All categories

3 years ago

51 please and thank you!

Mathematics

1 answer:

SpyIntel [72]3 years ago

3 0

Check the picture below.

Bearing in mind that above sea level units are positive and below it are negative.

You might be interested in

Expresion 2 5+4X2+6-2X2-1 Insert parentheses in Expression 2 so that it has a value of 19. Then show why your expression has a v

solong [7]

(5+4+6-2)×2×2-1
(9-4)×2×2-1
5×2×2--1
10×2-1
20-1
19

3 0

3 years ago

Reflections DO NOT preserve __________. This means the preimage (original) and the image(new) are NOT FACING the same way.

Sonbull [250]

It c okokokokokokokokokok

5 0

3 years ago

There are 7 animals a farmer's field.

sergey [27]

Answer:

2 cows and 5 chickens.

Step-by-step explanation:

NOTE: This will be solved using simultaneous equations.

Let x = cows and y = chickens

1. form two equations

4x + 2y = 18 ... (1)

x + y = 7 .... (2)

2. make y (or x) the subject of one equation

x + y = 7

y = 7 - x ... (3)

3. Substitute (3) into (1)

4x + 2( 7 - x) = 18

4. solve for x

4x + 14 - 2x = 18

2x = 4

x = 2

5. Now substitute x value into equation (2)

x + y = 7

2 + y = 7

y = 5

∴ cows (x) = 2

chickens (y) = 5

6 0

3 years ago

A bedroom door in the house has the same dimension as the front door but the length is 30 inches rather than 36 inches how much

arsen [322]

Answer:

6/5

Step-by-step explanation:

Calculate the volume by multiplying the measured length and width of the space together, then multiply the result by the height of the room. From the example, 10 * 25 feet = 250 square feet, and 5 * 10 feet = 50 square feet.

5 0

2 years ago

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

3 years ago