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2 years ago

Can you please help I have no idea how to I’m stuck

Mathematics

1 answer:

Scrat [10]2 years ago

5 0

Report this answered this on accident ........

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Factorise 3+81x³y³ step by step explanation

Delicious77 [7]

Answer:

3(1+27x^3,y^3)

Step-by-step explanation:

3 is a GCF between the two terms, remove it and any other GCFs, in this case the term 3 doesn’t have any x’s or y’s. the only thing that can be removed from this equation is 3.

7 0

2 years ago

PLEASE HELP!! WILL MARK BRAINLIEST!!

katen-ka-za [31]

Answer:

C. Yes Devon should have evaluated f(3).

Step-by-step explanation:

She substituted the wrong value. For x - 3 to be a factor f(3) must equal 0.

f(3) = 3^3 + 3^2 - 10(3) + 8 = 27 + 9 - 30 + 8

= 14, so x- 3 is not a factor anyway but she did substitute the wrong value.

7 0

2 years ago

Read 2 more answers

Find the minimum y-value on the graph of y=f(x)<br><br> F(x) = 7x^2 + 7 - 6

Kisachek [45]

Answer:

The minimum y-value is

$y=-\frac{31}{4}$ or $y=-7.75$

Step-by-step explanation:

we have

$f(x)=7x^{2}+7x-6$

This is the equation of a vertical parabola open up

The vertex is the minimum y-value on the graph

Convert the equation into vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+6=7x^{2}+7x$

Factor the leading coefficient

$f(x)+6=7(x^{2}+x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)+6+1.75=7(x^{2}+x+0.25)$

$f(x)+7.75=7(x^{2}+x+0.25)$

Rewrite as perfect squares

$f(x)+7.75=7(x+0.5)^{2}$

$f(x)=7(x+0.5)^{2}-7.75$ ------> equation in vertex form

The vertex is the point (-0.5,-7.75)

therefore

The minimum is the point (-0.5,-7.75)

The minimum y-value is $y=-7.75=-7\frac{3}{4}=-\frac{31}{4}$

see the attached figure to better understand the problem

6 0

2 years ago

A tangent to the curve y=6x-x² cuts the x axis at point P. Find the coordinates of P. Help me calc the coordinates please

Ilya [14]

We know that the line is tangent at the point (2,8)

The derivative of the function at x=2 is

$f'(x) = 6-2x \implies f'(2) = 6-4=2$

So, the tangent line passes through the point (2,8) and has slope 2. The equation is

$y-8 = 2(x-2) \iff y = 2x+4$

This line crosses the x axis where y=0:

$0 = 2x+4 \iff 2x = -4 \iff x = -2$

6 0

2 years ago

Can someone please explain to me how to solve this?

vfiekz [6]

Answer:

Step-by-step explanation:

√(x^5) = x^(5/2) = (x^2)√x

√9x = (√9)(√x) = (3)√x

which means: ((x^2)√x) + ((3)√x)

since both have a √x, you can factor it out. for example...if you look at A, if you apply the distributive property you can see it expands back out. hmu if you need a different. explanation

3 0

3 years ago