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aliina [53]
4 years ago
9

A rabbit can run short distances at a rate of 35miles per hour.a fox can run short distances at a rate of 21 miles per half hour

.which animal is faster,and by how much?
Mathematics
2 answers:
photoshop1234 [79]4 years ago
7 0
Since the fox's rate is based on half an hour and the rabbit's rate is based on a whole hour, you would need to create an equation that is spanned on the same amount of time. 

To do this you would, simply, multiply the fox's rate by 2.

21 x 2 = 42
\frac{1}{2} x 2 = 1

42 mi/1 h

A fox can run 7 miles per hour faster than a rabbit can. 

I hope this helps!
 
GenaCL600 [577]4 years ago
3 0
The fox is faster. This means that while the rabbit can run at a rate of 35 miles per hour, the fox runs 42 miles per hour. Therefore, the fox runs 7 more miles per hour than the rabbit.
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The answers to that would be x=-2
5 0
4 years ago
Please solve the following quadratic equation and show your work: x^2 -x =30
saveliy_v [14]
Rewriting the equation as a quadratic equation equal to zero:
     x^2 - x - 30 = 0
We need two numbers whose sum is -1 and whose product is -30. In this case, it would have to be 5 and -6. Therefore we can also write our equation in the factored form
     (x + 5)(x - 6) = 0
Now we have a product of two expressions that is equal to zero, which means any x value that makes either (x + 5) or (x - 6) zero will make their product zero.
     x + 5 = 0  => x = -5
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8 0
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
A member was charged 260$usd and then applied for financial aid and received a 30% discount . They reach out and ask for the 30%
sergiy2304 [10]

Answer:

The total charge with the discount is $182

The refund is $78

Step-by-step explanation:

Given data

Charge= $260

Discount=30%

Let us find the amount of the 30%

=30/100*260

=0.3*260

=$78

The new charge after the discount is

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The total charge with the discount is $182

The refund is $78

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3 years ago
Help ya girl out please
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Answer: 2,592











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