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2 years ago

2.) 3(10x + 9) - 9(3x - 5) SHOW WORK PLEASE

Mathematics

2 answers:

BigorU [14]2 years ago

6 0

Answer:

Let's simplify step-by-step.

3(10x+9)−9(3x−5)

Distribute:

=(3)(10x)+(3)(9)+(−9)(3x)+(−9)(−5)

=30x+27+−27x+45

Combine Like Terms:

=30x+27+−27x+45

=(30x+−27x)+(27+45)

=3x+72

avanturin [10]2 years ago

5 0

Answer:

3x+72

Step-by-step explanation:

Let's first simplify the equation 3(10x+9)−9(3x−5)

You want to distribute first

=(3)(10x)+(3)(9)+(−9)(3x)+(−9)(−5)

=30x+27+−27x+45

And then combine like terms:

=30x+27+−27x+45

=(30x+−27x)+(27+45)

=3x+72 (answer)

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2 years ago

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tamaranim1 [39]

Answer:

$\rm\displaystyle (x + 2{)}^{2} + (y + 5 {)}^{2} = {1}^{}$

$\rm\displaystyle (x - 7{)}^{2} + (y + 3{)}^{2} = { 7 }$

Step-by-step explanation:

<h3>QUESTION-1:</h3>

we are given the center and the redious of a circle equation

remember that,

$\rm\displaystyle E _{c} :(x - h {)}^{2} + (y - k {)}^{2} = {r}^{2}$

where (h,k) is the centre coordinate and r is redious of the circle

given that, h=-2 , k=-5 and r=1

Thus substitute:

$\rm\displaystyle (x - ( - 2){)}^{2} + (y - ( - 5) {)}^{2} = {1}^{2}$

simplify:

$\rm\displaystyle (x + 2{)}^{2} + (y + 5 {)}^{2} = {1}^{}$

<h3>QUESTION-2:</h3>

Likewise

$\rm\displaystyle E _{c} :(x - h {)}^{2} + (y - k {)}^{2} = {r}^{2}$

given that,h=7,k=-3 and r=√7

substitute:

$\rm\displaystyle (x - (7{))}^{2} + (y - ( - 3){)}^{2} = { \sqrt{7} ^{2} }$

simplify:

$\rm\displaystyle (x - 7{)}^{2} + (y + 3{)}^{2} = { 7 }$

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2 years ago

Identify the variation as direct, inverse, joint or combined. A = r²

myrzilka [38]

Answer : Direct Variation

Question :

Identify the variation as direct, inverse, joint or combined. A = $\pi r^2$

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2 years ago

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