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Harrizon [31]
3 years ago
5

Which is the best unit for measuring the weight of an elephant?

Mathematics
1 answer:
OLEGan [10]3 years ago
8 0
The best unit of measuring when weighing an elephant would be

b. tons
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Please answer this question, I will mark brainlest
adoni [48]

Answer:

B: 12

Step-by-step explanation:

length of one side =150

length of one path =12.5

Number of patches= 150 / 12.5

Number of patches = 12

hope it helps

please mark me as brainliest

5 0
2 years ago
Read 2 more answers
Confusion...help please
kifflom [539]

Given:

In triangle KLM, KL = 123 cm and measure of angle K is 35 degrees.

To find:

The length of the side KM to the nearest tenth of a centimeter.

Solution:

In a right angle triangle,

\cos \theta =\dfrac{Base}{Hypotenuse}

In the given right triangle KLM,

\cos K=\dfrac{KM}{KL}

\cos (35^\circ)=\dfrac{KM}{123}

0.819152=\dfrac{KM}{123}

Multiply both sides by 123.

0.819152\times 123=KM

100.755696=KM

KM\approx 100.8

The measure of side KM is 100.8 cm.

Therefore, the correct option is (2).

8 0
2 years ago
For ΔABC, ∠A = 2x - 2, ∠B = 2x + 2, and ∠C = 5x. If ΔABC undergoes a dilation by a scale factor of 1 2 to create ΔA'B'C' with ∠A
Zolol [24]
<h3>Answer:</h3>

A) ∠A = ∠A' = 38° and ∠B = ∠B' = 42°

<h3>Explanation:</h3>

The sum of angles in ∆ABC is 180°, so ...

... (2x -2) + (2x +2) + (5x) = 180

... 9x = 180

... x = 20

and the angles of ∆ABC are ∠A = 38°, ∠B = 42°, ∠C = 100°.

___

The sum of angles of ∆A'B'C' is 180°, so ...

... (58 -x) +(3x -18) +(120 -x) = 180

... x +160 = 180

... x = 20

and ∠A' = 38°, ∠B' = 42°, ∠C' = 100°.

_____

The values of angle measures of ∆ABC match those of ∆A'B'C', so we can conclude ...

... A) ∠A = ∠A' = 38° and ∠B = ∠B' = 42°

8 0
3 years ago
a child should no longer play in the kid zone area when they reach 4'4" tall joe is 54" tall is he allowed to play in the kid zo
Andreyy89

Answer: No.

Step-by-step explanation: Joe is 4'6" tall. Convert 54" to feet by dividing by 12. 54/12=4.5. Half of a foot is 6 inches.

6 0
3 years ago
Find all solutions of the equation in the interval [0, 2pi).
natali 33 [55]

Answer:

Step-by-step explanation:

Begin by squaring both sides to get rid of the radical. Doing that gives you:

sin^2x=1-cosx

Now use the Pythagorean identity that says

sin^2x =1-cos^2x and make the replacement:

1-cos^2x=1-cosx. Now move everything over to one side of the equals sign and set it equal to 0 so you can factor:

1-cos^2x+cosx-1=0 and then simplify to

cosx-cos^2x=0

Factor out the common cos(x) to get

cosx(1-cosx)=0 and there you have your 2 trig equations:

cos(x) = 0 and 1 - cos(x) = 0

The first one is easy enough to solve. Look on the unit circle and see where, one time around, where the cos of an angle is equal to 0. That occurs at

x=\frac{\pi }{2},\frac{3\pi}{2}

The second equation simplifies to

cos(x) = 1

Again, look to the unit circle and find where the cos of an angle is equal to 1. That occurs at π only.

So, in the end, your 3 solutions are

x=\frac{\pi}{2},\pi,\frac{3\pi}{2}

8 0
3 years ago
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