days each day is a news day

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27-4(26-2×3)to the 2nd power

Svetlanka [38]

First do the ones in the parentheses. So, 2x3=6. 26-6=20. 20^2=400. Now you have 27-4x400. 4x400=1600. 27-1600=-1573.

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3 years ago

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What is the value of x in <br> 150-35x=45

Annette [7]

Your answer is x = 3 because if you take 150 and subtract it by both sides:
150 - 35x = 45
-150 -150

you have to cross out both 150 and -150, take 45 and subtract 150 which will give you -105.
-35x = -105

take -35 and divide by both sides. Cross out the two -35's. Take -105/-35 which your answer is x = 3. Hopes this helps. :)

~Shadow

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3 years ago

If Maliyah won 25 games and llost 10 what would be the ratio for the games lost to total

Alla [95]

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10:35

Step-by-step explanation:

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3 years ago

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3 years ago

A cyclist travels at distance of 400 meters in 120 seconds towards school, calculate his speed. (Show your work)

DochEvi [55]

we know that

The scalar magnitude of the velocity vector is the speed. The speed is equal to

$Speed=\frac{distance}{time}$

in this problem we have

$distance=400\ m \\time=120\ sec$

substitute in the formula

$Speed=\frac{400}{120}$

$Speed=3.5\frac{m}{sec}$

therefore

<u>the answer Part a) is</u>

the speed is equal to $3.5\frac{m}{sec}$

<u>Part b) </u>Find the velocity

we know that

<u>Velocity </u>is a vector quantity; both magnitude and direction are needed to define it

in this problem we have

the magnitude is equal to the speed

$magnitude=3.5\frac{m}{sec}$

$direction=North\ East\ (NE)$

therefore

<u>the answer Part b) is</u>

the velocity is $3.5\frac{m}{sec}\ North\ East\ (NE)$

Part c)

we know that

the acceleration is equal to the formula

$a=\frac{V2-V1}{t2-t1}$

in this problem we have

$V2=0 \\V1=3.5\frac{m}{sec}$

$t2=15\ sec\\t1=0$

substitute in the formula

$a=\frac{0-3.5}{15-0}$

$a=-\frac{3.5}{15}\frac{m}{sec^{2}}$

$a=-\frac{7}{30}\frac{m}{sec^{2}}$

therefore

<u>the answer Part c) is</u>

the acceleration is $-\frac{7}{30}\frac{m}{sec^{2}}$

This is an example of negative acceleration

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3 years ago

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