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Well I just get some math problems from the AMC8 or AMC10 to do when I don't get the problem because I try to practice my stamina and accuracy

I'm not sure how I got good at math because I learned 2-1=1 when I was 3 years old and 1+1=2 when I was 2 and a half I began to like math after that and I worked hard on math but I'm not sure about how to help but just persevere(don't give up) and you do great or maybe you'll you might even have a cahnce to go to MATHCOUNTS and win a first place?

4/5*1195=956 bcause we divide by 5 to get 239 for the dnominator and then multiply by 4 for th numerator to get 956

Sorry, I do not really have any great tips.

I think one of the troubles is that maths classes are often too hard. 

If something appears to be unachievable it cannot be fun.

Try to concentrate on the basics. Get success with those. Build up your confidence.

If you had a worthwhile tutor that is how they would be helping you to achieve.  Building confidence works wonders for ultimate success. Without proper guidance, I expect this will be difficult for you to achieve so that is why I said I really do not have any great tips for you.  But maybe it will give you something useful to think about. 

Like so:

I'll let you take it from here.

Four-fifths of 1195?

Hello Guest!

\(\frac{4}{5}\times 1195=\frac{2\cdot 2\cdot 5\cdot 239}{5}=2^2\cdot 239=\color{blue}956\)

  !

I need help with this math question with how to solve answer:

The percentage voters who voted in this election was 55%, up 179 the percentage that voted at the last election.

Hello Danette!

Last time voted x %.

Then applies

\(\frac{x}{100}\cdot 179=55\%\\ x=\frac{100\cdot55\%}{179}\)

\(x=30.73\ \%\)

The percentage that voted at the last election is 30.73%.

  !

The perimeter of a rhombus is 48 and the sum of the lengths of its diagonals is 26.
What is the area of the rhombus?

Hello Guest!

A diagonal is x.

\((\frac{x}{2})^2+(\frac{26-x}{2}^2)=(\frac{48}{4})^2\)

\(\frac{x^2}{4}+\frac{676-52x+x^2}{4}=\frac{2304}{16}=\frac{576}{4}\\\ x^2+676-52x+x^2=576\\ 2x^2-52x+100=0\\x^2-26x+50=0\)

\(x=13\pm\sqrt{169-50}\\ x=13\pm\sqrt{119}\)

\(x_1=23.91\\x_2=2.091\)

\(A=4\times (\frac{x}{2}\cdot\frac{26-x}{2})/2\\ A=4\times \frac{2.901\cdot (26-2.901)}{2\cdot 2\cdot2}\)

\(A=25\)

\(A=4\times \frac{x(26-x)}{2\cdot 2\cdot 2}\\ A=4\times \frac{23.91\cdot (26-23.91)}{2\cdot 2\cdot 2}\)

\(A=25\)

  !

Sorry for the number of typos! [20 * 21] / 2 =210. There are other minor typos. Unfortunately, I cannot "edit" them, but the solution is correct.

At n = 19 dropping them and eating are DEAD EVEN at 190 points each. Therefore, the least number of froods for which dropping will earn more points than eating then, is when n = 20, at which number [20 x 21] = 210 points for dropping them versus 20 * 10 =200 points for eating them.

Yes, I do admit I have my flaws.

Im not the best at math.

Please understand that I do try to give helpful hints.

if you feel that giving more elaborate hints or just the answer I will do so.

thank you so much! the youtube video was pretty helpful :)

I think it's wrong

ok here 

The area of the entire large rectangle is (x+7)(x+5). The hole is (2x-3)(x-2). Not including the area of the hole, the area of the large rectangle would be (x+7)(x+5)-(2x-3)(x-2).

(x+7)(x+5)-(2x-3)(x-2)

=x^2+12x+35-2x^2-7x+6

=-x^2+19x+29

This is simple. x^2-4x+1 is not factorable so we plug it into the quadratic equation.

\(x = \frac{-b \pm \sqrt{b^2-4ac}} {2a}\\ x=\frac{4\pm\sqrt{(-4)^2-4(1)(1)}}{2(1)}\\ x=\frac{4\pm\sqrt{12}}{2}\\ x=\frac{4\pm2\sqrt{3}}{2}\\ x=2\pm\sqrt{3}\)

Cube the two roots.

\((2+\sqrt{3})^3+(2-\sqrt{3})^3\\ =\boxed{52}\)

We first convert 4x+2y=8 into slope intercept form.

4x+2y=8

2y=-4x+8

y=-2x+4

Parallel lines have the same slope, so -2 y=-2x will remain the same for the parallel line's equation. However, the y intercept will not and we let it be the variable b. We plug in the given point, (0,1).

y=-2x+b

1=0+b

b=1

So the equation is y=-2x+1.

Hi there, I'd like to help but can you please show the figures. It's easy, just take a screenshot, and select the "Image" button here in the text box and upload it. Thanks!

Create an equation and solve.

\(2x+10=\frac{1}{2}(5x+30)\\ 2x+10=\frac{5x+30}{2}\\ 4x+20=5x+30\\ x=-10\)

1. The answer is 2010 because finding the inverse is finding the f(x) that results in 2010 when plugged into x^2+3x^2+3x+1. When we get the inverse, we then have f(inverse), which is plugging it straight back into x^2+3x^2+3x+1. Then the result is the same.

2. f(f(1)

=f(1/1+2)

=f(1/3)

=1/(1/3+2)

=1/7/3

=3/7

We subtract the area of the bigger square from the area of the smaller square.

(x+7)(x+5) = x^2+12x+35

The larger rectangle has an expanded area of 2x^2 -7x + 6.

We subtract (2x^2 - 7x + 6) - (x^2+12x+35)

This gives us x^2 - 19x - 29

By the Pythagorean Theorem, the hypotenuse is 6*sqrt(11).