Point D is located on MV .The coordinates of D are (0,−3/4).What fraction of the distance from M to V is MD?

Rasek [7]

Answer is a

*The correct answer is IJ = JK

•explanation:

Given:
AB is the perpendicular bisector of
IK. ⇒ AB divides the line segment IK in two equal parts i.e. IJ = JK and the angle formed at the point of intersection J is 90° ⇒ ∠AJI = 90°. In ΔAIJ, By angle sum property of a triangle ∠AJI + ∠AIJ + ∠IAJ = 180° ( But ∠AJI = 90° ) ∠AIJ + ∠IAJ = 90° ⇒ ∠IAJ < 90° So, ∠IAJ is not a right angle. Its not given IK is a perpendicular bisector so AJ = BJ need NOT be true. As A does not lie on the line IK so A can not be the mid point of IK.

•Hence, we conclude the correct statement is IJ = JK

6 0

3 years ago

The regions bounded by the graphs of y=x2 and y=sin2x are shaded in the figure above. What is the sum of the areas of the shaded

Alex Ar [27]

Answer:

The sum of the area of the shaded regions = 0.248685

Explanation:

The sum of the area of the shaded region is given as follows;

The point of intersection of the graphs are;

y = x/2

y = sin²x

∴ At the intersection, x/2 = sin²x

sinx = √(x/2)

Using Microsoft Excel, or Wolfram Alpha, we have that the possible solutions to the above equation are;

x = 0, x ≈ 0.55 or x ≈ 1.85

The area under the line y = x/2, between the points x = 0 and x ≈ 0.55, A₁, is given as follows

1/2 × (0.55)×0.55/2 ≈ 0.075625

The area under the line y = sin²x, between the points x = 0 and x ≈ 0.55, A₂, is given using as follows;

$\int\limits {sin^n(x)} \, dx = -\dfrac{1}{n} sin^{n-1}(x) \cdot cos(x) + \dfrac{n-1}{n} \int\limits {sin^{n-2}(x)} \, dx$

Therefore;

$A_2 = \int\limits^{0.55}_0 {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_0 ^{0.55}$

∴ A₂ =1/2 × ((0.55 - sin(0.55)×cos(0.55)) - (0 - sin(0)×cos(0)) ≈ 0.0522

The shaded area, $A_{1 shaded}$ = A₁ - A₂ = 0.075625 - 0.0522 ≈ 0.023425

Similarly, we have, between points 0.55 and 1.85

A₃ = 1/2 × (1.85 - 0.55) × 1/2 × (1.85 - 0.55) + (1.85 - 0.55) × 0.55/2 = 0.78

For y = sin²x, we have;

$A_4 = \int\limits^{1.85}_{0.55} {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_{0.55} ^{1.85} \approx 1.00526$

The shaded area, $A_{2 shaded}$ = A₄ - A₃ = 1.00526 - 0.78 ≈ 0.22526

The sum of the area of the shaded regions, ∑A = $A_{1 shaded}$ + $A_{2 shaded}$

∴ A = 0.023425 + 0.22526 = 0.248685

The sum of the area of the shaded regions, ∑A = 0.248685

8 0

2 years ago

PLEASE SAY IF THEY ARE VALID OR INVALID! WILL MARK BRAINLIEST IF CORRECT ANSWERS!

viktelen [127]

Answer:

a. Valid

b. Invalid

c. Valid

Explanation:

Follow me and Mark me brainliest please

6 0

3 years ago

Santa's Workshop has three elves who coach all of the other elves, and each one is proficient as a master elf in every category

lutik1710 [3]

Answer:

what!!! that is IMPOSSIBLE!!!!!! yoy should ask a tutor

7 0

2 years ago

What allowed Japan’s population to grow

tatyana61 [14]

Answer:

This increase was directly related to slow but steady urban growth; the development of Hokkaido, Tōhoku, and southern Kyushu; and the introduction of commercial agriculture. In 1897, when industrialization first began, the population numbered more than 42 million.

Explanation:

5 0

2 years ago

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