All categories

3 years ago

Multiply 6 × 1/5 show work

2 answers:

8 0

Well first you have to add a 1 under the 6. So that's 6/1×1/5 and you multipy across. You multipy 1×6 and 5×1. That equals 6/5. Then you divide 6 and 5 so, your final answer is 1.2. Hope this helped you.

Aleksandr [31]3 years ago

4 0

6 x 1/5 = 1 1/5 or 1.2

You might be interested in

7) I am saving up for a video game. I currently have $28. If this is 30% of the

Snezhnost [94]

Answer:

93.3333333333 I think

7 0

2 years ago

Help Me with this question please

artcher [175]

Answer:

Step-by-step explanation:

The abs of y is 29 because 29 is positive

51 - 29 = 22

4 0

3 years ago

Read 2 more answers

The number of bees that visit a plant is 600 times the number of years the plant is alive, where t represents the number of year

zavuch27 [327]

C for sure, lock it down

7 0

3 years ago

MARSVECTORCALC6 3.4.020. My Notes A rectangular box with no top is to have a surface area of 64 m2. Find the dimensions (in m) t

ioda

Answer:

We would have

$l =w =\frac{8\sqrt{3}}{3} \\h = \frac{8\sqrt{3}}{6}$

where " l " is length, " w" is width and "h" is height.

Step-by-step explanation:

Step 1

Remember that

Surface area for a box with no top = $lw+2lh+2wh = 64$

where " l " is length, " w" is width and "h" is height.

Step 2.

Remember as well that

Volume of the box = $l*w*h$

Step 3

We can now use lagrange multipliers. Lets say,

$F(l,w,h) = lwh$

and

$g(l,w,h) = lw+2lh+2wh = 64$

By the lagrange multipliers method we know that

$\nabla F = \lambda \nabla g$

Step 4

Remember that

$\nabla F = (wh,lh,lw)$

and

$\nabla g = (w+2h,l+2h , 2w+2l)$

So basically you will have the system of equations

$wh = \lambda (w+2h)\\lh = \lambda (l+2h)\\lw = \lambda (2w+2l)$

Now, remember that you can multiply the first eqation, by "l" the second equation by "w" and the third one by "h" and you would get

$lwh = l\lambda (w+2h)\\\\lwh = w\lambda (l+2h)\\\\lwh = h\lambda (2w+2l)$

Then you would get

$l\lambda (w+2h) = w\lambda (l+2h) = h\lambda (2w+2l)$

You can get rid of $\lambda$ from these equations and you would get

$lw+2lh = lw+2wh = 2wh+2lh$

And from those equations you would get

$l = w =2h$

Now remember the original equation

$lw+2lh+2wh = 64$

If we plug in what we just got, we would have

$l^{2} + l^{2} + l^2 = 64 \\3l^{2} = 64 \\l = w = \frac{8\sqrt{3} }{3} \\h = \frac{8\sqrt{3} }{6}$

7 0

3 years ago

Please answer this if you can thank you :))

krek1111 [17]

<u>Answer:</u>

The variable that has the highest power is considered to be the degree of polynomials in an algebraic equation.

A column:

1) $4x^3$ .

The degree is 3.

2) $x^2+4$ .

The degree is 2.

3) $x-2$

The degree is 1.

B column:

1) $2x^2+1$

The degree is 2.

2) $x^2-x$

The degree is 2.

3) $x^3-5x^2+1$

The degree is 3.

A×B columns:

While Multiplying two terms in a equation, if the variables are same then multiply the constant value and sum the exponent value.

1) $(4x^3)(2x^2+1)$ .

= $8x^5+4x^3.$

The degree is 5.

2) $(x^2+4)(x^2-x)$ .

= $x^4-x^3+4x^2-4x$ .

The degree is 4.

3) $(x-2)(x^3-5x^2+1)$ .

$=x^4-5x^3+x-2x^3+15x^2-3.\\=x^4-7x^3+15x^2+x-3.$

The degree is 4.

3 0

3 years ago