<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B8%20%2B%20%20%5Csqrt%7B2%7D%20i%7D%7B3%20-%204i%7D%20" id="TexFormula1" title=" \

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3 years ago

frac{8 + \sqrt{2} i}{3 - 4i} " alt=" \frac{8 + \sqrt{2} i}{3 - 4i} " align="absmiddle" class="latex-formula">
please help me

Mathematics

1 answer:

Alexus [3.1K]3 years ago

5 0

Easy...
the root of 2 Is approximately 1.4 to 12343910 to the power of five to the six of the 7×3 which equals to five and then you have to divided by 3.3-40+ -1 so you had to do 8+1.4239 110 but she was to 9.1 9.423910 and then divided by -1 which to your power you get your mom God he

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6.3 divided by 2.5 HELP ME PLZZZZZZZZZZZZZZZ

Vadim26 [7]

Answer:

6.3/2.5=2.52

Step-by-step explanation:

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3 years ago

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Need step-by-step. Please answer correctly. This determines what grade I get. I will mark brainliest to the first person who ans

tatyana61 [14]

Answer:

3√3

Step-by-step explanation:

For the problem shown here, your answer 3√3 is correct.

When there is a radical by itself in the denominator, you multiply numerator and denominator by a radical that results in the product being rational. For a square root, that will usually be the same square root:

$\dfrac{9}{\sqrt{3}}=\dfrac{9}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{9\sqrt{3}}{3}=\boxed{3\sqrt{3}}$

If the problem has a sum in the denominator involving a square root, then you multiply numerator and denominator by the conjugate of that sum (the sum with the sign changed). This uses the special product "difference of squares" to eliminate the radical term.

<u>Example</u>:

$\dfrac{9}{2-\sqrt{3}}=\dfrac{9}{2-\sqrt{3}}\cdot\dfrac{2+\sqrt{3}}{2+\sqrt{3}}=\dfrac{9(2+\sqrt{3})}{2^2-(\sqrt{3})^2}=\dfrac{18+9\sqrt{3}}{4-3}\\\\=\boxed{18+9\sqrt{3}}$

It is easy to demonstrate that none of the offered choices for this problem has the same value as 9/√3.

9/√3 ≈ 5.196. Offered choices have values of about 4.798, 1.732, 6.681, 23.196 -- none even close.

Please discuss this question with your teacher.

3 0

3 years ago

Problem: A piece of farmland is a watering plot of land in a circular pattern. A sewer line needs to be installed from a new hou

Musya8 [376]

The entry and exit points of (2, 3), and (12, 6), and 200 ft. extension of the sprinkler system gives;

(1) The sewer line crosses the farmland at (6.53, 4.36), and (8.48, 4.9)

(2) The longest installable sprinkler system is approximately 172.4 feet

<h3>How can the points where the line crosses the farmland be found?</h3>

1. The slope of the sewer line is found as follows;

m = (6 - 3)/(12 - 2) = 3/10 = 0.3

The equation of the sewer line can be expressed in point and slope form as follows;

(y - 3) = 0.3×(x - 2)

y = 0.3•x - 0.6 + 3

y = 0.3•x + 2.4

The equation of the circumference of the sprinkler can be expressed as follows;

(x - 8)² + (y - 3)² = 2²

Therefore;

(x - 8)² + (0.3•x + 2.4 - 3)² = 2²

Solving gives;

x= 6.53, or x = 8.48

y = 0.3×6.53 + 2.4 = 4.36

y = 0.3×8.48 + 2.4 = 4.9

Therefore;

The sewer line crosses the farmland at (6.53, 4.36), and (8.48, 4.9)

2. When the farmland does not cross the sewer line, we have;

sewer line is tangent to circumference of farmland

Slope of radial line from center of the land is therefore;

m1 = -1/0.3

Equation of the radial line to the point the sewer line is tangent to the circumference is therefore;

y - 3 = (-1/0.3)×(x - 8)

Which gives;

y = (-1/0.3)×(x - 8) + 3

The x-coordinate is therefore;

0.3•x + 2.4 = (-1/0.3)×(x - 8) + 3

x ≈ 7.5

y = 0.3 × 7.5 + 2.4 ≈ 4.65

The longest sprinkler system is therefore;

d = √((7.5 - 8)² + (4.65 - 3)²) ≈ 1.724

Which gives;

The longest sprinkler system is 1.724 × 100 ft. ≈ 172.4 ft.

Learn more about the equation of a circle here:

brainly.com/question/10368742

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Hector needs to pour gravel over an area shaped like the parallelogram below. How many square feet will Hector cover with gravel

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Answer:

Step-by-step explanation:

16 x 12 = 192

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7/25 because you can divide everything. Y 3

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