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So far, I got -3, 5, and -11

what answers did you find?

Did you proof-read your question? Sould it be 338 or 336? There is no such thing as 338 being the difference between two consecutive odd perfect squares !.

Hi Guest!

In one hour, R, S, and T can do 1/4 of the job. 

In one hour, S and T can do 1/5 of the job. 

So R in one hour can do 1/20 of the job (1/4 - 1/5)

So if in one hour, R can do 1/20, then she needs 20 hours to do the whole job. 

I hope this helped. :)))

=^._.^=

For these problems, I'd highly recomend graphing them out. Personally, I love using desmos.

 https://www.desmos.com/calculator/npy869vaxj  Here's the link of the plotted triangle. 

The base is 5, and the height is 2. 

So the area is 5*2/2 = 5. 

I hope this helped. :)))

=^._.^=

Hi Guest!

number of girls = g

number of boys = 40 - g

5g = 4(40 - g) + 65

Try solving for the number of girls yourself.  :)))

Here's the answer if you still needed it: 

5g = 160 - 4g + 65

9g = 225

g = 25

So the number of girls is 25, meaning the number of boys is 15 (40-25)

Our final answer is 10 (25-15)

I hope this helped. :))

=^._.^=

Hi Guest,

The formula is Base*Height/3. 

Let's have the base be EHA, which the area is 5*6/2 = 15. 

That would make the height EF, which is 4. 

15*4/3 = 20.

I hope this helped. :)))

=^._.^=

Hi Guest!

one is 1

two thirds is 2/3

So if we want a mixed number, we can just combine them, 1 2/3. 

https://www.splashlearn.com/math-vocabulary/algebra/mixed-number

This is a link of mixed numbers. 

Keep practicing math, and you'll improve :))

=^._.^=

Thank you for your help! I managed to find a couple of solutions but I'm not sure if I'm missing any lol. 

.

There are 360 degrees in a rotation and you have 60 of them in this arc

so

arc length =  $\frac{60}{360}*circumference\\ =\frac{60}{360}*2\pi*15$

Please do not ask multiple questions on the same post.

You can repost Q2 on a new post, and delete if from here.     (So long as no one has offered an answer on it yet)

Question 1:

1. Find all integers a for which the polynomial ax^2-8x+3=0 has an integer root.

the roots are

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {8 \pm \sqrt{64-12a} \over 2a}\\ x=\frac{4\pm \sqrt{16-3a} }{a}$

16-3a    will have to be a perfect square

a could equal  0, 4, 5

They are the only positive a values that could work

a=0 is no good - can't divide by 0

a=4  roots are fractional

a=5  one of the roots will be 1  so that works

So   a=5 is the only positive value of a that will have an integer root.   There might be some negative values of a that work though....

Your question doesn't make sense, (not to me anyway)  I think something is missing.

Hi Teja2,

It is nice to meet you.

It is easier for me to answer myself than it is for me to check your answer.

So that is the only reason I am answering here

$4^{10}*8^{20}*16^{30}\\ =2^{20}*2^{60}*2^{120}\\ =2^{200}$

there you go, if you got it wrong then I did too.   

Give yourself a point!

The fifth term of an arithmetic progression is three times the second term,and the third term is 10.find the 20th term?

Der fünfte Term einer arithmetischen Folge ist dreimal so groß wie der zweite Term und der dritte Term ist 10. Finden Sie den 20. Term?

Hello Guest!

$\displaystyle a_{1}=a_{1},\ a_{2}=a_{1}+d,\ a_{3}=a_{1}+2d,\ a_{4}=a_{1}+3d, a_5=a_1+4d\dots$

${\displaystyle a_{i}=a_{1}+(i-1)\cdot d\quad } \ (explicit\ formula,\ explizite\ Formel)$

$a_5=(a_1+d)\cdot 3=a_1+(5-1)\cdot d\ (specification\ 1 )\\ a_3=a_1+(3-1)\cdot d=10\ (specification\ 2 )$

$3a_1+3d=a_1+4d\\ d=2a_1$

$a_1+2d=10\\ d=\dfrac{10-a_1}{2}$

$2a_1=\dfrac{10-a_1}{2}\\ 4a_1=10-a_1$

$\large a_1=2$

$d=\dfrac{10-a_1}{2}\\ d=\dfrac{10-2}{2}\\ $

$\large d=4$

$a_{i}=a_{1}+(i-1)\cdot d\\ a_{20}=2+(20-1)\cdot 4\\$

$\large a_{20}=78$

  !

1. You mention the method to be used for the solution, Vieta.

If the roots of the cubic are a, b and c, then

$\displaystyle (x-a)(x-b)(x-c) = 0, \\ \text{expanding,} \\ x^{3}-x^{2}(a+b+c)+x(ab+bc+ca)-abc=0.$

So, for the given cubic,

$\displaystyle a+b+c=3\dots \dots(1) \\ ab+bc+ca=-15/2 \dots\dots(2)\\ abc=3/2\dots \dots (3).$

It's just follow your nose after that.

$\displaystyle \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} =\frac{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}{a^{2}b^{2}c^{2}}.$

The denominator is easy, just substitute (3).

For the numerator it looks like you have to make use of (2), so,

$\displaystyle (ab+bc+ca)^{2}=a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2a^{2}bc+2ab^{2}c+2abc^{2}\\= a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2abc(a+b+c),$

and now it's just substitution from (1), (2) and (3).

$\displaystyle\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right)\\ \displaystyle\left(\frac{1}{b}+\frac{1}{a}\right)\left(\frac{1}{b}-\frac{1}{a}\right)\\ = \displaystyle\left(\frac{1}{b}\right)^2-\left(\frac{1}{a}\right)^2\\$

For the smallest value you want the absolute value of  b to be as big as possible,  and the absolute value of a to be as small as possible but neither can be 0

b=3,  a= -1

1/9  -  1 = -8/9

12 / 9 = 32 / EH          EH = 24      ==>  EH = UV

UV / VT = UE / EH                 24 / VT = 32 / 24

VT = 18

When a tangent and a secant, two secants, or two tangents intersect outside a circle then the measure of the angle formed is one-half the positive difference of the measures of the intercepted arcs.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

1/2(257 - 103) = 7x + 14     

x = 9

In circle H with m \angle GHJ= 90m∠GHJ=90 and GH=20 units, find the length of arc GJ.

Hello Guest!

$C=2\pi\cdot\overline{GH}\\\angle GHJ=90^\circ\\ \overline{GH}=20$

$(GJ)=\dfrac{2\pi \cdot \overline{GH}\cdot 90^\circ}{360^\circ}=\dfrac{2\pi \cdot 20\cdot 90^\circ}{360^\circ}=31.42$

  !

To get the minimum value, you take the smallest value for each variable.  So the minimum value is (1/(-5)) + 1/1)(1/(-5) - 1/1) = 24/25.

They won  4 out of (4+3=7)

4/7 * 154 = won games=............ games

     the rest they lost......

2 *  (7x+14)     =   257

x = 16.357

3453.74 = 1187 ( 1 + r/4)^12*4      solve for r

2.90964 = (1+r/4)⁴⁸

log (2.90964) = 48 log (1+r/4)

log (2.90964) / 48 =   log (1+r/4)     

1.0225 = 1 + r/4

r/4 = .0225      r = 9%  annual

4^(10)*8^(20)*16^(30)

= (2^2)^10 * (2^3)^20 * (2^4)^30

= 2^(2 * 10) * 2^(3 * 20) * 2^(4 * 30)

= 2^20 * 2^60 * 2^120

= 2^(20 + 60 + 120)

= 2^200

This is the first time I have posted an answer here. Please let me know if I'm doing this right :)