rhombus and a square have one and the same side of 6 cm. The area of the rhombus is 4/5 of the area of the square. Find the height of the rhombus.

Answers

Answer 1

Answer: Consider this option:
1. area_rombus=a*h, where a=6 - the length of the side, h - height.
h=area_rombus/a.
2. area_sq=a², where a=6 - the length of the square.
area_sq=36, area_rombus=4/5 *36=28.8.
3. according to the item 1 h=area_rombus/a=28.8/6=4.8.

answer: 4.8

Answer 2

Answer:

Answer:

4.8 cm

Step-by-step explanation:

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Yukio says the scale from DEF to ABC is 3:1. Is Yukio Correct? Explain

Answers

Answer: Yes, Yukio is correct.

Step-by-step explanation:

Assuming that Triangle DEF and ABC have the same angles (they do because they are right-angled), we can take the length from the larger triangle (DEF) and divide it by the length of the smaller triangle (ABC).

Length of DEF = 6cm

Length of ABC = 2cm

= 6/2

= 3

Proves that scale of DEF to ABC is 3:1

A regular hexagon is shown below. Find the value of x.
(11x + 21)°

Answers

The value of {x} for the hexagon shown is 9.

What is Hexagon?

A hexagon is a polygon with 6 equal sides. The perimeter of a hexagon is the sum of all the six lengths. The measure of the interior angle of a hexagon is 120 degrees.

Given is to find the value of {x} in the Hexagon shown below.

The measure of the internal angle of a hexagon is 120 degrees. So, we can write -

(11x + 21) = 120

11x = 120 - 21

11x = 99

x = 9

Therefore, the value of {x} for the hexagon shown is 9.

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Answer:

x=9

Step-by-step explanation:

The expression is for the interior angle of the hexagon; one interior angle is equal to .

Since an interior angle of a hexagon measures 120°, we have the equality

Now it is just a matter of solving for

11x=120-21

11x=99

x=99/11

x=9

a.Find a linear approximation of 4√14 . b. Justify visually whether your approximation is an over- or underestimate.

Answers

Answer:

Step-by-step explanation:

Let f(x) be the function

$\bf f(x)=√(x)$

A linear approximation of f is the Taylor polynomial of degree one:

$\bf f(x)\approx f(a)+f'(a)(x-a)$

Taking a = 16, and given that

$\bf f'(x)=\displaystyle(1)/(2√(x))$

we get

$\bf f(x)\approx f(16)+f'(16)(x-16)=√(16)+\displaystyle(1)/(2√(16))(x-16)=4+\displaystyle(x-16)/(8)$

$\bf √(14)=f(14)\approx 4+\displaystyle(14-16)/(8)=3.75\Rightarrow\n\n\Rightarrow 4√(14)\approx 4(3.75)=15$

Since 16 > 14, we can deduce that this is an overestimate.

The number of minutes that Samantha waits to catch the bus is uniformly distributed between 0 and 15 minutes. What is the probability that Samantha has to wait less than 4.5 minutes to catch the bus? a. 10%
b. 3%
c. 20%
d. 30%

Answers

Answer:

option d. 30%

Step-by-step explanation:

Data provided in the question:

$a=0$

$b=15$

Now,

The probability density function of the uniform distribution is given as:

$f(x)=(1)/(15-0)$

$f(x)=(1)/(15), 0<x<15$

Therefore,

The required probability will be

$P(X<4.5)=\int_(0)^(4.5) f(x) d x$

$P(X<4.5)=\int_(0)^(4.5) (1)/(15) d x$

$P(X<4.5)=\left((x)/(15)\right)_(0)^(4.5)$

$P(X<4.5)=\left((4.5)/(15)-(0)/(15)\right)$

$P(X<4.5)=0.3$

= 0.3 × 100%

= 30%

Hence,

The answer is option d. 30%

Final answer:

In a uniform distribution, all values have an equal chance of occurring. To find the probability of Samantha waiting less than 4.5 minutes, you divide 4.5 by the total time interval of 15 minutes. The answer is 30%.

Explanation:

The question is about the concept of uniform distribution in probability, which means that each value within the given range has an equal chance of being drawn. In this case, Samantha can wait anywhere between 0 and 15 minutes, so each minute has an equally likely chance of being the waiting time. To find the probability that Samantha has to wait less than 4.5 minutes, we simply divide this time interval by the total time interval. Hence, the calculation is 4.5 / 15 = 0.30 or 30%. So, the answer to the question of the probability that Samantha has to wait less than 4.5 minutes to catch the bus is 30%.

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Given the force field F, find the work required to move an object on the given oriented curve. F = (y, - x) on the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9) The amount of work required is ____ (Simplify your answer.)

Answers

Answer:

Step-by-step explanation:

We want to compute the curve integral (or line integral)

$\bf \int_(C)F$

where the force field F is defined by

F(x,y) = (y, -x)

and C is the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9).

We can write

C = $\bf C_1+C_2$

where

$\bf C_1$ = line segment from (1, 5) to (0, 0)

$\bf C_2$ = line segment from (0, 0) to (0, 9)

so,

$\bf \int_(C)F=\int_(C_1)F+\int_(C_2)F$

Given 2 points P, Q in the plane, we can parameterize the line segment joining P and Q with

r(t) = tQ + (1-t)P for 0 ≤ t ≤ 1

Hence $\bf C_1$ can be parameterized as

$\bf r_1(t) = (1-t, 5-5t)$ for 0 ≤ t ≤ 1

and $\bf C_2$ can be parameterized as

$\bf r_2(t) = (0, 9t)$ for 0 ≤ t ≤ 1

The derivatives are

$\bf r_1'(t) = (-1, -5)$

$\bf r_2'(t) = (0, 9)$

and

$\bf \int_(C_1)F=\int_(0)^(1)F(r_1(t))\circ r_1'(t)dt=\int_(0)^(1)(5-5t,t-1)\circ (-1,-5)dt=0$

$\bf \int_(C_2)F=\int_(0)^(1)F(r_2(t))\circ r_2'(t)dt=\int_(0)^(1)(9t,0)\circ (0,-9)dt=0$

In consequence,

$\bf \int_(C)F=0$

Final answer:

The work done by the force field F = (y, -x) along the given path is -5 Joules. This was calculated by breaking the path into two segments and calculating the work done for each segment.

Explanation:

To calculate the work done by the force field F = (y, -x) when moving an object along a specific path, we utilize the concept of the line integral or the dot product of the force and the displacement vector. We can break down the given path into two line segments and solve each separately.

The first segment is from (1, 5) to (0, 0). Only the x component of the displacement is non-zero here, the force is F = (5, -1). Thus the work done on this segment is given by W = F.d = Fd cos θ = -(5 N)(1 m)(cos(180)) = -5 J, where J stands for Joules, the unit of work or energy.

The second segment is from (0, 0) to (0, 9). The force and displacement are perpendicular so the work done is 0.

By adding the work done on these two segments, we arrive at the total work done: W_total = -5 J + 0 J = -5 J