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Tasya [4]
3 years ago
14

Which of the following represents the value of the missing side? Round to the nearest tenth.

Mathematics
1 answer:
dlinn [17]3 years ago
3 0
C because it is to the nearest tenth I think
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Graph this line using the slope and y-intercept: y=5x+2
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You can use Desmos, or any online graph-er to graph the line.

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Which is the simplified form of the expression?
Scorpion4ik [409]

Answer:

1 / q^30.

Step-by-step explanation:

[(p^2)(q^5)]^-4 * [(p^-4)(q^5)]^-2

Using the law (a^b)^c =  a^bc :-

=  p^-8 * q^-20 * p^8 * q^-10

= p^(-8+8) * q^(-20-10)

= p^0 * q^-30

= 1 * q^-30.

= 1 / q^30.

8 0
3 years ago
How much it will take for a bag to travel 5 meters across the floor of it is traveling at 1 m/s
tester [92]
Time = \frac{distance}{speed}

The distance the bag will travel is 5 meters, and the speed it is traveling at is 1 m/s. Plug these into the formula for time.

\frac{5 m}{1 m/s} = 5 s
5/1 = 5 and the m (meters) cancel out, so you are left with s, seconds.

It will take the bag 5 seconds to travel across the floor.
5 0
3 years ago
Can pls someone help me with my homework
Wewaii [24]

Answer:

is there a formula in the notes if so put it in then it might help out

Step-by-step explanation:

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3 years ago
13. The least common multiple of two non-zero integers a and b is the unique positive integer m such that (i) m is a common mult
Vlad [161]

Answer:

[a,b] divides n

Step-by-step explanation:

Let us denote the least common multiple of a and b [a,b]=m.

We want to prove that m divides n, where n is a multiple of a and b.

We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:

(1) ... n=mq+r, where 0<r<m

As n is a multiple of a and b, there exists s and t integers such that:

sa=n and tb=n

Same thing happens to m as it is the least common multiple, there exists u and v such that:

ua=m and vb=m

So (1) has the following form:

n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and

n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r

So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.

So this means that \frac{ab}{m} is and integer.

As m= vb, then \frac{m}{b} is an integer, lets say \frac{m}{b}=v; and as m=ua, then \frac{m}{a}=u.

So \frac{ab}{m}v=\frac{ab}{m}\frac{m}{b}=a, so \frac{ab}{m} divides a; on the other hand, \frac{ab}{m}u=\frac{ab}{m}\frac{m}{a}=b, so \frac{ab}{m} divides b. From this we can conclude that \frac{ab}{m} is a common divisor of a and b.

4 0
3 years ago
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