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3 years ago

Order the numbers from least to greatest.

Mathematics

2 answers:

Mars2501 [29]3 years ago

7 0

Answer:

4.446 then 4.464 then 5.228

Step-by-step explanation:

Hope this HELPS :D

padilas [110]3 years ago

7 0

4.446, 4.464, 5.228. in

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Hey guys- need help in this one

oksano4ka [1.4K]

Answer:

Step-by-step explanation:

$x=5+i\sqrt{7} , y = 5-i\sqrt{7}\\ x^2-2xy+y^2=(x-y)^2\\ (x-y)^2=(2i\sqrt{7} )^2\\4.i^2.7\\-28$

8 0

2 years ago

Read 2 more answers

What is the expression of one less than the quotient of four and a number x

Maru [420]

The expression is: (4/x) -1

7 0

3 years ago

Read 2 more answers

A right square pyramid with base edges of length $8\sqrt{2}$ units each and slant edges of length 10 units each is cut by a plan

seropon [69]

Answer:

32 $unit^{3}$

Step-by-step explanation:

Given:

The slant length 10 units
A right square pyramid with base edges of length 8 $\sqrt{2}$

Now we use Pythagoras to get the slant height in the middle of each triangle:

$\sqrt{10^{2 - (4\sqrt{2} ^{2} }) }$ = $\sqrt{100 - 32}$ = $\sqrt{68}$ units

One again, you can use Pythagoras again to get the perpendicular height of the entire pyramid.

$\sqrt{68-(4\sqrt{2} ^{2} )}$ = $\sqrt{68 - 32}$ = 6 units.

Because slant edges of length 10 units each is cut by a plane that is parallel to its base and 3 units above its base. So we have the other dementions of the small right square pyramid:

The height 3 units
A right square pyramid with base edges of length 4 $\sqrt{2}$

So the volume of it is:

V = 1/3 *3* 4 $\sqrt{2}$

= 32 $unit^{3}$

5 0

4 years ago

Determine if the lim f(x) exists using the graph below.. If it does, find its value. If it does not,x 1explain why

Leno4ka [110]

The limit of a function f at a point a exists if:

$\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)$ $\begin{gathered} \text{ From the given image, the value of the left limit of f at x=1 } \\ \lim_{x\to a^-}f(x)\lt1 \end{gathered}$

Also,

$\begin{gathered} \text{ From the given image, the value of the right limit of f at x=1 } \\ \lim_{x\to a^+}f(x)\gt1 \end{gathered}$

Therefore,

$\lim_{x\to1^-}f(x)\lim_{x\to1^+}f(x)$

Hence, the limit does not exist

8 0

1 year ago

How do I find the arc??

olganol [36]

A circle is 360° all the way around; therefore, if you divide anarc's degree measure by 360°, you find the fraction of the circle's circumference that the arc makes up. Then, if you multiply the length all the way around the circle (the circle's circumference) by that fraction, you get the length along thearc.

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3 years ago