All categories

3 years ago

Leahs age is twice her cosins age. If they add thier ages together, they get 36. How old are leah and her cosin

Mathematics

1 answer:

Art [367]3 years ago

6 0

Leah is 24 and her cousin is 12. Since 24 is double of 12, thats your answer and it makes the most sense.

You might be interested in

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

2 years ago

I will give brainlest if the answer is right Enter the missing numbers in the boxes to complete the table of equivalent ratios.

rjkz [21]

Answer:

2 , 10 , and 15 in order of missing spaces, up to down

Step-by-step explanation:

You find the ration 8/20 and find out if the ration is simplifiable and turn it to 2/5 which you multiply the rest of the answers too. Hope this answers your question :)

4 0

3 years ago

Complete the table shown to the right for the population

Bas_tet [7]

Answer:

2,1

Step-by-step explanation:

5 0

2 years ago

Need help with math problem if do get 5 star

11Alexandr11 [23.1K]

<h3><u>Answer</u><u>:</u></h3>

The point ( 22 , 23 ) lies in Ist quadrant

<h3><u>Explanation</u><u>:</u></h3>

The intersection of x and y axis divides the coordinate plane into 4 sections. These four sections are called quarrants. These quadrants are named as Roman numerals I, II, III and IV quadrant. The start with the top right corner and move in anti clockwise direction .

In a x y plane , both the values of x and y are positive in Ist quadrant

3 0

2 years ago

1. Divide 6x^4 + 2x^3 – 6x^2 – 14x – 1 by 3x + 1 by using synthetic division or long division. Show all

iris [78.8K]

      2x^3 - 2x - 4

3x + 1 6x^4 + 2x^3 - 6x^2 - 14x - 1
      6x^4 + 2x^3

       -6x^2 - 14x - 1
   -6x^2 - 2x

   -12x - 1
   -12x - 4

   3

3x + 1 is not a factor of the dividend because, dividing the dividend with 3x + 1 gives a remainder.

5 0

3 years ago