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4 years ago

IF U SMART HELP ME!!

Mathematics

1 answer:

Anna11 [10]4 years ago

7 0

Answer:

1.63

Step-by-step explanation:

you round up if the 5 and up

you round down if the 4 and down

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What is the missing constant term in the perfect square that starts with x^2+2x?

Natasha_Volkova [10]

Answer:

+ 1

Step-by-step explanation:

Given

x² + 2x

Using the method of completing the square

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

= x² + 2(1)x + 1²

= x² + 2x + 1

= (x + 1)² ← perfect square

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3 years ago

An anion with a -2 charge is found in which Group?

IgorLugansk [536]

Answer:

A and B any atom or group or atoms with a negative charge

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3 years ago

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Identify the vertex of the graph. Tell whether it is a minimum or maximum.

melisa1 [442]

It’s B you got to see the lowest part of the graph and you will call it mini except if the graph was going down you will call that point maximum.

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3 years ago

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Determine the measure of each missing angle

Drupady [299]

Answer:

w=y=86

Step-by-step explanation:

A triangle has a total angle of 180 degrees

w=y because the triangle is isoceles meaning that two sides are the same length (they are marked with a line through them)

180-94=86 degrees=w+y

w+y so 86/2=43 degrees

w=y=86

is that all the missing angles?

8 0

4 years ago

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

3 years ago