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tamaranim1 [39]
3 years ago
6

6. What is the value of n? 3/8n - 1/3 = 5/6

Mathematics
2 answers:
Brrunno [24]3 years ago
8 0

Answer:

n= 3 1/9

Step-by-step explanation:

3/8n= 5/6 + 1/3

3/8n = 7/6

n = 7/6  ÷ 8/3

n=28/9

n= 3 1/9

Murrr4er [49]3 years ago
6 0

Answer:

I think it's 3 1/9

Step-by-step explanation:

3/8n-1/3=5/6

×3. ×8. ×4

9/24n-8/24=20/24

+8/24. +8/24

9/24n=28/24

÷9/24. ÷9/24

n=28/24÷9/24

n=28/24×24/9

n=672/216

n=28/9

n=3 1/9

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Dafna11 [192]
Correct answer is 33.1 m.

Use The definition of trigonometric functions. The sine of an angle is the quotient of the opposite side and hypotenuse.

sin{65^o}= \frac{30}{x} 
\\x= \frac{30}{sin{65^o}}
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3 0
3 years ago
On Monday a team of street sweepers cleaned 1/3 of the city block. Tuesday, the team cleaned 1/2 as many blocks as on Monday. Ho
IrinaVladis [17]

Answer:

1/2 half

Step-by-step explanation:

(1/3)/2 = 1/6

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6 0
2 years ago
Please help!!!! question is/.........
Andrew [12]
Area of circle = 3.14 x 6^2 = 3.14 x 36 = 113.04

1/2 circle = 113.04 / 2 = 56.52

Area of rectangle = 8 x 6 = 48

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answer
104.52 ft^2
8 0
3 years ago
Find the vertical and horizontal asymptote if they exist
tresset_1 [31]

Hello!

Vertical asymptotes are determined by setting the denominator of a rational function to zero and then by solving for x.

Horizontal asymptotes are determined by:

1. If the degree of the numerator < degree of denominator, then the line, y = 0 is the horizontal asymptote.

2. If the degree of the numerator = degree of denominator, then y = leading coefficient of numerator / leading coefficient of denominator is the horizontal asymptote.

3. If degree of numerator > degree of denominator, then there is an oblique asymptote, but no horizontal asymptote.

To find the vertical asymptote:

2x² - 10 = 0

2(x² - 5) = 0

(x - √5)(x + √5) = 0

x = √5 and x = -√5

Graphing the equation, we realize that x = -√5 is not a vertical asymptote, so therefore, the only vertical asymptote is x = √5.

To find the horizontal asymptote:

If the degree of the numerator < degree of denominator, then the line, y = 0 is the horizontal asymptote.

Therefore, the horizontal asymptote of this function is y = 0.

Short answer: Vertical asymptote: x = √5 and horizontal asymptote: y = 0

7 0
3 years ago
Find the area of an equilateral triangle with apothem 7 cm. Round to the nearest whole number
Yuliya22 [10]
In <span>equilateral triangle all sides are equal =x,
from  right triangle that  is formed by </span>apothem , height 7*3= 21
hypotenuse /opposite leg = sin angle,
hypotenuse is a side of triangle =x, opposite leg is apotheme =21,
angle in equilateral triangle =60⁰
21/x= sin 60, x*sin60=21, x=21/sin60, x=21/(√3/2), x=42/√3
Area of triangle =1/2 x*x*sin angle

Area of triangle =1/2 *42/√3*42/√3*sin 60=1/2*(42²/3)*(√3/2) ≈ 255 cm²



this is just second way to do this problem, that I did at first (either way is correct)
<span> apothem divides one side by half,
so we get small right triangle with sides x hypotenuse, x/2 is one leg , and 21 is another leg
</span>by Pythagorean theorem 
x²=(x/2)² +21²
x²-x²/4=441
3x²/4=441
3x²=441*4=1764
x²=1764/3
x=42/√3
Area of triangle =1/2*base *height =1/2*42/√3 *21≈255 cm²
8 0
3 years ago
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