1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
gayaneshka [121]
2 years ago
11

Can someone help 12 points on the line

Mathematics
1 answer:
ANTONII [103]2 years ago
7 0

Answer:

x=6

Step-by-step explanation:

We can write this as a ratio

x        9

---- = --------

10       15

Using cross products

15x = 10*9

15x = 90

Divide by 15

15x/15 = 90/15

x =6

You might be interested in
A triangular prism and two nets are shown:
leonid [27]
The net for a triangular prism consists of 
2 identical triangles, here all nets have sides 5,12, and 13.
3 rectangles, all with a common dimension: the height H=14 of the prism.

Each of the 3 rectangles should have dimensions
H=14 plus one of the following:
each of 5, 12, 13 corresponding to one side of the base (triangle).

So the dimensions of the rectangles are
5x14, 12x14, 13x14

And dimensions of the triangles are
5,12 and 13.

The total surface area is therefore
(5+12+13)*14 + 2*(5*12/2)
=420+60
=480.

There is only one net that satisfies this condition.

8 0
3 years ago
Xây dựng khoảng ước lượng đối xứng cho tỷ lệ tổng thể, trình bày bài toán tìm cỡ mẫu tối thiểu trong ước lượng khoảng cho tỷ lệ
Nikolay [14]
So basically you just
7 0
2 years ago
For which distributions is the median the best measure of center?
Stolb23 [73]

Answer:

The answer would be 2nd top choice and 1st bottom choice

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
What is the length of BC in the right triangle below?
Neko [114]

Answer and Step-by-step explanation:

<u>How to solve:</u>

Use the Pythagorean Theorem to find the length of side <em>BC</em>.

<u>Pythagorean Theorem Formula:</u>

<u />A^2+B^2=C^2<u />

<u />

Side AB will be A, side AC will be B, and side BC will be C.

<u>Plug in values.</u>

8^2+15^2 = C^2\\\\\\64 + 225 = C^2\\\\289 = C^2\\Square.root.both.sides.\\\\\sqrt{289} = C\\17 = C\\\\\\

<u>This means the answer is A. 17.</u>

<u></u>

<u>I hope this helps!</u>

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

5 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 <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
Other questions:
  • PLEASE HELP!!!!!!!!!!!!!
    14·1 answer
  • Emily made 75% of the baskets that she attempted if she may nine baskets how many attempts did you make?
    15·1 answer
  • Doreen schmidt is a chemist. She needs to prepare 36 ounces of a 10% shydrochloric acid solution. Find the amount of 18% solutio
    11·1 answer
  • Which expression are solution to the equation 2.4y=13.75
    13·1 answer
  • Farmer Ellie has another circular field for crops. This one has a diameter of 35 meters. What is the area of the field?
    14·1 answer
  • An equation is shown below:
    7·1 answer
  • Can someone please help me, this is for summer school and I need fast replies, I have 29 other question like these, please help.
    10·1 answer
  • My sister is crying because this is the only thing that can get her grade up I tried to help but it isn’t making any sense so pl
    10·1 answer
  • Please help problems below!!!
    10·1 answer
  • Can someone help me with this
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!