1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
olga2289 [7]
3 years ago
8

2)

Mathematics
2 answers:
AlexFokin [52]3 years ago
7 0

Answer: she has spent $60 dollars and she has 180 more to go

Step-by-step explanation:

Bess [88]3 years ago
5 0

Answer:

75%

Step-by-step explanation:

You said she raised 25% so that leave 75% left because adding both 75 and 25 together makes 100%. If you do 100 minus 25 you'll get 75%.

You might be interested in
Solve and check:8a-4=25+6(a-8)
OlgaM077 [116]
8a - 4 = 25 + 6 (a - 8)

(distribute the 6)

8a - 4 = 25 + 6a - 48

(combine like terms 25 - 48)

8a - 4 = 6a - 23

( subtract 6a from both sides)

2a - 4 = -23

(add 4 to both sides)

2a = -19

(divide both sides by 2)

a = - 9.5

////CHECK/////

8 (-9.5) - 4 = 25 + 6 ( -9.5 -8)

(plug each side into a calculator)

-80 = -80

FINAL ANSWER:

a = -9.5
5 0
2 years ago
I need help with this
Ksenya-84 [330]

Answer:

6.7-(-2.5)=9.2                  6.7+(-2.5)=4.2                          Neither

2.5 + 6.7                          6.7 - 2.5                                  2.5 - 6.7

                                       -2.5 + 6.7                                 -2.5 - 6.7

                                                                                        -6.7 + 2.5

7 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
erma4kov [3.2K]

Answer:

\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}

Step-by-step explanation:

Given

5 tuples implies that:

n = 5

(h,i,j,k,m) implies that:

r = 5

Required

How many 5-tuples of integers (h, i, j, k,m) are there such thatn\ge h\ge i\ge j\ge k\ge m\ge 1

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed:  This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

^{n}C_r => ^{n + 4}C_5

^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}

^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}

Expand the numerator

^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}

<u><em>Solved</em></u>

6 0
2 years ago
If 588 digits were used to number the pages of a book, how many pages are in this book?
andre [41]

Answer:

294 pages are in this book

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
I need to simplify this
Korvikt [17]
Hi there!

3 \sqrt{54} + 2 \sqrt{24} =
First we split up the square root into two parts.

3 \times \sqrt{9} \times \sqrt{6} + 2 \times \sqrt{4} \times \sqrt{6} =
Now we calculate the value of the square roots which have an integer as a solution

3 \times 3 \times \sqrt{6} + 2 \times 2 \times \sqrt{6} =
Multiplying the integers gives us our next step.

9 \sqrt{6} + 4 \sqrt{6} =
And finally we add up the roots.

13 \sqrt{6}
3 0
3 years ago
Other questions:
  • PLLLLLLLLLZZZZ HHHHHHHEEEEEEEEELLLLLLPP
    11·1 answer
  • WHAT IS THE CORRECT SCALE FACTOR?
    11·1 answer
  • Mrs.Feinstein recorded the weight, in pounds, recorded of each of her 8 great-grandchildren. Her data are shown.
    11·1 answer
  • Can someone please help me with this ​
    14·1 answer
  • What would the exponential growth model be for the given situation?
    9·1 answer
  • 2.5 + 3(5-2.1p) = -13.37
    13·2 answers
  • Solve: 0 = 2x^2 + 16x
    8·1 answer
  • "the quotent of a number and 4" i dont wanna fail math
    12·1 answer
  • For each graph,
    9·1 answer
  • Which of these is true and why
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!