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2 years ago

6

quadritlateral with two separate pairs of equal sides and my four sides cannot be equal, my equal sides are next to each other.

What am

1 answer:

Anarel [89]2 years ago

3 0

Check the picture below, notice those three quadrilaterals, bearing in mind that a square and rhombus both have equal sides all around.

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The density of silver is 10.5 g/cm3. find the volume of a silver bar if its mass is 950 g

Mademuasel [1]

950 g * 1 cm^3/10.5g = 950/10.5 cm^3 = 90.476 cm^3

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3 years ago

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ella [17]

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2 years ago

Please help me answer number 4! And if you can, please explain!

Leno4ka [110]

If Nathan does 9 minutes of push ups for 3 days, that's 9 divided by 3, which is 3. (9/3=3) So nathan does 3 minutes of push ups each day, so just multiply each day by 3 to find the correlation between the two.

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2 years ago

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3 years ago

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Multiply. Check picture.

VashaNatasha [74]

The answer is $3x^4-13x^3-x^2-11x+6$ .

Solution:

Use algebraic identity: $a^m\times a^n=a^{m+n}$

For example: $x^2\times x=x^{2+1}=x^3$

Given expression $(x^2-5x+2)$ and $(3x^2+2x+3)$ .

To multiply these equations.

$(x^2-5x+2)\times(3x^2+2x+3)$

$=x^2(3x^2+2x+3)-5x(3x^2+2x+3)+2(3x^2+2x+3)$

$=(3x^4+2x^3+3x^2)+(-15x^3-10x^2-15x)+(6x^2+4x+6)$

$=3x^4+2x^3+3x^2-15x^3-10x^2-15x+6x^2+4x+6$

Combine like terms together.

$=3x^4+(2x^3-15x^3)+(3x^2-10x^2+6x^2)-15x+4x+6$

$=3x^4-13x^3-x^2-11x+6$

$(x^2-5x+2)\times(3x^2+2x+3)=3x^4-13x^3-x^2-11x+6$

Hence the answer is $3x^4-13x^3-x^2-11x+6$ .

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3 years ago