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2 years ago

Help plzzzzzzzzzzz the question in the pic and I will give brainliest to the right answer

Mathematics

2 answers:

Lelechka [254]2 years ago

8 0

Answer:1/3

Step-by-step explanation:

exis [7]2 years ago

8 0

Answer:

the answer is A hope this helps you

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B12x%7D%7B2x-3x-10%7D%20-%20%5Cfrac%7Bx%7D%7B%28x-5%29%7D" id="TexFormula1" title="\f

uysha [10]

Answer:

Step-by-step explanation:

$\frac{12x}{2x-3x-10}-\frac{x}{x-5}=\frac{12x}{-x-10}-\frac{x}{x-5}\\\\=\frac{12x}{-(x+10)}-\frac{x}{x-5}\\\\=\frac{-12x}{x+10}-\frac{x}{x-5}\\\\=\frac{-12x*(x-5)}{(x+10)*(x-5)}-\frac{x*(x+10)}{(x-5(x+10)}\\\\=\frac{-12x*x-5*(-12x)}{(x+10)(x-5)}-\frac{x*x+x*10}{(x-5)(x+10)}\\\\=\frac{-12x^{2}+60x}{(x+10)(x-5)}-\frac{x^{2}+10x}{(x-5)(x+10)}\\\\=\frac{-12x^{2}+60x-(x^{2}+10x)}{(x+10)(x-5)}\\\\=\frac{-12x^{2} + 60x - x^{2}-10x}{x^{2}+5x-50}\\\\=\frac{-13x^{2}+50x}{x^{2}+5x-50}$

8 0

2 years ago

Read 2 more answers

Please answer this correctly

Klio2033 [76]

Answer:

x =1

Step-by-step explanation:

Assuming the figures are similar, we can use ratios to solve

x 2

---- = -----

4 8

Using cross products

8x = 2*4

8x = 8

Divide by 8

8x/8 = 8/8

x=1

6 0

2 years ago

A kilogram of seawater contains 1/40 kg of salt. How much salt is there in two buckets of seawater, which contain 500 liters of

never [62]

Two buckets of seawater contain $13\frac{28}{40}$ kg of salt.

Step-by-step explanation:

Amount of salt in 1 kg water = $\frac{1}{40}\ kg$

It is given that;

1 liter = $1\frac{12}{125} = \frac{137}{125}\ kg$

500 liters = $500*\frac{137}{125} = \frac{68500}{125}$

500 liters = 548 kilogram

As,

1 kg of seawater = $\frac{1}{40}\ kg$ of salt

548 kilograms = $\frac{1}{40}*548$

548 kilograms = $13\frac{28}{40}$ kilograms of salt

Two buckets of seawater contain $13\frac{28}{40}$ kg of salt.

Keywords: unit rate, multiplication

Learn more about multiplication at:

#LearnwithBrainly

5 0

2 years ago

Find domain of y=rad20-6x

Evgesh-ka [11]

Domain, on an "even root" context, means, an even root cannot have a negative radicand, since say for example $\bf \sqrt{-25}\ne -5\qquad why?\implies (-5)(-5)=25$

so... you end up with an "imaginary value"

so... for the case of 20-6x
"x", the domain, or INPUT
can afford to have any value, so long it doesn't make the radicand negative

to check for that, let us make the expression to 0, and see what is "x" then

$\bf 20-6x=0\implies 20=6x\implies \cfrac{20}{6}=x\implies \cfrac{10}{3}=x$

now, if "x" is 10/3, let's see $\bf \sqrt{20-6\left( \frac{10}{3} \right)}\implies \sqrt{20-20}\implies \sqrt{0}\implies 0$

now, 0 is not negative, so the radicand is golden

BUT, if "x" has a value higher than 10/3, the radicand turns negative
for example $\bf x=\frac{11}{3}
\\\\\\

\sqrt{20-6\left( \frac{11}{3} \right)}\implies \sqrt{20-22}\implies \sqrt{-2}$

so.. that's not a good value for "x"

thus, the domain, or values "x" can safely take on, are all real numbers from 10/3 onwards, or to infinity if you wish

5 0

2 years ago

Please help!

ch4aika [34]

Answer:

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers