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Oliga [24]
1 year ago
6

I dont understand how to do this

Mathematics
2 answers:
Aneli [31]1 year ago
8 0

Answer:

x ≈ 38.61

Step-by-step explanation:

using the tangent ratio in the right triangle

tan47° = \frac{opposite}{adjacent} = \frac{x}{35} ( multiply both sides by 35 )

35 × tan47° = x , then

x ≈ 37.53 ( to 2 dec. places )

Sidana [21]1 year ago
5 0
Greeting to you! My name is Cecille and I’m here to answer that question

For more explanation, please see the attachment

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Scientific Notation is made up of two number parts. The second part is a power of base 10.
Aneli [31]

Answer:

Step-by-step explanation:

True

7 0
3 years ago
Find a formula for the nth term
Tomtit [17]

Answer:

an =a1+(n−1)d

Step-by-step explanation:

4 0
2 years ago
To run a mile, Jamal must run 4 laps around the track. His goal is to run 3 miles. Jamal has run 9 laps so far.
vodka [1.7K]
Don’t really know what your question is but 4 laps times 3 miles is 12 laps. He’s got 3 more laps to finish his goal.

Pro tip: Include the actual question
7 0
3 years ago
Read 2 more answers
What is the value of x? <br><br> Enter your answer in the box.<br><br> x =___
Zanzabum

Answer:

x = 12

Step-by-step explanation:

The diagram shows an equilateral triangle (because all angles are congruent). All sides in equilateral triangle are congruent, so

4x-10=5x-22=3x+2

Solve the equation

4x-10=5x-22\\ \\4x-10-4x=5x-22-4x\\ \\-10=x-22\\ \\-10+22=x-22+22\\ \\12=x

Check this value:

4x-10=4\cdot 12-10=48-10=38\ units \\ \\5x-22=5\cdot 12-22=60-22=38\ units\\ \\3x+2=3\cdot 12+2=36+2=38\ units

Thus, x=12

8 0
3 years ago
The total source voltage in the circuit is 6-3i V. What is the voltage at the middle source
Bezzdna [24]

Answer:

<em>The voltage at the middle source is</em> (2-4\mathbf{i})\ V

Step-by-step explanation:

<u>Voltage Sources in Series</u>

When two or more voltage sources are connected in series, the total voltage is the sum of the individual voltages of each source.

The figure shown has three voltage sources of values:

2 + 6\mathbf{i}

a + b\mathbf{i}

2 - 5\mathbf{i}

The sum of these voltages is:

V_t=4+a+(6+b-5)\mathbf{i}

Operating:

V_t=4+a+(1+b)\mathbf{i}

We know the total voltage is 6-3\mathbf{i}, thus:

4+a+(1+b)\mathbf{i}=6-3\mathbf{i}

Equating the real parts and the imaginary parts independently:

4+a=6

1+b=-3

Solving each equation:

a = 2

b = -4

The voltage at the middle source is (2-4\mathbf{i})\ V

5 0
2 years ago
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