Find the coordinates of the midpoint of each segment formed by the given points, (2,4) and (-4,4)

fenix001 [56]

Answer:

the triangle must be assumed to be a right triangle
AB = 148.4 yards
segment AB is shorter than the curved path

Step-by-step explanation:

Part 1: The main assumption required is the one required to use the hint. The triangle must be assumed to be a right triangle.

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Part 2: The approximate distance AB is the hypotenuse of the assumed right triangle, so will satisfy the Pythagorean theorem. For distances in yards, ...

AB² = 80² +125² = 22025

AB = √22025 ≈ 148.4 . . . . . yards

The approximate length across the pond is 148.4 yards.

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Part 3: A straight line is the shortest path between two points. Segment AB is a straight line path, so the length of segment AB will be shorter than any curved path between A and B.

4 0

4 years ago

4. If the sides of a square measure 9V3 units, then find the length of the diagonal.

Nitella [24]

Answer:

Step-by-step explanation:

To find the diagonal, use Pythagorean theorem,

diagonal² = side² + side²

= (9√3)² + (9√3)²

= 9²(√3)² + 9²(√3)²

= 81*3 + 81*3

= 243 + 243

= 486

diagonal = √486 = 22.05 units

7 0

3 years ago

Find the value of x. A.75 B.25 C.35 D.105

Tcecarenko [31]

Answer:

B. 25

Step-by-step explanation:

$3x \degree = 75 \degree \\ (alternate \: \angle \: s) \\ \\ 3x = 75 \\ \\ x = \frac{75}{3} \\ \\ x = 25$

8 0

3 years ago

Pls help asap If g(x) is a linear function such that g(-3) = 2 and g(1) = -4, find g(7)

PilotLPTM [1.2K]

Answer:

G(x) = F(x) = 1/2x

Step-by-step explanation:

8 0

3 years ago

Let R be the region bounded by

loris [4]

a. The area of $R$ is given by the integral

$\displaystyle \int_1^2 (x + 6) - 7\sin\left(\dfrac{\pi x}2\right) \, dx + \int_2^{22/7} (x+6) - 7(x-2)^2 \, dx \approx 9.36$

b. Use the shell method. Revolving $R$ about the $x$ -axis generates shells with height $h=x+6-7\sin\left(\frac{\pi x}2\right)$ when $1\le x\le 2$ , and $h=x+6-7(x-2)^2$ when $2\le x\le\frac{22}7$ . With radius $r=x$ , each shell of thickness $\Delta x$ contributes a volume of $2\pi r h \Delta x$ , so that as the number of shells gets larger and their thickness gets smaller, the total sum of their volumes converges to the definite integral

$\displaystyle 2\pi \int_1^2 x \left((x + 6) - 7\sin\left(\dfrac{\pi x}2\right)\right) \, dx + 2\pi \int_2^{22/7} x\left((x+6) - 7(x-2)^2\right) \, dx \approx 129.56$

c. Use the washer method. Revolving $R$ about the $y$ -axis generates washers with outer radius $r_{\rm out} = x+6$ , and inner radius $r_{\rm in}=7\sin\left(\frac{\pi x}2\right)$ if $1\le x\le2$ or $r_{\rm in} = 7(x-2)^2$ if $2\le x\le\frac{22}7$ . With thickness $\Delta x$ , each washer has volume $\pi (r_{\rm out}^2 - r_{\rm in}^2) \Delta x$ . As more and thinner washers get involved, the total volume converges to

$\displaystyle \pi \int_1^2 (x+6)^2 - \left(7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \pi \int_2^{22/7} (x+6)^2 - \left(7(x-2)^2\right)^2 \, dx \approx 304.16$

d. The side length of each square cross section is $s=x+6 - 7\sin\left(\frac{\pi x}2\right)$ when $1\le x\le2$ , and $s=x+6-7(x-2)^2$ when $2\le x\le\frac{22}7$ . With thickness $\Delta x$ , each cross section contributes a volume of $s^2 \Delta x$ . More and thinner sections lead to a total volume of

$\displaystyle \int_1^2 \left(x+6-7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \int_2^{22/7} \left(x+6-7(x-2)^2\right) ^2\, dx \approx 56.70$

7 0

2 years ago