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Ber [7]
3 years ago
15

In the triangle below, which of the following best describes DH ?

Mathematics
2 answers:
hodyreva [135]3 years ago
7 0

Answer:

B an angle bisector

jeka57 [31]3 years ago
6 0

Answer:

B. Angle bisector.

Step-by-step explanation:

The answer is angle bisector because DH is dividing/separating an angle.

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The answer is 9x^3-6x^2+3x
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A scientist is studying certain germs. She places 3 germs in a special solution that will help the germs grow. The number of ger
vivado [14]
3*2 is 6 6*2 is 12 12*2 is 24 24*2 is 48 48*2 is 96 96*2 is 192 and 192*2 is 384
7 0
3 years ago
Can someone help me. Please
Lerok [7]

Step-by-step explanation:

a. the angle B in triangle ABC:

180-47=133°

angle C in ABC:

180-(40+133)=7°

using sine rule:

a/sin A= b/sin B= c/sin C

50/sin 7= BC/sin 40

BC= (50*sin40)/ sin7

=263.7 m

b.sin 47= opposite/hypotenuse

sin 47= CD/263.7

CD= sin 47*263.7

= 192.9

6 0
3 years ago
Jake is a factory manager for a company that produces batteries. Occasionally, some are found to be defective. Based on previous
Aleksandr [31]

Based on the estimated number of defective batteries and the number of batteries that were actually defective, the percent error of Jake's estimate is 20%

<h3>How to find the percent error?</h3>

The percent error of a problem can be found by the formula:
= (Expected results - Actual results) / 100%

Expected results = 12 defective batteries

Actual results = 10 defective batteries

Jake's percentage error is therefore:

= (12 - 10) / 100%

= 2 / 10 x 100%

= 0.2 x 100%
= 20%

In conclusion, the percent error of Jake's estimate is 20%.

Find out more on percent error at brainly.com/question/5493941

#SPJ1

7 0
1 year ago
What is the repulsive force between two pith balls that are 13.0 cm apart and have equal charges of −24.0 nC?
daser333 [38]

Answer:

The repulsive force is 3.067\times10^{-4}N.

Step-by-step explanation:

Consider the provided information.

The coulomb's law to calculate the repulsive force: F=\frac{kQ_1Q_2}{r^2}

Where the value of k is 9.00×10⁹ Nm²/C²

Substitute the respective values in the above formula.

F=\frac{9\times10^9\frac{N\cdot m^2}{c^2} \times(-24\times10^{-9}C)^2}{[(13 cm)(\frac{1m}{100cm} )]^2}

F=\frac{9\times10^9\frac{N\cdot m^2}{c^2} \times(-24\times10^{-9}C)^2}{(0.13 m)^2}

F\approx0.0003067N

F=3.067\times10^{-4}N

Hence, the repulsive force is 3.067\times10^{-4}N.

8 0
3 years ago
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