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Reil [10]
1 year ago
13

Which list of fractions is in order from smallest to largest A. 3/4 3/5 3/10 3/12. B. 5/3 3/4 3/12 3/10 C. 3/10 3/5 3/4 3/12. D.

3/12 3/10 3/5 3/4
Mathematics
1 answer:
Pepsi [2]1 year ago
5 0

A. 3/4 = 0.75 3/5 = 0.6 3/10 = 0.3 3/12 = 0.25

Right order

3/12 < 3/10 < 3/5 < 3/4

The right answer is letter D.

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What is the domain and range of f(x)= |x+6|
andreev551 [17]

as the function is polynomial domain exist for all real number ie (-infinity to + infinity) but range exist (0 to +infinity ) due to modulus negetive range do not exist

4 0
3 years ago
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Express 2x^2+8x+3 in the form 2(x+p)^2+q
e-lub [12.9K]

Answer:

2(x + 2)² - 5

Step-by-step explanation:

Given

2x² + 8x + 3

To obtain the required form use the method of completing the square.

The coefficient of the x² term must be 1, thus factor 2 out of 2x² + 8x

= 2(x² + 4x) + 3

add/subtract ( half the coefficient of the x- term)² to x² + 4x

= 2(x² + 2(2)x + 4 - 4) + 3

= 2(x + 2)² - 8 + 3

= 2(x + 2)² - 5

with p = 2 and q = - 5

4 0
3 years ago
Which combinations of transformations will always produce congruent figures? check all that apply
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You didnt provide the selections, but some answers would most likely be rotations, translations, and reflections.
4 0
3 years ago
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A storage company must design a large rectangular container with a square base. The volume is 6750ft3. The material for the top
GuDViN [60]
Ответ 163.54
Ниже расписано описание

4 0
3 years ago
A soccer ball is kicked from 3 feet above the ground. the height,h, in feet of the ball after t seconds is given by the function
dexar [7]
Here's our equation.

h=-16t+64t+3

We want to find out when it returns to ground level (h = 0)

To find this out, we can plug in 0 and solve for t.

0 = -16t+64t+3 \\ 16t-64t-3=0 \\ use\ the\ quadratic\ formula\ \frac{-b\±\sqrt{b^2-4ac}}{2a}  \\ \frac{-(-64)\±\sqrt{(-64)^2-4(16)(-3)}}{2*16}  = \frac{64\±\sqrt{4096+192}}{32}

= \frac{64\±\sqrt{4288}}{32} = \frac{64\±8\sqrt{67}}{32} = \frac{8\±\sqrt{67}}{4} = \boxed{\frac{8+\sqrt{67}}{4}\ or\ 2-\frac{\sqrt{67}}{4}}

So the ball will return to the ground at the positive value of \boxed{\frac{8+\sqrt{67}}{4}} seconds.

What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!

\frac{1}2(\frac{8+\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(2+\frac{\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(4) = 2

h=-16t^2+64t+3 \\ h=-16(2)^2+64(2)+3 \\ \boxed{h=67}


8 0
4 years ago
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