All categories

3 years ago

#15 only please. thanks.

Mathematics

1 answer:

Sonbull [250]3 years ago

7 0

Easy,

plotting it on a graph...

(-2,3) (4,3) (4,-4) (-2,-4)

Now, if done correctly, when you count the "points" it should be lengthwise

6 units, and width would be 7

6*7=42

The Area is 42 units squared.

You might be interested in

Calculate the protein intake of Miss Exercise who weighs 112 lb. Remember: (0.45)(w-lb) = (w-kg) and the recommended protein in

gogolik [260]

50.4 is your answer there for it would be 50.4

5 0

3 years ago

Read 2 more answers

Make n the subject of the formula: M=3n

natta225 [31]

Divide each side by 3. ----- n=M/3 .

4 0

3 years ago

What is eight divided by two sevenths

Vilka [71]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Help me solve (b) in this quadrilateral

Savatey [412]

Answer:

b = 87°

Step-by-step explanation:

In order to answer this question, we need to utilise an important angle fact which is <em>angles in a quadrilateral add up to 360° </em>

Using the information we can set up an equation to find the value of b

→ Form equation

63 + 140 + 70 + b = 360

→ Simplify

273 + b = 360

→ Minus 273 from both sides isolate b

b = 87°

3 0

2 years ago

Read 2 more answers

Solve the following recurrence relation: <br> <img src="https://tex.z-dn.net/?f=A_%7Bn%7D%3Da_%7Bn-1%7D%2Bn%3B%20a_%7B1%7D%20%3D

-Dominant- [34]

By iteratively substituting, we have

$a_n = a_{n-1} + n$

$a_{n-1} = a_{n-2} + (n - 1) \implies a_n = a_{n-2} + n + (n - 1)$

$a_{n-2} = a_{n-3} + (n - 2) \implies a_n = a_{n-3} + n + (n - 1) + (n - 2)$

and the pattern continues down to the first term $a_1=0$ ,

$a_n = a_{n - (n - 1)} + n + (n - 1) + (n - 2) + \cdots + (n - (n - 2))$

$\implies a_n = a_1 + \displaystyle \sum_{k=0}^{n-2} (n - k)$

$\implies a_n = \displaystyle n \sum_{k=0}^{n-2} 1 - \sum_{k=0}^{n-2} k$

Recall the formulas

$\displaystyle \sum_{n=1}^N 1 = N$

$\displaystyle \sum_{n=1}^N n = \frac{N(N+1)}2$

It follows that

$a_n = n (n - 2) - \dfrac{(n-2)(n-1)}2$

$\implies a_n = \dfrac12 n^2 + \dfrac12 n - 1$

$\implies \boxed{a_n = \dfrac{(n+2)(n-1)}2}$

4 0

2 years ago