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3 years ago

Hey ASAPPP PLS 80 points

Mathematics

2 answers:

Klio2033 [76]3 years ago

4 0

Answer:

using pythagoras theorem

a^2+b^2=c^2

8^2+x^2=(x+2)^2

64+x^2=x^2+2*x*2+2^2

64+x^2=x^2+4x+4(here x^2 of left hand side and x^2 of right hand side gets cancel)

64=4x+4

64-4=4x

60/4=x

15=x

Step-by-step explanation:

alexdok [17]3 years ago

4 0

Answer is up here, give them brainliest they did a great job ^^

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I have been trying so hard to find this and I cant

vladimir2022 [97]

Answer:

the answer is 46cm

Step-by-step explanation:

5 0

3 years ago

Which set of points contains the solutions to the inequality y ≤ 8x – 3? A. {(–3,–17), (4,11), (7,19)} B. {(3,22), (2,3), (8,27)

Nady [450]

Just to clarify, here are the answer choices:

A. {(–3,–17), (4,11), (7,19)}

B. {(3,22), (2,3), (8,27)}

C. {(4,29), (–6,–58), (7,19)}

D. {(–2,–18), (4,37), (5,15)}

And all of the points in the set must satisfy y ≤ 8x – 3.

The only way is to go through each answer choice, eliminating each that is wrong.

A) -17 ≤ -3*8-3

-17 ≤ -21 ✘

B) 22 ≤ 3*8-3

22 ≤ 21 ✘

C) a) 29 ≤ 8*4-3

29≤29 ✔

b) -58 ≤ 8*(-6)-3

-58 ≤ -49 ✔

c) 19 ≤ 7*8-3

19 ≤ 53 ✔

It becomes clear that the answer is C. But just to make sure, we must check D to make sure C is really the answer.

D) -18 ≤ 8(-2)-3

-18 ≤ -19 ✘

That means that C is the answer.

8 0

3 years ago

An airplane leaves an airport at 9:00 p.M. With a heading of 270 and a speed of 610 mph. At 10:00 p.M. The pilot changes the he

timama [110]

Answer:

2330.51 miles

Step-by-step explanation:

Given that the speed of the airplane = 610 mph.

The airplane leaves an airport at 9:00 P.M. with a heading of 270 degrees and at 10:00 P.M., the pilot changes the heading to 310 degrees.

So, for 1 hour the plane is heading at 270 degrees and for 3 hours, from 10:00 p.m to 1:00 a.m, the plane is heading at 310 degrees as shown in the figure.

As, distance = time x speed, so

The distance covered at 270 degrees, $d_1 = 1\times610=610$ miles.

The distance covered at 310 degrees, $d_2 = 3\times610=1830$ miles.

Total distance covered, d, is the magnitude of the sum of vectors $\vec{d_1}$ and $\vec{d_2}$ as shown in the figure.

The angle between the vectors $\vec{d_1}$ and $\vec{d_2}$ , $\theta=310-270=40$ degree.

Magnitude of sum of the vectors \vec{d_1} and \vec{d_2},

$d= \sqrt{|\vec{d_1}|^2+|\vec{d_2}|^2+2|\vec{d_1}|\;|\vec{d_2}|\cos\theta} \\\\\Rightarrow d=\sqrt{610^2+1830^2+2|\vec{d_1}|\;|\vec{d_2}|\cos\(40^{\circ})}$