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3 years ago

Which expression is equivalent to ? I need help quick please!!!

Mathematics

1 answer:

emmasim [6.3K]3 years ago

8 0

Hi!

Your answer is the third option!

I hope this helps!

Feel free to give brainliest of you think I deserve it!

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Please help me... Question is in the picture

zmey [24]

The answer is “122“ 180-58 = 122

5 0

3 years ago

The line of symmetry of the parabola whose equation is y = ax^2 -4x + 3 is

alex41 [277]

Answer:

B. -1

Step-by-step explanation:

If we have a parabola whose equation is:

$y=ax^{2} +bx+c$

The line of symmetry is calculated as:

$x=\frac{-b}{2a}$

Now, we have the equation $y=ax^{2}-4x+3$ and the line of symmetry is $x=-2$

Where:

$b=-4\\c=3$

So, we can replace $b$ by -4 and $x$ by -2 and solve for $a$ using the following equation as:

$x=\frac{-b}{2a}\\-2=\frac{-(-4)}{2a}\\-2(2)a=4\\-4a=4\\a=-1$

It means that the equation of the parabola is equal to:

$y=-1x^{2}-4x+3$

7 0

3 years ago

Cory is building an enclosure with recycled cardboard for her collection of model cars. Her design is a shown below.

UkoKoshka [18]

Answer:

B.25

Explanation:

The figures are similar. This means that corresponding sides have the same ratio.

This helps us set up a proportion.

We know that ML corresponds with DC, which gives us the ratio

8/40

EF will correspond with NH; this gives us the ratio

5/x

The proportion is then

8/40 = 5/x

Cross multiplying, we have

8(x) = 40(5)

8x = 200

Divide both sides by 8:

8x/8 = 200/8

x = 25

5 0

4 years ago

Read 2 more answers

Divide 2x^4 -5x^3+3-1 by 2x -1 <br> please help and please show the steps on how you did it

MaRussiya [10]

The answer below (I need 20 characters to respond so just ignore the words between the parenthesis on this response)

4 0

4 years ago

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

3 0

3 years ago