Two angles are comlementary. the first angle measures x and the second angle measures 3x-10. find x?

Mathematics

1 answer:

ryzh [129]3 years ago

6 0

X=25

4x-10=90
Add 10 to both sides
4x=100
Divide both sides by 4

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90 PTS HELP

nadya68 [22]

Answer:

$\frac{(\frac{3}{5}*1000) }{6}$

Step-by-step explanation:

First, let's find out how many words Maria has left to write. If she has already finished 2/5 of her essay, she has 3/5ths left to write. We can do 3/5 times 1000 to find out how many words are left.

$\frac{3 }{5} * \frac{1000}{1}$

$\frac{3000}{5}$

$600$

Thus, there are 600 words left to write.

Now to find the amount of time, we can find how take to write 600 words at a pace of 6 words per minute by dividing 600 by 6.

$\frac{600}{6}$

$100$

It will take 100 minutes for Maria to write the rest of her essay. Now let's think about an expression that tells us this information. It sounds like your questions is supposed to have options, but I doesn't seem like you've posted them, no worries however.

So first, the amount she has left to write.

$(\frac{3}{5} * 1000)$

We know this is 600, but we don't have to simplify it for the expression.

Now thinking back to how we found the time, we can divide this '600' by 6.

$\frac{(\frac{3}{5}*1000) }{6}$

This is the expression that shows the amount of time it will take to write the rest of the essay.

$\frac{((1-(\frac{2}{5})*1000) }{6}$ is also a valid solution.

3 0

3 years ago

Read 2 more answers

What are the steps to solve the quadratic equation x^2+4x-6=0

makvit [3.9K]

$METHOD\ 1:\\\\Use:\\\\(*)\qquad(a+b)^2=a^2+2ab+b^2\\---------------------\\\\x^2+4x-6=0\qquad\text{add 6 to both sides}\\\\x^2+4x=6\\\\x^2+2(x)(2)=6\qquad\text{add}\ 2^2\ \text{to both sides}\\\\\underbrace{x^2+2(x)(2)+2^2}_{(*)}=6+2^2\\\\(x+2)^2=6+4\\\\(x+2)^2=10\Rightarrow x+2=\pm\sqrt{10}\qquad\text{subtract 2 from both sides}\\\\\boxed{x=-2-\sqrt{10}\ \vee\ x=-2+\sqrt{10}}$

$METHOD\ 2:\\\\\text{use the quadratic formula:}\\\\ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta>0,\ \text{the equation has two solutions}\ x=\dfrac{-b\pm\sqrt\Delta}{2a}\\\\if\ \Delta=0,\ \text{then the equation has one solution:}\ x=\dfrac{-b}{2a}\\\\if\ \Delta$

$x^2+4x-6=0\\\\a=1,\ b=4,\ c=-6\\\\\Delta=4^2-4(1)(-6)=16+24=40>0\\\\\sqrt\Delta=\sqrt{40}=\sqrt{4\cdot10}=\sqrt4\cdot\sqrt{10}=2\sqrt{10}\\\\x_1=\dfrac{-4-2\sqrt{10}}{2(1)}=\dfrac{-4}{2}-\dfrac{2\sqrt{10}}{2}=-2-\sqrt{10}\\\\x_2=\dfrac{-4+2\sqrt{10}}{2(1)}=\dfrac{-4}{2}+\dfrac{2\sqrt{10}}{2}=-2+\sqrt{10}$