Find the area, in square units, of ABC plotted below.
A(0,7)
B(7,-2)
D(2, -3)
C(-3,-4)

Find The Area, In Square Units, Of ABC Plotted Below.A(0,7)B(7,-2)D(2, -3)C(-3,-4)

Answers

Answer 1
I used a different methods :use the 3 vertices points that was around the triangle , (I WOULD NOT use point D )
So A=52 square units
And the brackets was absolute value
Find The Area, In Square Units, Of ABC Plotted Below.A(0,7)B(7,-2)D(2, -3)C(-3,-4)
Answer 2

The area of the triangle ABC for the considered triangle plotted in the considered image is 66.5 sq. units approximately.

What is the distance between two points ( p,q) and (x,y)?

The shortest distance(length of the straight line segment's length connecting both given points) between points ( p,q) and (x,y) is:

[tex]D = \sqrt{(x-p)^2 + (y-q)^2} \: \rm units.[/tex]

The coordinates of the points A, B, C, and D in the given figure are:

A(0, 7)B( 7,-2)C(-3, -4)D(2, -3)

Finding the length of the line segments AD and CB, which will be the distance between A and D, and C and B respectively.

Thus, we get:

Length of line segment AD = |AD| = distance between A and D = [tex]\sqrt{(0-7)^2 + (7-(-2))^2} = \sqrt{7^2 + 9^2} = \sqrt{130} \: \rm units.[/tex]

Similarly, we get:
|CB| = [tex]\sqrt{(-3-7)^2 + (-4-(-2))^2} = \sqrt{10^2 + 6^2} = \sqrt{136} \: \rm units.[/tex]

If we take CB as the base of ABC triangle, then as AD is perpendicular on CB, and touches the peak of the triangle ABC from its base, so AD is height of the triangle.

Thus, as we know know that:

Height's measurement of ABC = |AD|= [tex]\sqrt{130} \: \rm units[/tex]Base length of ABC = |BC| = [tex]\sqrt{136} \: \rm units[/tex]

Thus, the area of the triangle ABC is:

[tex]A = \dfrac{base \times height}{2} = \dfrac{\sqrt{130} \times \sqrt{136}}{2} \approx 66.5 \: \rm unit^2[/tex]

Thus, the area of the triangle ABC for the considered triangle plotted in the considered image is 66.5 sq. units approximately.

Learn more about distance between two points here:

brainly.com/question/16410393

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Answer:

z1(z2 + z3) = z1 z2 + z1 z3 ⇒ verified down

Step-by-step explanation:

∵ z1 = 4 + 3i

∵ z2 = 3 - 2i

∵ z3 = i + 5

Find the left side

∵ z1(z2 + z3) = (4 + 3i)[3 - 2i + i + 5]

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→ Remember that i² = -1

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→ Add the like terms

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z1(z2 + z3) = 35 + 20i

Find the right side

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∴ z1 z2 + z1 z3 = [(4)(3)+(4)(-2i)+(3i)(3)+(3i)(-2i)] + [(4)(i)+(4)(5)+(3i)(i)+(3i)(5)]

∴ z1 z2 + z1 z3 = [12 -8i + 9i -6i²] + [4i + 20 + 3i² + 15i]

→ Replace i² by -1

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∴ z1 z2 + z1 z3 = [12 -8i + 9i + 6] + [4i + 20 - 3 + 15i]

→ Add the like terms

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∵ The right side = 35 + 20i

∴ The left side = the right side

z1(z2 + z3) = z1 z2 + z1 z3

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Answer:

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Step-by-step explanation:

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G
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Answer:

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Step-by-step explanation:

They are not vertical angles this is what vertical angles look like ><

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Step-by-step explanation:

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

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7, 0.8, 1/2, -0.8, -1/10, -2, -3,

Step-by-step explanation:

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Step-by-step explanation:

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Which statement describes the inverse of m(x) = x^2 – 17x?


a. The domain restriction x ≥ StartFraction 17 Over 2 EndFraction results in m–1(x) =StartFraction 17 Over 2 EndFraction minus StartRoot x + StartFraction 289 Over 4 EndFraction EndRoot .


b. The domain restriction x ≥ StartFraction 17 Over 2 EndFraction results in m–1(x) =StartFraction 17 Over 2 EndFraction + StartRoot x + StartFraction 289 Over 4 EndFraction EndRoot .


c. The domain restriction x ≥ Negative StartFraction 17 Over 2 EndFraction results in m–1(x) =StartFraction 17 Over 2 EndFraction minus StartRoot x + StartFraction 289 Over 4 EndFraction EndRoot .


d.The domain restriction x ≥ Negative StartFraction 17 Over 2 EndFraction results in m–1(x) =StartFraction 17 Over 2 EndFraction + StartRoot x + StartFraction 289 Over 4 EndFraction EndRoot .

Answers

Given:

The function is

[tex]m(x)=x^2-17x[/tex]

To find:

The inverse of the given function.

Solution:

We have,

[tex]m(x)=x^2-17x[/tex]

Substitute m(x)=y.

[tex]y=x^2-17x[/tex]

Interchange x and y.

[tex]x=y^2-17y[/tex]

Add square of half of coefficient of y , i.e., [tex]\left(\dfrac{-17}{2}\right)^2[/tex] on both sides,

[tex]x+\left(\dfrac{-17}{2}\right)^2=y^2-17y+\left(\dfrac{-17}{2}\right)^2[/tex]

[tex]x+\left(\dfrac{17}{2}\right)^2=y^2-17y+\left(\dfrac{17}{2}\right)^2[/tex]

[tex]x+\left(\dfrac{17}{2}\right)^2=\left(y-\dfrac{17}{2}\right)^2[/tex]        [tex][\because (a-b)^2=a^2-2ab+b^2][/tex]

Taking square root on both sides.

[tex]\sqrt{x+\left(\dfrac{17}{2}\right)^2}=y-\dfrac{17}{2}[/tex]

Add [tex]\dfrac{17}{2}[/tex] on both sides.

[tex]\sqrt{x+\left(\dfrac{17}{2}\right)^2}+\dfrac{17}{2}=y[/tex]

Substitute [tex]y=m^{-1}(x)[/tex].

[tex]m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}[/tex]

We know that, negative term inside the root is not real number. So,

[tex]x+\left(\dfrac{17}{2}\right)^2\geq 0[/tex]

[tex]x\geq -\left(\dfrac{17}{2}\right)^2[/tex]

Therefore, the restricted domain is [tex]x\geq -\left(\dfrac{17}{2}\right)^2[/tex] and the inverse function is [tex]m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}[/tex].

Hence, option D is correct.

Note: In all the options square of [tex]\dfrac{17}{2}[/tex] is missing in restricted domain.

Answer:

C on edu

Step-by-step explanation:

Luca had a large container of yogurt with 1 and StartFraction 5 Over 8 EndFraction pounds of yogurt left in it. If a serving of yogurt is StartFraction 3 Over 8 EndFraction of a pound, to the nearest whole serving, how many servings of yogurt are in the container?

Answers

Answer:There are 4 serving in the container.

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

Took Edge 2020 Fraction Multiplication and Division Test

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