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tatiyna
3 years ago
15

Please help!!!!! Please

Mathematics
2 answers:
konstantin123 [22]3 years ago
8 0

Answer:

96,144,192

Step-by-step explanation:

well it's easy

Dimas [21]3 years ago
7 0
So you just multiply 48 by 2 then by 3 and then by 4 resulting to a=96, b=144, c=192
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Marie's dad drives an SUV that uses 2.5 litres of fuel for every 20 km driven. He drives to work 20 km each way, 5 days a week.
mash [69]

Answer: 12.5 litres

Step-by-step explanation:

The SUV takes 2.5 litres for every 20 km driven.

The trip to work is 20 km which means that going to work costs Marie's father 2.5 litres every day in fuel.

In a week, there are 5 working days.

The amount of fuel used per week is therefore:

= 2.5 * 5

= 12.5 litres

5 0
3 years ago
A bug was running along a number line at a speed of 11 units per minute. It never changed its direction. If at 7:15 it was at th
barxatty [35]

Answer:

155 Units

Step-by-step explanation:

Rate of Bug (given) = 11 units PER MINUTE

It never changed direction, so it was going in positive direction (assume).

In 7:15 pm (evening), it was at Point 100,

We want the point at which it was at 7:20 pm.

7:20pm - 7:15pm = 5 minutes

So, time passed 5 minutes. It's rate is 11 units PER MINUTE, so in 5 mins:

11 * 5 = 55 units

Assuming he is going in positive direction, the bug will be at:

100 + 55 = 155 Units

8 0
3 years ago
Read 2 more answers
Write 625 using exponents in as many ways as you can?
Arlecino [84]

Answer:

See below

Step-by-step explanation:

625 = 25^2, or (5^2)^2 = 5^4

8 0
3 years ago
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the length of a chord of a circle is 24 cm long and is 5 cm away from the center of the circle what is the radius of the circle​
iogann1982 [59]

Answer:A chord with length 24 cm at 5 cm from the center gives a radius of 13 cm, using a half-length of 12 cm and the distance of 5 cm in Pythagoras’s theorem to find the radius.

Using the same radius and Pythagoras's theorem again, a chord 10 cm from the circle with radius 13 cm has a half-length of sqrt(69), or a full length of double that, so 2*sqrt(69).

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4 0
3 years ago
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Lim x → 0 sin3x)/5x^3 -4
JulsSmile [24]
\bf \lim\limits_{x\to 0}\ \cfrac{sin^3(x)}{5x^3}-4
\\ \quad \\\\ \cfrac{sin^3(x)}{5x^3}-4\implies \cfrac{[sin(x)]^3}{5x^3}-4
\\ \quad \\
 

\cfrac{[sin(x)]^3}{x^3}\cdot \cfrac{1}{5}-4
\\ \quad \\
thus
\\ \quad \\
\lim\limits_{x\to 0}\cfrac{[sin(x)]^3}{x^3}\cdot \lim\limits_{x\to 0}\cfrac{1}{5}-4\qquad \boxed{recall \qquad \lim\limits_{x\to 0} \cfrac{sin(x)}{x}\implies 1}
\\ \quad \\
1\cdot \cfrac{1}{5}-4
8 0
3 years ago
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