A) x²-4x+3=0 prove in factorization form

Mathematics

1 answer:

Marina CMI [18]2 years ago

8 0

Answer: (x-1) (x-3)

Step-by-step explanation: x²-3x-x+3

x²-4x+3

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What is<br> as a decimal rounded to 3 decimal places?

puteri [66]

“Three decimal places” is the same as “the nearest thousandth.” So, for example, if you are asked to round 3.264 to two decimal places it means the same as if your are asked to round 3.264 to the nearest hundredth. Some questions, like the example below, will ask you to “show your answer correct to two decimal places."

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2 years ago

What method should I use to solve this?

tekilochka [14]

Answer:

$9.10

Step-by-step explanation:

$455 \div 50 = 9.1$

455 ÷ 50 = 9.1. The student tickets cost $9.10. (I think)

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2 years ago

HEEEEEELLLLLLLLPPPPPPPPPPPPP-2/3+1/3

motikmotik

Answer:

the total will be 1

Step-by-step explanation:

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2 years ago

Assume that the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder. Based on this assumption,

kompoz [17]

If the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder, then its volume is

$V_{flask}=V_{sphere}+V_{cylinder}.$

Use following formulas to determine volumes of sphere and cylinder:

$V_{sphere}=\dfrac{4}{3}\pi R^3,\\ \\V_{cylinder}=\pi r^2h,$

wher R is sphere's radius, r - radius of cylinder's base and h - height of cylinder.

Then

$V_{sphere}=\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi \left(\dfrac{4.5}{2}\right)^3=\dfrac{4}{3}\pi \left(\dfrac{9}{4}\right)^3=\dfrac{243\pi}{16}\approx 47.71;$
$V_{cylinder}=\pi r^2h=\pi \cdot \left(\dfrac{1}{2}\right)^2\cdot 3=\dfrac{3\pi}{4}\approx 2.36;$
$V_{flask}=V_{sphere}+V_{cylinder}\approx 47.71+2.36=50.07.$

Answer 1: correct choice is C.

If both the sphere and the cylinder are dilated by a scale factor of 2, then all dimensions of the sphere and the cylinder are dilated by a scale factor of 2. So

R'=2R, r'=2r, h'=2h.

Write the new fask volume:

$V_{\text{new flask}}=V_{\text{new sphere}}+V_{\text{new cylinder}}=\dfrac{4}{3}\pi R'^3+\pi r'^2h'=\dfrac{4}{3}\pi (2R)^3+\pi (2r)^2\cdot 2h=\dfrac{4}{3}\pi 8R^3+\pi \cdot 4r^2\cdot 2h=8\left(\dfrac{4}{3}\pi R^3+\pi r^2h\right)=8V_{flask}.$