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4 years ago

5x-2y=3 in point-slope form

Mathematics

1 answer:

noname [10]4 years ago

6 0

2y = 5x - 3
y =( 5x-3 )/2

slope = 5
x = 3/5
y = 3/2

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Drag each expression to show whether it is equivalent to 3 time times 9 or to 2x3x9

horrorfan [7]

Answer:

The answers to your questions are given below

Step-by-step explanation:

A. 3 × 9 = 27

B. 2 × 3 × 9 = 54

Next, we shall determine which option will give the same result obtained above.

9 × 3 = 27

9 × 3 × 2 = 54

3 × (4 + 5) = 3 × 9 = 27

(2 × 3) × 9 = 6 × 9 = 54

Therefore,

3 × 9 = 27

9 × 3 = 27

3 × (4 + 5) = 27

2 × 3 × 9 = 54

9 × 3 × 2 = 54

(2 × 3) × 9 = 54

SUMMARY

3 × 9 >>>>>>>>> 2 × 3 × 9

9 × 3 >>>>>>>>> 9 × 3 × 2

3 × (4 + 5) >>>>> (2 × 3) × 9

6 0

3 years ago

Circumference and Area questions look at the picture

Helga [31]

Note that the formula for the circumference of a circle is πd, while the formula for the area of a circle is πr².

π≈3.14

A. C=πd

Simply plug in the numbers into the formula.

Diameter=Radius*2

17*2=34

C=34(π)

B. (π)(5²)

Plug in the numbers into the formula. Remember that half of the diameter is the radius.

C. (π)(4.5)

There are two possible formulas that could be used to calculate the circumference of a circle: πd and 2πr.

The expression above is simply multiplying the circle's radius times pi. Therefore, it is not a method that could be used to find the circumference of a circle.

D. (π)(6.5²)

Remember that the formula for calculating the area of a circle is πr².

Half of the diameter is 6.5 (13.5/2=6.5). 6.5 cm. is the radius. Now just plug the numbers into the formula.

(π)(6.5²)

Therefore, the last answer choice is the correct answer.

5 0

3 years ago

Find the length of the radius of the circle whose perimeter is xcm and the area is

Jobisdone [24]

$\bf \stackrel{\textit{perimeter}}{\textit{circumference}}\textit{ of a circle}\\\\ C = 2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=x \end{cases}\implies x = 2\pi r\implies \boxed{\cfrac{x}{2\pi }=r} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ A=x \end{cases}\implies x = \pi \left( \boxed{\cfrac{x}{2\pi }} \right)^2\implies x = \pi \cdot \cfrac{x^2}{2^2\pi^2}$

$\bf x = \cfrac{x^2}{4\pi }\implies 4\pi x = x^2\implies \cfrac{4\pi x}{x}=x\implies \blacktriangleright 4\pi = x\blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{since the radius is}}{r = \cfrac{x}{2\pi }}\implies r =\cfrac{4\pi }{2\pi }\implies \blacktriangleright r = 2\blacktriangleleft$