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Shkiper50 [21]
3 years ago
13

The equation of a straight line is 3y + 9x -12 =0. Use the equation to find the

Mathematics
1 answer:
Debora [2.8K]3 years ago
3 0

Answer:

-3

Step-by-step explanation:

Divide through the equation by the leading coefficient to obtain:

y+3x-4=0

y=4-3x

and our gradient is -3

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17. What expression is equivalent to log(200) - log (2)? Calculate the answer.
IRINA_888 [86]

Answer: 2

Step-by-step explanation:

Recall from the laws of Logarithms:

Log a - Log b = Log ( a/b )

That means

Log 200 - Log 2 = Log ( 200/2)

= Log 100 , which could be written as

Log 10^{2}

Recall from laws of Logarithms:

Log a^{b} = b Log a

Therefore:

Log10^{2} = 2 Log 10

Also from law of Logarithm

Log 10 = 1

Therefore 2 Log 10 = 2 x 1

= 2

3 0
3 years ago
He ride a bike for 15 miles oer hour how many miles did he ride
Klio2033 [76]
Yes it’s 15, Thats cool man
7 0
3 years ago
The formula for the volume V of a cylinder is V=πr2h, where r is the radius of the base and h is the height of the cylinder. Sel
Alja [10]

Answer:

see explanation

Step-by-step explanation:

Given

V = πr²h ( isolate h by dividing both sides by πr² )

\frac{V}{\pi r^2} = h

When V = 36π , r = 3 , then

h = \frac{36\pi }{\pi (3)^2} = \frac{36\pi }{9\pi } = 4 cm

4 0
3 years ago
Read 2 more answers
Part 3: Write the equation of and graph an ellipse.
tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

The center is halfway between vertices, at (4, -6).

It is also halfway between foci.

:::::

The vertices are vertically aligned, so the parabola is vertical.

General equation for a vertical ellipse:

 (y-k)²/a² + (x-h)²/b² = 1

with

 center (h,k)

 vertices (h,k±a)

 co-vertices (h±b,k)

 foci (h,k±c), c² = a²-b²

Apply your data and solve for h, k, a, and b.

center (h,k) = (4, -6)

h = 4

k = -6

vertices (4,-6±a) = (4,-6±9)

a = 9

foci (4,-6±c) = (4,-6±5√2)

b² = a² - c² = 9² - (5√2)² = 31

b = √31

The equation becomes

 (y+6)²/81 + (x-4)²/31 = 1

:::::

length of major axis = 2a = 18

length of minor axis = 2b = 2√31

8 0
3 years ago
Item 17 The amount of chlorine in a swimming pool varies directly with the volume of water. The pool has 2.5 milligrams of chlor
zvonat [6]
8000 gal *\frac{ 1 L}{0.264 gal}*\frac{2.5 mg}{1 L} = 75,757 mg

Rounding to nearest thousand gives 76,000 mg of chlorine
7 0
3 years ago
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