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2 years ago

Find the tangent of the angle in between the lines 2x+3y–5=0 and 5x=7y+3?

Mathematics

1 answer:

tino4ka555 [31]2 years ago

7 0

Answer:

-29/31

Step-by-step explanation:

We are given;

The equations;

2x+3y–5=0 and 5x=7y+3

We are required to determine the tangent of the angle between the two lines;

We need to know that;

When an equation is written in the form of, y = mx + c

Then, tan θ = m , where θ is the angle between the line and the x-axis.

Therefore, we can find the tangent of the angle between each line given and the x-axis.

2x+3y–5=0

we first write it in the form, y = mx + c

We get, y = -2/3x + 5/3

Thus, tan θ₁ = -2/3

5x=7y+3

In the form of y = mx + c

We get; y = 5/7x - 3/7

Thus, tan θ₂ = 5/7

Using the formula, θ = tan^-1((m1-m2)/(1+m1m2)) , where θ is angle between the two lines.

Thus, the tangent of the angle between the two lines will be;

tan θ = ((m1-m2)/(1+m1m2))

= ((-2/3-5/7)/(1 + (-2/3 × 5/7)))

= -29/21 ÷ 31/21

= -29/31

Thus, the tangent of the angle between the two lines is -29/31

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Which of these is NOT a terminating decimal? A) 1.88 B) 1.875 C) 0.88 D) 0.8

slamgirl [31]

Answer:

Step-by-step explanation:

8 0

2 years ago

7 - 3 = -x + 1 Solve for X.

Sati [7]

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5 0

2 years ago

Read 2 more answers

If 1m = 3.3 ft, how many cubic feet is the Kuroshio Sea tank? Volume is 9,450. Show your calculations!

Naddika [18.5K]

Answer:

339604.65 cubic feet

Step-by-step explanation:

We are given that 1 m is equal to 3.3 ft.

1 m = 3.3 ft

This means that if we find the volume of a cube of 1 m, we have its volume as:

$1^3 = 1 m^3$

In the same way, the volume of a cube of 3.3 ft (1 m) is:

$3.3^3 = 35.937 ft^3$

This means that:

$1m^3 = 35.937 ft^3$

The Kuroshio sea tank has volume of $9450 m^3$ . Its volume in cubic feet is therefore:

$1m^3 = 35.937 ft^3\\\\9450 m^3 = 9450 * 35.937 = 339604.65 ft^3$

The volume of the tank is 339604.65 cubic feet.

3 0

2 years ago

Write in simplest form: <img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B24a%5E%7B10%7Db%5E%7B6%7D%7D" id="TexFormula1" ti

Charra [1.4K]

Answer:

$2a^3b^2\sqrt[3]{3a}$

Step-by-step explanation:

Use the following rules for exponents:

$a^m*a^n=a^{m+n}\\\\\sqrt[3]{x^3}=x$

Simplify 24. Find two factors of 24, one of which should be a perfect cube:

$8*3=24\\\\2^3=8$

Insert:

$\sqrt[3]{2^3*3a^{10}b^6}$

Now split the exponents. Split 10 into as many 3's as possible:

$10=3+3+3+1$

Insert as exponents:

$\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^6}$

Split 6 into as many 3's as possible:

$6=3+3$

Insert as exponents:

$\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^3*b^3}$

Now simplify. Any terms with an exponent of 3 will be moved out of the radical (rule #2):

$2\sqrt[3]{3*a^3*a^3*a^3*a^1*b^3*b^3}\\\\\\2*a*a*a\sqrt[3]{3*a^1*b^3*b^3}\\\\\\2*a*a*a*b*b\sqrt[3]{3*a^1}$

Simplify:

$2a^3b^2\sqrt[3]{3a}$

:Done

6 0

2 years ago

Consider the continuous random variable x, which has a uniform distribution over the interval from 110 to 150. The probability t

Virty [35]

Answer:

The probability that x will take on a value between 120 and 125 is 0.14145

Step-by-step explanation:

For uniform distribution between a & b

Mean, xbar = (a + b)/2

Standard deviation, σ = √((b-a)²/12)

For 110 and 150,

Mean, xbar = (150 + 110)/2 = 130

Standard deviation, σ = √((150-110)²/12 = 11.55

To find the probability that x will take on a value between 120 and 125

We need to standardize 120 & 125

z = (x - xbar)/σ = (120 - 130)/11.55 = - 0.87

z = (x - xbar)/σ = (125 - 130)/11.55 = - 0.43

P(120 < x < 125) = P(-0.87 < x < -0.43)

We'll use data from the normal probability table for these probabilities

P(120 < x < 125) = P(-0.87 < x < -0.43) = P(z ≤ -0.43) - P(z ≤ -0.86) = 0.33360 - 0.19215 = 0.14145

Hope this Helps!!!

3 0

2 years ago