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2 years ago

Find the area of quadrilateral with vertices at points (3,0)(4,5)(-1,4)(-2,-1)

Mathematics

2 answers:

marissa [1.9K]2 years ago

6 0

Step-by-step explanation:

$Let the points be A(3,0),B(4,5),C(−1,4) and D(−2,−1)Area of a rhombus =21×(diagonal1×diagonal2)Hence, the distance between the points A(3,0) and C(−1,4)=(−1−3)2+(4−0)2=16+16=32And, Distance between the points B(4,5) and D(−2,−1)BD=(−2−4)2+ (−1−5)2=36+36=72So, Area of the rhombus =21×32×72=21×32×72=21×4×8×36×2=21×2×68×2=62×4×2=6×2×2=24 sq units$

guapka [62]2 years ago

4 0

Answer: 20

Step-by-step explanation:

Let 's complete our quadrilateral to a rectangle ; and subtract from it the area highlighted in red

$\displaystyle \rm S_{\diamond}=6\cdot 5 =30 \\\\ S_{ABCD}=30-(5:2+5:2+5:2+3:2+1\cdot1)=30-(7,5+1,5+1)=\\\\30-10=20 \\\\ S_{ABCD}=20$

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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

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

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

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

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

<u><em>Solved</em></u>

6 0

2 years ago

What is the equation of a line that is perpendicular to −x+2y=4 and passes through the point (−2, 1) ? Enter your equation in th

ANTONII [103]

Perpendicular lines have slopes that multiply to get -1
y=mx+b is the slope intercept equation
m=slope

so get into y=mx+b form
-x+2y=4
solve for y
add x both sides
2y=x+4
divide by 2 both sides
y=(1/2)x+2
the slope is 1/2

perppendicular line slope multiplies to -1
1/2 times what=-1
times 2/1 both sides
what=-2/1=-2

y=-2x+b
find b
we are given a point is (-2,1)
when x=-2, y=1
1=-2(-2)+b
1=4+b
minus 3 both sides
-3=b

so the equation is
y=-2x-3

your teacher might want it in standard form (ax+by=c) so
add 2x both sides
2x+y=-3 is an equation

so y=-2x-3 is correct (slope intercept form)
2x+y=-3 is also correct (standard form)

6 0

2 years ago

Write the quadratic equation in standard form. (x - 3) 2 = 0

Pani-rosa [81]

2 x -6 equals 0. add 6 to both sides. divide both sides by two. x equals 3

8 0

2 years ago

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Joe runs 9.25 times around a track in 1, 125.803 seconds. If one lap around the track is 502.3 meters,

musickatia [10]

Step-by-step explanation:

Speed = Distance÷Time