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4 years ago

What is the area of a polygon with verticies of A(0,1) B(0,5) C(4,5) D(6,1)

1 answer:

5 0

It will help to graph it, so that you can see the shape. However, you can also just solve it by looking at the vertices. If you graph it, you'll see that the shape is a trapezoid. The two bases are 4 and 6, while the height is 5 - 1 = 4. So, we solve by using the equation for the area of a trapezoid: (6 + 4) * 4 * .5 = 20 units squared.

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If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

erma4kov [3.2K]

Answer:

$\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Step-by-step explanation:

Given

5 tuples implies that:

$n = 5$

$(h,i,j,k,m)$ implies that:

$r = 5$

Required

How many 5-tuples of integers $(h, i, j, k,m)$ are there such that $n\ge h\ge i\ge j\ge k\ge m\ge 1$

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed: This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

$^{n}C_r => ^{n + 4}C_5$

$^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}$

$^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}$

Expand the numerator

$^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Solved

6 0

3 years ago

Solve the simultaneous equation y-2x+1=0 and 4x^2+3y^2-2xy=7

Lapatulllka [165]

The first equation is $y-2x+1=0$

$y=2x-1$ (Equation 1)

The second equation is $4x^{2}+3y^{2}-2xy=7$ (Equation 2)

Putting the value of x from equation 1 in equation 2.

we get,

$4x^{2}+3(2x-1)^{2}-2x(2x-1)=7$

$4x^{2}+3(4x^{2}+1-4x)-4x^{2}+2x=7$

by simplifying the given equation,

$12x^{2}-10x-4=0$

$6x^{2}-5x-2=0$

Using discriminant formula,

$D=b^{2}-4ac$

$D=25-4 \times 6 \times -2 = 73$

Now the formula for solution 'x' of quadratic equation is given by:

$x=\frac{-b+\sqrt{D}}{2a} and x=\frac{-b-\sqrt{D}}{2a}$

$x=\frac{5+\sqrt{73}}{12} and x=\frac{5-\sqrt{73}}{12}$

Hence, these are the required solutions.

8 0

3 years ago

2p^2+10=0 Please help!

ohaa [14]

$2p^2+10=0\ \ \ |-10\\\\2p^2=-10\ \ \ |:2\\\\p^2=-5$
No real solutions, because for all real numbers the square of number is greater or equal 0.

7 0

3 years ago

Solve : 3[3[3]-2] square 2 -3[6-4] square 2 please help me

Hoochie [10]

What are the brackets for? o and im srry for asking but could u mark this "answer" brainliest?? i just need one more to lvl up Im srry ik this isnt an answer but it would mean the world to me if u did
:D

5 0

3 years ago

U need question problem 2 to solve problem 3

Sergio [31]

Answer:

You are correct

Step-by-step explanation:

Start with 1 1/2. This can be made into an improper fraction which is 3/2

Now multiply both top and bottom of 3/2 by 5

(3*5)/(2 * 5) = 15 / 10

16/10 is just slightly bigger than 15/10

3 0

3 years ago

Read 2 more answers