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so, if there were x adults and 2x children, we have
7x + 3*2x = 1248

GuestJan 18, 2022

+189

I got a=60 and b=3/10 and so a+5b=60+5*(3/10).

When you simplify that, a+5b=103/2.

ImDecentAtMathJan 18, 2022

+189

-2

C: (16,5)

ImDecentAtMathJan 18, 2022

+118658

Thanks Ticon6 ;)

MelodyJan 18, 2022

There are 22 ways.

GuestJan 18, 2022

The answer is 1200 pounds.

GuestJan 18, 2022

The probability is 1/6.

GuestJan 18, 2022

+189

-2

Already solved here: https://web2.0calc.com/questions/geometry_92608

ImDecentAtMathJan 18, 2022

+189

-2

We can say z=a+bi, so z^2 is a^2-b^2+2abi. For our original equation to be true, a^2-b^2=2 and ab=1. Try solving the problem from there.

ImDecentAtMathJan 18, 2022

+189

-2

cphil answered your question at this link: https://web2.0calc.com/questions/quadratic-equations_12

ImDecentAtMathJan 18, 2022

+189

-2

Since triangleEFC is iscoceles, AF+FC=AF+FE=25. This is also true for DE with triangleBDE also being iscoceles and AD+DB=AD+DE=25.

The perimeter of the paralellogram is (AD+DE)+(AF+FE) and we know that AD+DE=25 and AF+FE=25, so the answer is 25+25=50.

Hope this helped.

ImDecentAtMathJan 18, 2022

+189

-1

It's not, thanks for trying to help though.

ImDecentAtMathJan 18, 2022

+189

-1

In my answer, I will assume the discriminant in the quadratic formula is 0.

The way to solve is:

b^2-4ac=0

b^2=4ac

b^2=4*2*82

b=+- sqrt(656)

b=+- 4sqrt(41)

ImDecentAtMathJan 18, 2022

The answer is 7/12*pi.

GuestJan 18, 2022

+189

-1

The first triangle has an area of 4sqrt3. We can find this by using the Pythagorean Theorem on the first triangle by dividing it into two equal part and get the height is sqrt(4^2-2^2) or 2sqrt3. This multiplied by 4 then divided by 2 is 4sqrt3. The next triangle we can solve it's side lengths by writing equation(s). In this case the equation(s) would be sqrt(s^2-4^2)=4-x and sqrt(2x^2)=s where s is the side length of the second triangle and x is the length of the long lines until the triangle's vertices on it(not the lines with 4's written next to them). we can simplify the second equation to s=x*sqrt2 and substitute. sqrt(x*sqrt2-16)=4-x. We can square both sides to get x*sqrt2-16=x^2-8x+16, this can be simplified to x^2-(8+sqrt2)*x+32=0. This is how much I will do. I think you can try to solve from here.

ImDecentAtMathJan 18, 2022

+189

-1

I think it would just be (1/5)^4 * (4/5)^2 since the problems aren't said to be different in any way so your answer would probably be 16/15625.

I might've gotten this wrong and if so, sorry, but idk.

ImDecentAtMathJan 18, 2022

+189

-1

First we can simplify each sqrt on it's own. sqrt(28x) is 2sqrt(7x), sqrt(15x) can't be simplified, and sqrt(20x) is 2sqrt(5x). We can now seperate the integers from the squares and rewrite the original equation as (2*2) *sqrt(7x)*sqrt(15x)*sqrt(5x). We can see that sqrt(15x) is sqrt(5x)*sqrt(3) and the third sqrt is sqrt(5x) so them multiplied is (sqrt(5x))^2*sqrt3 where the sqrt gets canceled by the power of 2. So now we can simplify that to (2*2*5x)*sqrt(7x)*sqrt(3) which finally is (20x)*sqrt(21x).

The answer is (20x)*sqrt(21x).

Hopefully this helps and you understand.

ImDecentAtMathJan 18, 2022

Thx so much for the help, I understand now.

GuestJan 18, 2022

+37091

70 % off means you pay only 30 % of the original price

.30 * 70 = $................

ElectricPavlovJan 18, 2022

For this problem, the new price is 30% of the regular price since 100%-70% = 30%. Now we figure out that 30% of $70 is (3/10) * $70= $21.

The answer is $21.

GuestJan 18, 2022

+37091

First calculate the value of cos x ( sin^2 + cos^2 =1)

then sin 2x = 2 sin x cos x

ElectricPavlovJan 18, 2022

+2668

Use the quadratic formula $ {-b \pm \sqrt{b^2-4ac} \over 2a}$

You will get two imaginary solutions: 31/12 + (sqrt143i)/12

and: 31/12 - (sqrt143i)/12

BuilderBoiJan 18, 2022

We can write a system of equations as a= pi*r^2 and a+10 = pi*(2r)^2. This is where a is the area and r is the radius. The new radius is 2r. Now we can simplify the second equation to a+10=4(pi*r^2). We can subtract the first equation from the second equation and get 10=3(pi*r^2) and r=sqrt(10/(3*pi)) so 2r = 2sqrt(10/(3*pi)).

You can blame me if I'm wrong but I think this is right.

GuestJan 18, 2022

x²-7x=38

Solve the equation for x by finding a, b, and c to plug into the quadratic formula.

All equations of the form ax²+bx+c = 0 can be solved using the quadratic equation! :D It will give two solutions, one being negative and the other being positive. Let's solve!

x²-7x=38

Subtract 38 from both sides

x²-7x -38 = 0

This looks like the ax²+bx+c = 0 formula. A, b, and c are 1, -7, and -38. plug them into the quadratic formula

$x = {-(-7) \pm \sqrt{7^2-4*1*(-38)} \over 2(1)}$

simplify to get $x = {7 \pm \sqrt{201} \over 2}$

separate into two equations, since quadratic equations always have 2 answers--one positive and one negative

Answer 1: $x = {7 + \sqrt{201} \over 2}$

Answer 2: $x = {7 - \sqrt{201} \over 2}$

HesperosJan 18, 2022

You can rewrite the equation as x^2-7x-38=0, then use the quadratic formula, which leads to the answers x=[7-sqrt(201)]/2 and x=[7+sqrt(201)]/2.

I am pretty sure this is how you do it.

GuestJan 18, 2022

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