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3 years ago

What is 1/5 multiplied by 4 in the simplest form?

Mathematics

2 answers:

natita [175]3 years ago

6 0

1/5 x 4 is going to be 4/5 because
1 x4 =4
5 1 =5
so, it equals 4/5!

soldi70 [24.7K]3 years ago

6 0

Answer:

4/5

Step-by-step explanation:

1/5 * 4 = 4/5

Hope this helped, and have a good day!

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Gala2k [10]

Not sure, but if you find out.... lmk. Bc this website is poor

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3 years ago

Read 2 more answers

Yolanda needs 5 pounds of ground beef to make lasagna for a family reunion. One package of ground beef weighs 2 1/2 pounds. anot

stealth61 [152]

You need to put the answers into improper fractions.
So 2 1/2 into 5/2
And 2 3/5 into 13/5
Now your fractions need the same denominator so multiply 13/5 by 2 and multiply 5/2 by 5
You should now have 25/10 and 26/10
Now turn them back into mixed fractions.
2 5/10
2 6/10
2+2=4
5/10 + 6/10 = 11/10 OR 1 1/10
4+ 1 1/10 = 5 1/10
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8 0

3 years ago

Two positive, consecutive, odd integers have a product of 143.

zhannawk [14.2K]

Answer:

<em>The answer is 13</em>

Step-by-step explanation:

Two positive and consecutive old numbers are x and x - 2.

=> x(x - 2) = 143

=> x^2 - 2x - 143 = 0

=> x^2 + 11x - 13x - 143 = 0

=> x(x + 11) - 13(x + 11) = 0

=> (x + 11)(x - 13) = 0

=> x = -11 (invalid)

or x = 13 (valid), the remaining number is 13 - 2 = 11

=> The two numbers are 11 and 13, and the greater number is 13.

Hope this helps!

7 0

4 years ago

Find the distance between 4 2/3 and −5 1/3 on a number line. Write your answer in the simplest form.

jeka94

Answer:

3/25

Step-by-step explanation:

5 0

4 years ago

Compute the second partial derivatives ∂2f ∂x2 , ∂2f ∂x ∂y , ∂2f ∂y ∂x , ∂2f ∂y2 for the following function. f(x, y) = 2xy (x2 +

blsea [12.9K]

Answer with step-by-step explanation:

We are given that a function

$f(x,y)=2xy(x^2+y^2)^2$

Differentiate partially w.r.t x

Then, we get

$\frac{\delta f}{\delta x}=2y(x^2+y^2)^2+8x^2y(x^2+y^2)=(x^2+y^2)(2x^2y+2y^3+8x^2y)=2(5x^2y+y^3)(x^2+y^2)$

Differentiate again w.r.t x

$\frac{\delta^2f}{\delta x^2}=2(10xy)(x^2+y^2)+4x(5x^2y+y^3)=20x^3y+20xy^3+20x^3y+4xy^3=40x^3y+24xy^3$

Differentiate function w.r.t y

$\frac{\delta f}{\delta y}=2x(x^2+y^2)^2+2xy\times 2(x^2+y^2)\times 2y$

$\frac{\delta f}{\delta y}=(x^2+y^2)(2x^3+2xy^2+8xy^2)=2(x^2+y^2)(x^3+5xy^2)$

Again differentiate w.r.t y

$\frac{\delta^2f}{\delta x^2}=2(2y)(x^3+5xy^2)+20xy(x^2+y^2)=4x^3y+20xy^3+20x^3y+20xy^3=24x^3y+40xy^3$

Differentiate partially w.r.t y

$\frac{\delta^2f}{\delta y\delta x}=2(2y(5x^2y+y^3)+(x^2+y^2)(5x^2+3y^2))=10x^4+36x^2y^2+10y^4$

$\frac{\delta^2f}{\delta y\delta x}=10x^4+36x^2y^2+10y^4$ $\frac{\delta^2f}{\delta x\delat y}=2(2x(x^3+5xy^2)+(3x^2+5y^2)(x^2+y^2))=10x^4+36x^2y^2+10y^4$

$\frac{\delta^2f}{\delta x\delat y}=10x^4+36x^2y^2+10y^4$

Hence, if f(x,y) is of class $C^2$ (is twice continuously differentiable), then the mixed partial derivatives are equal.

i.e $\frac{\delta^2f}{\delta y\delta x}=\frac{\delta^2f}{\delta x\delta y}$

8 0

4 years ago