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2 years ago

Anyone know the answer i don’t need the steps or the work just the answer. ty!

Mathematics

1 answer:

NeX [460]2 years ago

7 0

Answer:

= x^4 y^5 / 3

Step-by-step explanation:

I got this answer = x^4 y^5 / 3

But my answer is not found directly

And I think they deliberately set up this arrangement to make us mess up

, but in your Options I think it's ( B )

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Can someone give me the answer?

Ulleksa [173]

24 is the correct answer because the question is about triangle similarity so if you divide 15 by 3 then you get 5 which is the base of the smaller triangle.

5 0

3 years ago

PLEASE HELP ME!!! ILL GIVE THE FIRST PERSON BRAINLIEST!!!

BartSMP [9]

Problem 2

Midpoint: Think 1/2. A midpoint cuts a line segment in 1/2 (in this question). That means that the left segment = the right segment. Remember: midpoint means 1/2.

LN is given as 14.

LM is 1/2 the distance of 14

LM = 1/2 * 14

LM = 7

Problem 3

If the midpoint = the 1/2 way point, the two halves are equal. Remember a midpoint divides the 2 parts into 2 EQUAL parts.

4a - 2 = 18 Add 2 to both sides

4a = 18 + 2

4a = 20

a = 20 /4

a = 5

Problem 4

Remember that midpoint means 1/2. That a midpoint cuts a segment into 2 equal segments

Equation

2n + 2 = 5n - 4

Solve

2n + 2 = 5n - 4 Add 4 to both sides

2n + 2 + 4 = 5n Subtract 2n from both sides.

6 = 5n - 2n

6 = 3n Divide both sides by 3

6/3 = n

n = 2

<u>Answer:</u> B

Problem 5

And again the whole line segment is divided into 2 equal parts.

<u>Equation</u>

6p - 12 = 4p Add 12 to both sides

6p = 12 + 4p Subtract 4p from both sides.

6p - 4p = 12

2p = 12 Divide by 2

p = 12/2

p = 6 <<<<< Answer

6 0

3 years ago

On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or

almond37 [142]

Answer:

a) $\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$ , b) $\theta = \frac{\pi}{4}$ , c) $y_{max} = 84.375\,m$ , $t = 3.75\,s$ .

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

$\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j$

The particular expression for the cannonball is:

$\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$

b) The components of the position of the cannonball before hitting the ground is:

$x = (90\cdot \cos \theta)\cdot t$

$0 = 90\cdot \sin \theta - 6\cdot t$

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

$0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)$

$0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x$

$0 = 4050\cdot \sin 2\theta - 6\cdot x$

$6\cdot x = 4050\cdot \sin 2\theta$

$x = 675\cdot \sin 2\theta$

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

$\frac{dx}{d\theta} = 1350\cdot \cos 2\theta$

$1350\cdot \cos 2\theta = 0$

$\cos 2\theta = 0$

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

$\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta$

$\frac{d^{2}x}{d\theta^{2}} = -2700$

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

$y = 45\cdot t - 6\cdot t^{2}$

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

$\frac{dy}{dt} = 45 - 12\cdot t$

$45-12\cdot t = 0$

$t = \frac{45}{12}$

$t = 3.75\,s$

Now, the second derivative is used to check if such solution leads to a maximum:

$\frac{d^{2}y}{dt^{2}} = -12$

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

$y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}$

$y_{max} = 84.375\,m$

7 0

3 years ago

If trapezoid ABCD was reflected over the y-axis, reflected over the x-axis, and rotated 180°, where would point A' lie?

Dmitrij [34]

A is the point (-4, 1)

i) reflection over the y-axis projects a point (a, b) to (-a, b):

(-4, 1) is projected to (4, 1),

ii) reflection over the x axis projects a point (a, b) to (a, -b):

(4, 1) is projected to (4, -1),

iii) rotation 180 degrees: projects (a, b) to (-a, -b):

so (4, -1) is projected to (-4, 1)

Answer: A'(-4, 1)

8 0

2 years ago

A farmer water 3/8 of a field what percentage is equivalent to the fraction of the field the farmer wanted?

Aleks [24]

I believe the correct answer is: 37%

4 0

3 years ago