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3 years ago

What is the area of a parallelogram whose vertices are A(−12, 2) , B(6, 2) , C(−2, −3) , and D(−20, −3) ?

2 answers:

7 0

I am finding the answer right now but there is a website that you can use for math that really helps. It's called Desmos.

lukranit [14]3 years ago

5 0

These have two sides with constant y values, i.e. two sides parallel to the x axis. We'll check they're the same length

AB=6 - -12 = 18

CD = -2 - -20 = 18 good

That's the base. The height is 2 - -3 = 5 so an area of 18(5)=90

Answer: 90

Even if we didn't know it was a nicely oriented parallelogram we can get the area with the shoelace formula, which says it's half the absolute value of the sum of the cross products of each side.

A(−12, 2) , B(6, 2) , C(−2, −3) , D(−20, −3)

B(6, 2) , C(−2, −3) , D(−20, −3) , A(−12, 2)

A = (1/2) | (-12)(2) - 2(6) + 6(-3) - 2(-2) + (-2)(-3) - (-3)(-20) + (-20)(2) - (-3)(-12) | = |-180|/2 = 90

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If h(x)= (f o g)(x) and h(x)= 4 √x+7, find g(x) if f(x) = 4 √x+1

stepladder [879]

(f o g)(x)=f(g(x))

so

h(x)=f(g(x))
$4\sqrt{x+7}=4\sqrt{g(x)+1}$
hmm, see the differences

x+7=g(x)+1

x+6=g(x)
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3 years ago

1 - 1/3, 2 - 4/6, 1 - 3/9, 2 - 1/2, please answer them

marusya05 [52]

#1 - 2/3
#2 - 1 1/3
#3 - 2/3
#4 - 1 1/2

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3 years ago

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ive been stuck on this question for a while :( i was absent for the first half of this lesson so i dont completely know what im

Deffense [45]

Answer:

(0, 9 ) and (- 4, 1 )

Step-by-step explanation:

To determine which ordered pairs lie on the graph substitute the x- coordinate of the point into the right side of the equation and compare the value obtained with the y- coordinate

(- 20, - 49 )

2x + 9 = (2 × - 20) + 9 = - 40 + 9 = - 31 ≠ - 49

(1, 10 )

2x + 9 = (2 × 1) + 9 = 2 + 9 = 11 ≠ 10

(0, 9 )

2x + 9 = (2 × 0) + 9 = 0 + 9 = 9 ← point lies on graph

(- 4, 1 )

2x + 9 = (2 × - 4) + 9 = - 8 + 9 = 1 ← point lies on graph

(- 3, 40 )

2x + 9 = ( 2 × - 3) + 9 = - 6 + 9 = 3 ≠ 40

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3 years ago

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Setler [38]

1 crawdad/shrimp i believe

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3 years ago

For each of 2 ≤ n ≤ 5, compute c(n, 0) + c(n, 2) + ⋅⋅⋅ + c(n, 2k), where 2k is largest even integer ≤ n, and compute c(n, 1) + c

Ksivusya [100]

Denote $C(n,m)=\dbinom nm$ . Then

$n=2\implies\displaystyle\binom20+\binom22=1+1=2$
$n=3\implies\displaystyle\binom30+\binom32=1+3=4$
$n=4\implies\displaystyle\binom40+\binom42+\binom44=1+6+1=8$
$n=5\implies\displaystyle\binom50+\binom52+\binom54=1+10+5=16$

In general, it would appear that

$\displaystyle\sum_{k=0}^{2\lfloor\frac n2\rfloor}\binom n{2k}=2^{n-1}$

On the other hand,

$n=2\implies\displaystyle\binom21=2$
$n=3\implies\displaystyle\binom31+\binom33=3+1=4$
$n=4\implies\displaystyle\binom41+\binom43=4+4=8$
$n=5\implies\displaystyle\binom51+\binom53+\binom55=5+10+1=16$

so that in general, we also get

$\displaystyle\sum_{k=0}^{2\lfloor\frac{n-1}2\rfloor+1}\binom n{2k+1}=2^{n-1}$

7 0

2 years ago