All categories

3 years ago

Select the inequality that models the problem.

1 answer:

3 0

Here is the answer for the given problem above.
We know that the formula for getting the perimeter is P =2W +2L.
Now, it is defined that <span>length of a rectangle is twice its width so this gives us L=2w. And the perimeters is less than 50 meters so it is P < 50.
So the answer for this would be 2(2w) + 2w <50. It would be the first option. Hope this answer helps.</span>

You might be interested in

Which statement describes what these four powers have in common?

ololo11 [35]

The answer to your question is,

all the powers have the value of 1 because the exponent is zero

-Mabel <3

3 0

4 years ago

Read 2 more answers

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

2. Let A, B and c be three towers. The bearing of B from A is 120° and angle BC is 68º. The distance from Tower A to Tower B is

g100num [7]

(i) The bearing of B from A and angle x are supplementary, then:

$\begin{gathered} 120\degree+x=180\degree \\ x=180\degree-120\degree \\ x=60\degree \end{gathered}$

(ii)

From the above diagram, the bearing angle of A from B is:

$120\degree+180\degree=300\degree$

(iii)

From the above diagram, the bearing angle of C from B is:

$180\degree+52\degree=232\degree$

(iv)

Applying the law of cosines with the sides AB = 145m and CB = 240 m, and the angle ABC = 68°, the distance AC is:

$\begin{gathered} AC^2=AB^2+CB^2-2\cdot AB\cdot CB\cdot cos\left(\angle ABC\right) \\ AC^2=145^2+240^2-2\cdot145\cdot240\cdot cos(68\degree) \\ AC^2=21025+57600-69600\cdot cos(68\operatorname{\degree}) \\ AC^2=52552.3811 \\ AC=\sqrt{52552.3811} \\ AC\approx229\text{ m} \end{gathered}$

6 0

1 year ago

Find the 34th term for this sequence -6, 24, 54, 84, …

Hatshy [7]

Please forgive me if this answers wrong but I think your answer will be 54, -6, 24

4 0

2 years ago

If $300 is invested at a rate of 6% per year and is compounded quarterly, how much will the investment be worth in 12 years?

MAXImum [283]

Answer:

$613.04

Step-by-step explanation:

<em>Compound Interest Formula:</em>

$A = P(1+\frac{r}{n})^{nt}$

n = number of compounds

t = time

r = interest rate

P = principle amount (original amount)

A = final amount

Since it's compounded quarterly, that means there will be 4 compounds per year, so n=4. The interest rate has to be converted to the decimal value, and this is done by simply dividing it by 100 to get r=0.06.

Plug Values into equation:

$A = 300(1+\frac{0.06}{4})^{12*4}$