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3 years ago

How many 3 1⁄2 lengths can be cut from a string measuring 105m?

Mathematics

2 answers:

Ann [662]3 years ago

8 0

Answer:

The answer is 30

Step-by-step explanation:

<h3><u>Given</u>;</h3>

3½ = 3.5

Now,

105 ÷ 3.5 = 30

Thus, The answer is 30

<u>-TheUnknownScientist 72</u>

valentina_108 [34]3 years ago

3 0

Answer:

101.5 or 101 1/2

Step-by-step explanation:

105/3.5 = 101.5

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What is the value of b<br> 2b + 8 − 5b + 3 = −13 + 8b − 5

Yuri [45]

Answer:

See below.

Step-by-step explanation:

2b + 8 − 5b + 3 = −13 + 8b − 5

Reorder like terms.

2b-5b+8+3=8b-13-5

Combine those like terms. Then solve.

-3b+11=8b-18

-11 -11

-3b=8b-29

-8b -8b

-11b=-29

/-11 /-11

b=2.64

-hope it helps

3 0

3 years ago

Read 2 more answers

Give the coordinates and quadrant of Point Z

Lorico [155]

Answer:

(9,10) and Quadrant I

Step-by-step explanation:

I hope this helps you :)

4 0

3 years ago

Problems that are not answered I need help with quickly!

Anika [276]

Whats the problem? just substitute. For example: 3x-5 if x=4 so what you do is this; 3(4)-5=7

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 that $n\ge h\ge i\ge j\ge k\ge m\ge 1$

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

3 years ago

Y = –x + 4

aliina [53]

Y = -x + 4.....so sub in -x + 4 in for y in the other equation

x + 2y = -8
x + 2(-x + 4) = -8
x - 2x + 8 = -8
x - 2x = -8 - 8
-x = -16
x = 16

y = -x + 4
y = -16 + 4
y = - 12

one solution (16,-12)

8 0

3 years ago

Read 2 more answers