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3 years ago

Please help! 20 points + brainliest

Mathematics

1 answer:

klemol [59]3 years ago

7 0

For the first one we divide every element by sqrt3

we get

q+2/sqrt3 =sqrt5/sqrt3

q=(sqrt5-2)/sqrt3

q=sqrt3*sqrt5-2*sqrt3/3

q=sqrt15-2sqrt3/3

which is the third option

For the second one

sqrt(2y-5) +4=8

first we move the 4 to the right

sqrt(2y-5)=4

we remove the square root by bringing to the power of two

2y-5=16

2y=21

y=21/2

y=10 1/2

third option

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Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

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2 years ago

1. <br><br> Name the postulate or theorem you can use to prove the triangles congruent.

zzz [600]

Answer:

phthagoras

Step-by-step explanation:

7 0

3 years ago

A straight line, L, is perpendicular to the straight line y=2x-1 and passes through the point (8,1). Find an equation of L

luda_lava [24]

Let's call that other line with the eqn. y=2x-1 line M.
This line is in slope-intercept form, which means that it is written in the form y=mx+b where m is the slope and b is the y-intercept.
This means that the slope of line M is 2.

Perpendicular lines have slopes which are opposite reciprocals.
(that is to say, if you flipped the fraction and changed the sign)

Of course, 2 isn't a fraction, but it's implied as 2/1.
The opposite reciprocal would then be -1/2.
Let's plug this into our slope-intercept form equation for line L.

y = mx + b
y = -1/2x + b

Of course, we still need to find that y-intercept. (y when x = 0)
To do this, we need to interpret the slope.
Slopes are rise over run, so m = -1/2 means a change of -1 in y = 2 in x.
Let's take a point we know is in our line, (8, 1).
To find that y-intercept, we want x to be 0.
To do this, we'd have to subtract 8 from x.
And according to our slope, this means adding 4 to y.
Our y-intercept is at (0, 5), with the value b that we use being just 5.

$\boxed{y=-\frac{1}2x+5}$

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2 years ago

The winner in a vote for class president received 3/4 of the 240 votes. How many votes did the winner receive

zubka84 [21]

The winner received 180 votes.

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3 years ago

How many grams of barium chloride are there in 0.200 moles

Anni [7]

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