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This question isn't hard at all, and only involves one key fact. 

First, we HAVE to know the formula 

Dilation Factor = Length of Image / Length of Pre-image

So, now, we just plug in some numbers and we are done!

Length of AB (image) = 30 units

\(5 = 30 / OA \)

\(OA = 30/5\)

\(OA = 6\)

So OA is basically just 6 units. 

Thanks! :)

This isn't hard to do, we just have to set up an equation. 

First, let s be the sidenlength of a equation. 

\(s^2 = 10s + 38 \\ s^2 - 10s = 38 \\ s^2 -10s + 25 = 38 + 25\)

\((s - 5)^2 = 63 \)

\(s - 5 = \sqrt{63}\)

\(s = \sqrt{63} + 5\)

The other possible equation would get a negative value, but a sidlength can't be negative, so we reject that possibility. 

So we have \(s = \sqrt{63} + 5\)

Thanks! :)

Not too bad of a question. 

\(\Delta EFG \cdot \frac{2h}{3}=18\\ \Delta EFG =\frac{3\cdot 18}{2h}\\ [{EFGH}]=\Delta EFG\cdot \frac{1}{3}h\\ [{EFGH}]=\frac{3\cdot 18}{2h}\cdot \frac{1}{3}h\\ [{EFGH}]=9\)

I think that we should start with factoring/finding a quadratic formula that we know has a root of 5 or -5 

we can get a factored version of a quadratic formula with roots of 5 and -5  ==> (x-5)(x+5) = we can get x^2-25 if we simplefy it 

so your anwser would be 

x^2 - 0x + (-25) 

thanks 

add -3x^2 to both sides 

add 9x to both sides 

add 4 to both sides

10x^2+10x-1 = 0 

we can plug into the quadratic formula : 

- we will get (-5+sqrt 35) / 10.  or (-5-sqrt 35) / 10

then you can plug these value into the other expression we can get 412 / 199

i think you should work on getting to know the quadratice equation more + simplfying equations in general

subtract 4x^2 from both sides 

- get 1x^2 

-add 17x on both sides = get 6x on the left side 

-subtract 6 from both sides = -2 on the right side 

if you do this you SHOULD get , 

x^2 + 6x - 2 = 0 

plug into quadratic formula (search up) : where ax^2 + bx + c is the formate ( a = 1.  b = 6  c = -2 )
we should get the solution: -3+sqrt(11) or -3-sqrt(11) and they represent a and b 

1/ (-3+sqrt(11))^3 + 1/ (-3-sqrt(11))^3 = 63/2 or 31.5

see if you can get it!

and someone please check my work

Let's first find u and v. Carrying all terms to one side and combining like terms, we get \(x^2-6x+13=0\)

Using the quadratic equation, we get \(x=3+2i,\:x=3-2i\), meaning we have \(u =3-2i\) and \(v =3+2i\).

Plugging these values into \(u/v^2 + v/u^2\), we get \(\frac{3-2i}{(3+2i)^2} + \frac{3+2i}{(3-2i)^2}\)

Simplifying the expression gets us the fraction \(-\frac{18}{169}\)

Thanks! :)

Combining all the like terms, we have \(2x^2-4x-68=0\). 

Using the qudratic equation to find x, we come up with \(x=1+\sqrt{35},\:x=1-\sqrt{35}\)

This problem isn't hard really, It's just moving all terms to one side and combining like terms, before using the quadratic equation to find x. 

Thanks! :)

The value of b in z^2 + bz + c = 0 is equal to the sum of the two roots. 

It's just 4+2i + (-5)+8i = -1+10i

Thanks! :)

Find WX.

\(h_{yz}=\sqrt{r^2-2^2}\\ h_{uv}=\sqrt{r^2-3^2}\\ h_{yz}-h_{uv}=2\\ \sqrt{r^2-2^2}-\sqrt{r^2-3^2}=2\\ r=\frac{\sqrt{145}}{4}=3.0104\ (WolframAlpha)\)

\(h_{yz}=2.25\\ h_{uv}=0.2500\\ h_{wx}=(2.25+0.250)/2=1.25\\ \overline{WX}=2 \sqrt{r^2-h_{wx}^2}=2\sqrt{3.0104^2-1.25^2}\\ \color{blue}\overline{WX}=5.4772\)

 !

The only solution works out to be  \(x=2\).

May 18 2024

https://web2.0calc.com/questions/geometry_18743#r1

 !

Find a + b.

\({\color{blue}m_{AB}=2}=\frac{y_b-y_a}{x_b-x_a}=\frac{b^2-a^2}{b-a}=\frac{(b+a)(b-a)}{b-a}\color{blue}=b+a\\ \color{blue}a+b=2\)

 !

x^2 - 2x + 1 = (x-1)^2

its just 1

Combine like terms

2x^2 - 32x + c

binomial square expansion says that (a+b)^2 = a^2 + 2ab + b^2

so a = sqrt2 * x

2ab = 2 * sqrt2 * x * sqrt(c)

-32x = 2 * sqrt2 * x * sqrt(c)

-16 = sqrt(2c)

256 = 2c

c = 128

x^2 + sx + 144 - 63 + x^2 - 25 - 18

combine like terms: 

2x^2 + sx + 38

if this is a square of a binomial then the first term must be (sqrt2 * x)

2x^2 + sx + 38 = (sqrt2 * x + [other term])^2

this other term has to be sqrt 38 because the contant term of the expression is 38

2x^2 + sx + 38 = (sqrt2 * x + sqrt38)^2

sx is 2ab according to binomial square expansion: (a+b)^2 = a^2 + 2ab + b^2

a = sqrt2 * x

b = sqrt38

so 2ab = sqrt76 * x

so s = sqrt76 = 2sqrt19

\(\color{black }Compute\ the\ slope\ of\ line\ segment\ \overline{PQ}.\)

\(Let\ \overline{ OQ}=\overline{OP} =5\\ f_c(x)=+\sqrt{5^2-x^2}\\ f_{oq}(x)=5x\\ f_{op}(x)=4x\)

\(\sqrt{25-x^2}=5x\\ 25-x^2=25x^2\\ x_q= \sqrt{\frac{25}{26}}=0.9806\\ y_q=\sqrt{\frac{25}{26}}\cdot 5=4.903\\ \sqrt{25-x^2}=4x\\ x_p=\sqrt{\frac{25}{17}}=1.2127\\ y_p=\sqrt{\frac{25}{17}}\cdot 4=4.8507\)

\(m_{PQ}=\frac{y_q-y_p}{x_q-x_p}=\frac{4.9029-4.8507}{0.9806-1.2127}\\ \color{blue }m_{PQ}=-0.2249\\ \color{blue }The\ slope\ of\ line\ segment\ \overline{PQ}\ is\ -0.2249.\)

 !

Let's solve this step-by-step to find the number of problems the mathematician solves the day he drinks coffee.

Normal Day:

Problems solved in a normal day = t (hours) * p (problems/hour) = pt problems

Coffee Day:

Hours worked with coffee = t - 11 hours

Problems solved per hour with coffee = 4p + 2 problems/hour

Let x be the number of problems solved on the coffee day. Then, we have x = (t - 11) * (4p + 2)

Twice as many problems:

The number of problems solved with coffee (x) is twice the number of problems solved on a normal day (pt).

This translates to the equation: x = 2 * pt

Solving for x:

Now we can substitute the expression for x from step 2 into the equation from step 3: (t - 11) * (4p + 2) = 2 * pt

Expanding the left side: 4pt + 2t - 22p - 22 = 2pt

Combining terms with p and t: 2pt + 2t - 22p - 2pt = -22 2t - 22p = -22 t - 11p = -11 (dividing both sides by 2)

Special Condition:

We are given that t and p are positive integers. From the equation t - 11p = -11, we can see that t must be a multiple of 11 (because it's equal to -11p plus another integer, -11).

Finding possible values of t and p:

We can try different integer values for p that are multiples of 7 (since 4p is always even and adding 2 makes it even as well).

For example:

If p = 7, then t = (11 * 7) - 11 = 70 (valid integer).

Other values of p might not result in an integer value for t.

Coffee Day Problems:

Plugging p = 7 and t = 70 (from our example) into the expression for problems solved with coffee (x = (t - 11) * (4p + 2)): x = (70 - 11) * (4 * 7 + 2) = 59 * 30 = 1770 problems

Therefore, on the day the mathematician drinks coffee, he solves 1770 problems (given p = 7 and t = 70).

Fold l is 8*sqrt(2) inches long.

The greatest common divisor of 957 and 1537 is 87.