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2 years ago

6

Can you please answer number 82

1 answer:

hichkok12 [17]2 years ago

8 0

A.) the answer is 2x+y+7
b.)the answer is 2x+7y-9
c.)the answer is 24x+48
d.)the answer is 24x+6
e.)the answer is 5x+9

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Multiply and simplify. 2x4yx⋅6xy2z3

OLEGan [10]

Answer:

The given expression $2x^4yx \times 6xy^2z^3$ on multiplying is $12x^6y^3z^3$

Step-by-step explanation:

Consider the given two expressions $2x^4yx$ and $6xy^2z^3$

We have multiply both expressions,

$2x^4yx \times 6xy^2z^3$

To multiply two terms first multiply constant numbers that is 6 × 2 = 12

For x , y and z apply property of exponent,

$x^a\cdot x^b=x^{a+b}$

Then power of x together will be,

$x^4\cdot x\cdot x=x^{4+1+1}=x^6$

Similarly for y powers,

$y \cdot y^2=y^{1+2}=y^3$

Since first term do not have any expression for z so it will remain same.

Thus, the given expression on multiplying become,

$2x^4yx \times 6xy^2z^3$ is $12x^6y^3z^3$

7 0

2 years ago

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Let r and s be the roots of x^2 - 6x + 2 = 0. Find (r - s)^2.

Sphinxa [80]

Answer: $(r-s)^{2} = 28$

Step-by-step explanation:

$x^{2} -6x+9=-2+9$
$(x-3)^{2}=7$
$(x-3)= -\sqrt{7}$ or $\sqrt{7}$ <-- there's no plus/minus sign.

$(x-3) = -\sqrt{7}$
$x = 3 -\sqrt{7}$ <-- root.

$(x-3) = \sqrt{7}$
$x = 3+\sqrt{7}$ <-- root.

$[(3-\sqrt{7})-(3+\sqrt{7})]^{2}$
$(3-\sqrt{7} -3 - \sqrt{7})^{2}$ <-- don't forget to distribute the negative.
$(-2\sqrt{7})^{2}$
$(4)(7)$
$28$

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2 years ago

Which is true regarding chords and diameters of circles?

Gelneren [198K]

Answer:

Both chords and diameters have two endpoints on a circle. Diameters must intersect the center of a circle.

Step-by-step explanation:

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2 years ago

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ⓌⒽⓎ ⓜⓊⓈⓉ ⓉⒽⒺ ⓇⒺⓜⒶⒾⓃⒹⒺⓇ ⒷⒺ ⓁⒺⓈⓈ ⓉⒽⒶⓃ ⓉⒽⒺ ⒹⒾⓋⒾⓈⓄⓇ

Lynna [10]

Answer:

The remainder is the number that is left over after dividing. It is always less than the divisor. The remainder is an essential part of the answer and is shown in the quotient preceded by the letter R.

Step-by-step explanation:

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2 years ago

Please help I’m in desperate need of it I will give brainliest

Romashka [77]

Answer:

I think its a student loan

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