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Bad White [126]

2 years ago

9

Suppose 255,113 people live in a city. Is it reasanoble to say that about 300,000 people live in the city? Use the number line t

o help you solve the problem. Explain.

1 answer:

motikmotik2 years ago

3 0

It is reasonable. Since 255,113 is closer to 300,000 than 200,000 on the number line, you can round the number to 300,000.

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An alien empire is considering taking over planet Earth, but they will only do so if the portion of rebellious humans is less th

8090 [49]

Step-by-step explanation:

How was the government able to make children go to residential school?

6 0

2 years ago

Give 1 pair of Vertical and 1 pair of Supplementary angles

mojhsa [17]

Solution:

Vertical angles are a pair of opposite angles formed by intersecting lines. re vertical angles. Vertical angles are always congruent.

These two angles (140° and 40°) are Supplementary Angles because they add up to 180°:

Notice that together they make a straight angle.

Hence,

From the image

The following pairs form vertical angles

$\begin{gathered} \angle1=\angle3(vertical\text{ angles)} \\ \angle2=\angle4(vertical\text{ angles)} \\ \angle5=\angle7(vertical\text{ angles)} \\ \angle6=\angle6(vertical\text{ angles)} \end{gathered}$

Hence,

One pair of the vertical angles is ∠1 and ∠3

Part B:

Two angles are said to be supplementary when they ad together to give 180°

Hence,

From the image,

The following pairs are supplementary angles

$\begin{gathered} \angle5+\angle6=180^0(supplementary\text{ angles)} \\ \angle5+\angle8=180^0(supplementary\text{ angles)} \\ \angle7+\angle8=180^0(supplementary\text{ angles)} \\ \angle6+\angle7=180^0(supplementary\text{ angles)} \\ \angle1+\angle2=180^0(supplementary\text{ angles)} \\ \angle1+\angle4=180^0(supplementary\text{ angles)} \\ \angle2+\angle3=180^0(supplementary\text{ angles)} \\ \angle3+\angle4=180^0(supplementary\text{ angles)} \end{gathered}$

Hence,

One pair of supplementary angles is ∠5 and ∠6

8 0

10 months ago

Which is greater? 0.006 or 0.1

frez [133]

Answer:

0.1

Step-by-step explanation:

In a problem like this the number that is more to the left is greater. As you can see 0.1 is more to the left

5 0

2 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 <em>main key</em> 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 <em>indeterminate form 0/0.</em> So , <em>Rationalizing</em> the <em>numerator</em> i.e multiplying both numerator and denominator by the <em>conjugate of numerator </em>

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

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

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 <em>need</em> to <em>eliminate (√x-2)</em> from the denominator somehow , or the limit will again be <em>indeterminate </em>,so if you think <em>carefully</em> as <em>I thought</em> after <em>seeing the question</em> i.e what if we <em>add 4 and subtract 4</em> in <em>numerator</em> ? 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 <em>numerator</em> from 2nd term , so that we can <em>take (√x-2) common</em> 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<em> (√x-2) common</em> 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 <em>radical</em> that makes our <em>limit again and again</em> <em>indeterminate</em> ;

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

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

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

Now , <em>putting the limit ;</em>

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

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

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

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

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

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

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

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

3 0

2 years ago

Read 2 more answers

You are ordering a pizza and soda for the big game. A deep dish pizza costs $10.25, and a two-liter of soda is $2.00. For each t

lions [1.4K]

10.25 + 2.00 = 12.25...cost of pizza and soda

12.25 - 1.50x = 4.75....with x being the number of touchdowns
-1.50x = 4.75 - 12.25...I subtracted 12.25 from both sides
-1.50x = - 7.50...now divide both sides by -1.50
x = -7.50 / -1.50
x = 6......They would have to score 6 touchdowns

5 0

3 years ago