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3 years ago

Ok last one lets get it!

Mathematics

2 answers:

hammer [34]3 years ago

4 0

Answer:

No yes no yes no

Step-by-step explanation:

RideAnS [48]3 years ago

3 0

Dallas: No
Landing: Yes
Louisville: No
Madison: Yes
Montgomery: No

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Rosa's test scores are 78, 92, 88, and 89. What must she score on her next test to have an average of 90? A) 95 B) 97 C) 99 D) 1

Morgarella [4.7K]

Answer: D

Step-by-step explanation:

Let x represent the score on her next test

$\frac{78 + 92 + 88 + 89 + x}{5} = 90$

$(5)\frac{78 + 92 + 88 + 89 + x}{5} = (5)90$

78 + 92 + 88 + 89 + x = 5(90)

347 + x = 450

-347 -347

x = 103

3 0

3 years ago

Help pleasee I don't know how to do this

andreyandreev [35.5K]

Answer:

Just factorize the given plynomials.

Step-by-step explanation:

like in first example 7x+49

7(x+7)

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2 years ago

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A grocery store has 5 pounds of gronola. one customer buys 2 3rds pounds of gronola and another buys 5 6ths pounds. after these

Kryger [21]

Answer:

3 1/2 Pounds

Step-by-step explanation:

2/3=4/6

4/6+5/6=9/6

30/6-9/6=21/6

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2 years ago

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Hey i need some help ill give you brainest 2. Ms. Juhal was making t-shirts. One of the designs had the instructions to make poi

liubo4ka [24]

a is the best way to find it if there is points involved so keep that in mind

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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