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4 years ago

No its 17 tests and 3 weeks

Mathematics

1 answer:

krok68 [10]4 years ago

8 0

Do u need help because ur question isn't clear

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The ratio of a dog and cat owners is four to 5 there are 48 dog owners how many cat owners

Verdich [7]

Answer:

Step-by-step explanation:

Let the number of dog owners and cat owners be 4x and 5x

4x = 48

x = 48/4

x = 12

Therefore 5x = 5×12

= 60

6 0

3 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

2. The actual tree on the left is 10 feet tall. The model tree on the right is 8 inches tall. What is the scale factor?

Lina20 [59]

Answer:

D. 1/15

Step-by-step explanation:

Actual tree = 10 ft = 12 × 10 = 120 in.

Model tree = 8 in.

Scale factor = model tree/actual tree

Scale factor = 8/120

Scale factor = 1/15

4 0

3 years ago

Nico owns 11 instructional piano books. Two are beginner books, six are intermediate books, and three are advanced books. If two

Solnce55 [7]

2 beginner, 6 intermediate, 3 advanced...total of 11

P(advanced) = 3/11
replaced
P(beginner) = 2/11

P(both) = 3/11 * 2/11 = 6/121 <=

8 0

4 years ago

Which expression is equivalent to (27.3-5.9)4?

makkiz [27]

Answer:

109.2-23.6 = 85.6

Step-by-step explanation:

use bodmas

B=bracket

O=of or multiplication

D=division

M=multiplication

A=addition

S=subtraction

so when you are opening a bracket you multipli the number in the bracket with the one outside the bracket

I.E

4×27.3=109.3

4×5.9=23.6

109-23.6=85.6

3 0

3 years ago

Read 2 more answers