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3 years ago

How many times greater are the underlined values in 880 with the 88 underlined

Mathematics

1 answer:

AveGali [126]3 years ago

6 0

Hundredths

Hope dis helps

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Given c=(2.4,0.45) and d = (7,-4) find the direction of 5c+4d

kakasveta [241]

Answer:

The direction of 5c + 4d is approximately 19.0° clockwise with

the positive part of x-axis

Step-by-step explanation:

* Lets talk about the direction of a vector

- If the vector is (x , y), then

# Its magnitude is √(x² + y²)

# Its direction is tan^-1 (y/x)

- The direction is the angle between the positive part of x-axis and

the vector

* Lets solve the problem

∵ c = (2.4 , 0.45) and d = (7 , -4)

- To find 5c + 4d , multiply the two coordinates of c by 5 and the two

coordinates of d by 4

# Multiply c by 5

∴ 5c = [5(2.4) , 5(0.45)]

∴ 5c = (12 , 2.25)

# Multiply d by 4

∴ 4d = [4(7) , 4(-4)]

∴ 4d = (28 , -16)

- Lets add 5c and 4d

∴ 5c + 4d = (12 , 2.25) + (28 , -16)

∴ 5c + 4d = [(12 + 28) , (2.25 + -16)]

∴ 5c + 4d = (40 , -13.75)

- The vector is in the 4th quadrant because the x-coordinate is

positive and the y-coordinate is negative

* Find the direction of 5c + 4d

∵ The direction is tan^-1 (y/x)

∴ The direction is tan^-1 (-13.75/40) = -18.97°

* The direction of 5c + 4d is approximately 19.0° clockwise with

the positive part of x-axis

3 0

4 years ago

Evaluate the limit

wel

We are given with a limit and we need to find it's value so let's start !!!!

${\quad \qquad \blacktriangleright \blacktriangleright \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}$

But , before starting , let's recall an identity which is the main key to answer this question

${\boxed{\bf{a^{2}-b^{2}=(a+b)(a-b)}}}$

Consider The limit ;

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}$

Now as directly putting the limit will lead to indeterminate form 0/0. So , Rationalizing the numerator i.e multiplying both numerator and denominator by the conjugate of numerator 

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}\times \dfrac{\sqrt{x}+\sqrt{3\sqrt{x}-2}}{\sqrt{x}+\sqrt{3\sqrt{x}-2}}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-\sqrt{3\sqrt{x}-2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}{(x^{2}-4^{2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

Using the above algebraic identity ;

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x})^{2}-(\sqrt{3\sqrt{x}-2})^{2}}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-(3\sqrt{x}-2)}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}+2}{\{(\sqrt{x})^{2}-2^{2}\}(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

Now , here we need to eliminate (√x-2) from the denominator somehow , or the limit will again be indeterminate ,so if you think carefully as I thought after seeing the question i.e what if we add 4 and subtract 4 in numerator ? So let's try !

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2+4-4}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(x-4)+2+4-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

Now , using the same above identity ;

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+6-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+3(2-\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

Now , take minus sign common in numerator from 2nd term , so that we can take (√x-2) common from both terms

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)-3(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

Now , take (√x-2) common in numerator ;

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)\{(\sqrt{x}+2)-3\}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

Cancelling the radical that makes our limit again and again indeterminate ;

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\cancel{(\sqrt{x}-2)}\{(\sqrt{x}+2)-3\}}{\cancel{(\sqrt{x}-2)}(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}+2-3)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-1)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

Now , putting the limit ;

${:\implies \quad \sf \dfrac{\sqrt{4}-1}{(\sqrt{4}+2)(4+4)(\sqrt{4}+\sqrt{3\sqrt{4}-2})}}$

${:\implies \quad \sf \dfrac{2-1}{(2+2)(4+4)(2+\sqrt{3\times 2-2})}}$

${:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{6-2})}}$

${:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{4})}}$

${:\implies \quad \sf \dfrac{1}{(4)(8)(2+2)}}$

${:\implies \quad \sf \dfrac{1}{(4)(8)(4)}}$

${:\implies \quad \sf \dfrac{1}{128}}$

${:\implies \quad \bf \therefore \underline{\underline{\displaystyle \bf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}=\dfrac{1}{128}}}}$

3 0

2 years ago

Read 2 more answers

What is the surface area of cone below Diameter AB=21 and distance from B to C =36

Vesnalui [34]

Answer:

1533.88261311522

Step-by-step explanation:

Let r be the radius of the (base/circle)

and L be the slant height of the cone

Formula ………………………………………………………………………………………………………

The surface area of a cone = the (curved/lateral) surface area + the base

$=\pi r^{2}\ \ \ +\ \ \ \pi Lr$

=========================================

$r=\frac{21}{2} =10.5$

L = BC = 36

$\text{surface area} =\pi \left( 10.5\right)^{2} +\pi \times 36\times \left( 10.5\right)$

$=346.360590058275 + 1187.52202305694$

$=1533.88261311522$

6 0

2 years ago

What time is 5 3/4 hours after 11:32 am?

Norma-Jean [14]

Answer:

5:17pm

Step-by-step explanation:

3 0

3 years ago

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C+-c=0 Identity Property Associative Property Communtative Property Inverse Property

Zanzabum

Answer:

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7 0

3 years ago