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3 years ago

I need help solving these equations....

Mathematics

1 answer:

Harlamova29_29 [7]3 years ago

7 0

6 is easy
7. never divide by a fraction, flip ot upside down and multiply, so multiply by 5 on both sides
8. subtract 3.45 on both sides
9. the answer is just 1

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Please could you find the answers to the questions in the attachment.

Fudgin [204]

( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ )we need 3 equations
1. midpoint equation which is ( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ ) when you have 2 points

2. distance formula which is D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height

so

x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=( $\frac{11+19}{2} , \frac{10+6}{2}$ )
midpoint=( $\frac{30}{2} , \frac{16}{2}$ )
midpoint= (15,8)

point x=(15,8)

y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=( $\frac{5+21}{2} , \frac{8+0}{2}$ )
midpoint=( $\frac{26}{2} , \frac{8}{2}$ )
midpoint=(13,4)

Y=(13,4)

legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= $\sqrt{(19-11)^{2}+(6-10)^{2}}$
D= $\sqrt{(8)^{2}+(-4)^{2}}$
D= $\sqrt{64+16}$
D= $\sqrt{80}$
D= $4 \sqrt{5}$
BC= $4 \sqrt{5}$

X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= $\sqrt{(13-15)^{2}+(4-8)^{2}}$
D= $\sqrt{(-2)^{2}+(-4)^{2}}$
D= $\sqrt{4+16}$
D= $\sqrt{20}$
D= $2 \sqrt{5}$
XY= $2 \sqrt{5}$

the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC= $4 \sqrt{5}$ and XY= $2 \sqrt{5}$

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= $\sqrt{(21-5)^{2}+(0-8)^{2}}$
D= $\sqrt{(16)^{2}+(-8)^{2}}$
D= $\sqrt{256+64}$
D= $\sqrt{320}$
D= $4 \sqrt{2}$
AD= $4 \sqrt{2}$

so we have
AD= $4 \sqrt{2}$
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
( $4 \sqrt{2}+4 \sqrt{5}$ ) times 1/2 times $2 \sqrt{5}$ =
( $4 \sqrt{2}+4 \sqrt{5}$ ) times \sqrt{5} [/tex] =
$4 \sqrt{10}+4*5$ = $4 \sqrt{10}+20$ =80 $\sqrt{10}$ =252.982

X=(15,8)
Y=(13,4)
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$
Area=80 $\sqrt{10}$ square unit or 252.982 square units

7 0

3 years ago

MUST ANSWER ALL PARTS!!!!

dsp73

Hello there,

The outlier of the data set is 92. Lets first find everything with the outlier.
Mean: 60.08
Median: 57
Mode: 52
Range:46

Now let's take out the outlier of the set and see the difference.
Mean: 54.27
Median:56
Mode: 52
Range: 15

So as we can see without the outlier everything is closer together. The outlier shouldn't be included because that makes the entire set screwed up. The data is more accurate without the outlier.

Hope this helps!

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3 years ago

Every orthogonal matrix is diagonalizable. O True O False

Simora [160]

Answer:

FALSE

Step-by-step explanation:

the correct answer is FALSE

A matrix is said to be orthogonal when it is multiplied by its own transpose then the matrix comes out to be Identity.

$A^T.A = I$

the basic condition for a matrix to be diagonalizable is that the matrix should be symmetric.

but an orthogonal matrix is not necessarily symmetric so, the statement is false.

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kiruha [24]

It simple ☺ idk how to help u. hope this ll help u understand. place ur pencil on B's dot n make it move 3 units to the right then 2 units to the up. same goes to A n C. u ll get the ans.

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