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2 years ago

PLEASE help please I really need this! I’ll Mark the brainiest ! :)

Mathematics

2 answers:

BaLLatris [955]2 years ago

8 0

(-6x, 4) now give me brainiest

zhannawk [14.2K]2 years ago

5 0

-6x,4 very easyyyyytificjdhcbfididnc

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sandra and her friend went to the candy store. each of them purchased a bag of jelly beans. sandra's bag weighed 1.25 pounds. He

Nuetrik [128]

You would subtract.

The question is asking who has more, so to find out, you need to take away the smaller amount from the bigger amount. (1.25-1.05)

So Sandra bought more candy. Hope this helps!

3 0

3 years ago

Read 2 more answers

Someone please be awesome and help me please :(

solong [7]

Answer:

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

$x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$

Step-by-step explanation:

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

They wanted to complete the square so they took the thing in front of x and divided by 2 then squared. Whatever you add in, you must take out.

$x^2+\frac{b}{a}x+(\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

Now we are read to write that one part (the first three terms together) as a square:

$(x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

I don't see this but what happens if we find a common denominator for those 2 terms after the square. (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.

$(x+\frac{b}{2a})^2+\frac{4ac}{4a^2}-\frac{b^2}{4a^2}=0$

They put it in ( )

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

I'm going to go ahead and combine those fractions now:

$(x+\frac{b}{2a})^2+(\frac{-b^2+4ac}{4a^2})=0$

I'm going to factor out a -1 in the second term ( the one in the second ( ) ):

$(x+\frac{b}{2a})^2-(\frac{b^2-4ac}{4a^2})=0$

Now I'm going to add (b^2-4ac)/(4a^2) on both sides:

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

I'm going to square root both sides to rid of the square on the x+b/(2a) part:

$x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}$

$x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}$

Now subtract b/(2a) on both sides:

$x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$

Combine the fractions (they have the same denominator):

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

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3 years ago

A display that uses bars to show data is an

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Histogram
A histogram is a graphic display of quantitative variables that uses bars to represent the frequency of the count of the data in each interval. A boxplot is a graphic display of quantitative data that demonstrates the five-number summary.

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3 years ago

I need help with linear functions can anybody help me?

NNADVOKAT [17]

I can help you! Just let me how what the questions are

7 0

3 years ago

Determine all critical points the the function:

romanna [79]

$\bf y=x^2-32\sqrt{x}\implies \cfrac{dy}{dx}=2x-32\cdot \cfrac{1}{2}x^{-\frac{1}{2}}\implies \cfrac{dy}{dx}=2x-\cfrac{32}{2\sqrt{x}}
\\\\\\
\cfrac{dy}{dx}=2x-\cfrac{16}{\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{2\sqrt{x^3}-16}{\sqrt{x}}\\\\
-----------------------------\\\\$

$\bf 0=\cfrac{2\sqrt{x^3}-16}{\sqrt{x}}\implies 0=2\sqrt{x^3}-16\implies 16=2\sqrt{x^3}
\\\\\\
8=\sqrt{x^3}\implies 2^3=\sqrt{x^3}\implies (2^3)^2=x^3\implies (2^6)^{\frac{1}{3}}=x
\\\\\\
2^{\frac{2}{1}}=x\implies \boxed{4=x}\\\\
-----------------------------\\\\
\textit{the other critical point at }\sqrt{x}=0\implies \boxed{x=0}$

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3 years ago