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Sveta_85 [38]
3 years ago
7

-8x-11x what is the answer

Mathematics
1 answer:
gregori [183]3 years ago
7 0

Answer:

-19x

Step-by-step explanation:

so you start with -8 and when you subtract 11 from -8 its the same as adding them but because your subtracting you'll decrease which make the answer -19x. (hopefully this made sense)

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Victor opens a savings account with $750. The account pays 6% interest, compounded every six months. If he makes no other deposi
Oksanka [162]
You would do 750 divided by 6% and that equals 125. So victor would deposit $125 after 18 months
6 0
3 years ago
Simplify this math problem show Your work
loris [4]

9514 1404 393

Answer:

  (p -9q)/(4p² +12pq)

Step-by-step explanation:

The least common denominator will be the product of the denominators.

  \dfrac{-3}{4p}+\dfrac{1}{p+3q}=\dfrac{-3(p+3q)+1(4p)}{(4p)(p+3q)}=\boxed{\dfrac{p-9q}{4p^2+12pq}}

5 0
3 years ago
Help me please I don’t know this
Afina-wow [57]
The answer is the number will have to be less than 11 which means that it is a. 9
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8 0
3 years ago
A parabola has a line of symmetry x = -5. The minimum value of the quadratic function that it represents is -7. Find a possible
Semmy [17]
The standard equation of a parabola is:

(y-k) = a(x - h)² , where h and k are the coordinates of the Vertex
 
Since the vertex is ( - 5 , -7), then :

(y +7) = a (x + 5)².
It's given that the vertex is a minimum, that means the coefficient a>0 and the parabola is open upward.

Depending on the value of a (always a>0) , the equation :

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8 0
3 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

4 0
3 years ago
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