All categories

3 years ago

What would be the answer of my fractions 9 1/2 + 4 3/5=

Mathematics

2 answers:

Nina [5.8K]3 years ago

8 0

Answer:

14 $\frac{1}{10}$ or $\frac{141}{10}$

Step-by-step explanation:

Turn your fractions into improper and make them have the same denominator

$\frac{95}{10}$ + $\frac{46}{10}$ = $\frac{141}{10}$

Leave it as is or simplify to 14 $\frac{1}{10}$

damaskus [11]3 years ago

8 0

Answer:

14 1/10

Step-by-step explanation:

Make it to improper fraction

9 1/2 would be 19/2

and 4 3/5 = 23/5

Then to add u have to make the denominators common:

19/2 + 23/5

Both goes common denominator with 20

To make it 20:

(19 × 10/2 ×10) + (23 ×4/5×4)

190 / 20 + 92 / 20

Add those (190 + 92) / 20

282 / 20

Make it to the simplest form:

141 / 10

Then make it mixed fraction:

14 1/10

You might be interested in

Janet has a bottle of milk which is of 500 ml . She spills 1/2 of it . How much did she spill ?

Roman55 [17]

Answer:

She spilled 250 ml

Step-by-step explanation:

5 0

4 years ago

Solve:<br><br><img src="https://tex.z-dn.net/?f=%5Cunderset%7Bx%5Crightarrow~3%7D%7B%5Clim%7D~%5Cdfrac%7B2x%5E2-18%7D%7Bx%5E2-3x

vodomira [7]

Hello, please consider the following.

$\displaystyle \lim_{x\rightarrow3}~\dfrac{2x^2-18}{x^2-3x} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x^2-3^2)}{x(x-3)} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x-3)(x+3)}{x(x-3)} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x+3)}{x} \\ \\ \\=\dfrac{2(3+3)}{3}\\ \\ \\=\dfrac{2*3*2}{3} =\Large \boxed{\sf \bf \ 4 \ }$

Thank you

8 0

3 years ago

While on a walk in the country, you pass a field full of horses and chickens. After a quick count, you determine there are 43 he

Irina-Kira [14]

Horse = x = 1 head and 4 feet

Chicken =y = 1 head and 2 feet

Horse + chicken = x + y = 43 ( total heads)

Horse = 43 - y

4x + 2y = 122

4(43-y) + 2y = 122

Simplify:

172-4y + 2y = 122

Combine like terms

172-2y = 122

Subtract 172 from both sides

-2y = -50

Divide both sides by-2

Y = 25

X + y = 43

X + 25 = 43

X = 18

There are 18 horses and 25 chickens.

4 0

3 years ago

A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?

Len [333]

Answer:

(a) $Selection = 6724520$

(b) $At\ most\ 12 = 6553976$

(c) $At\ most\ 8 = 6066720$

(d) $At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

Step-by-step explanation:

Given

$Colors = 8$

$Balloons = 28$ --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

$n => 28 + 8-1$ $= 35$

$r = 28$

$Selection = ^{35}^C_{28$

$Selection = \frac{35!}{(35 - 28)!28!}$

$Selection = \frac{35!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29}{7!}$

$Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}$

$Selection = \frac{33891580800}{5040}$

$Selection = 6724520$

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

$n => 15 + 8 -1 = 22$

$r = 15$

Selection of at least 13 =

$At\ least\ 13 = ^{22}C_{15}$

$At\ least\ 13 = \frac{22!}{(22-15)!15!}$

$At\ least\ 13 = \frac{22!}{7!15!}$

$At\ least\ 13 = 170544$

Ways of selecting at most 12 =

$At\ most\ 12 = Total - At\ least\ 13$ --- Complement rule

$At\ most\ 12 = 6724520- 170544$

$At\ most\ 12 = 6553976$

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

$n => 19+ 8 -1 = 26$

$r = 19$

Selection of at least 9 =

$At\ least\ 9 = ^{26}C_{19}$

$At\ least\ 9 = \frac{26!}{(26-19)!19!}$

$At\ least\ 9 = \frac{26!}{7!19!}$

$At\ least\ 9 = 657800$

Ways of selecting at most 8 =

$At\ most\ 8 = Total - At\ least\ 9$ --- Complement rule

$At\ most\ 8 = 6724520- 657800$

$At\ most\ 8 = 6066720$

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

$n =6+6-1 = 11$

$r = 6$

Selection =

$^{11}C_6 = \frac{11!}{(11-6)!6!}$

$^{11}C_6 = \frac{11!}{5!6!}$

$^{11}C_6 = 462$

Using inclusion/exclusion rule of two sets:

$Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 6724520 - 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

3 0

3 years ago

Pythagorean triples are given by the formulas x2 - y2, 2xy, and x2 + y2. Use the formulas for the Pythagorean triples to prove w

Arte-miy333 [17]

Let $k$ be an integer. Suppose there is a triangle with legs of length 16 and $2k+1$ . Then by the Pythagorean theorem, the length of the hypotenuse should be

$\sqrt{16^2+(2k+1)^2}=\sqrt{4k^2+4k+257}$

The formulas for Pythagorean triples say that if the legs are integers, then so must be the hypotenuse, because if $x=16$ and $y=2k+1$ are integers, then so are $x^2-y^2$ , $2xy$ , and $x^2+y^2$ .

However, $4k^2+4k+257$ is not a perfect square trinomial, which means for any integer $k$ , the length of the hypotenuse is not an integer, so such a triangle doesn't exist.

4 0

3 years ago