PLS HELP, WILL GIVE BRAINLIEST

Answers

Answer 1

Answer: 11x-2 that’s the answer

Answer 2

Answer: the answer should be 11x-2

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What is the product? assume x>0 (√3x + √5) (√15x +2√30

Answers

The first step for solving this expression is to multiply the parenthesis.
$√(45) x^(2) + 2√(90) x + √(75)x + 2√(150)$
Simplify the first radical.
$3√(5) x^(2) + 2√(90) x + √(75)x + 2√(150)$
Simplify the second radical.
$3√(5) x^(2) + 2X3√(10) x + √(75)x + 2√(150)$
Simplify the third radical.
$3√(5) x^(2) + 2X3√(10) x + 5√(3)x + 2√(150)$
Simplify the final radical.
$3√(5) x^(2) + 2X3√(10) x + 5√(3)x + 10√(6)$
Lastly,, calculate the product of 2 × 3 $√(10)$ x to get your final answer.
$3√(5) x^(2) + 6√(10) x + 5√(3)x + 10√(6)$
Let me know if you have any further questions.
:)

Answer: B: 3x√5+6√10x+5√3x+10√6x

Step-by-step explanation:

Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, if the two machines together produce 5 4 w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets.A. 4B. 6C. 8D. 10E. 12

Answers

C. Hope this helps :)

Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3

Answers

$(tan^2x)/(1+cot^2x)+(cot^2x)/(1+tan^2x)=sec^2x\ cosec^2x-3\n\nL=(tan^2x(1+tan^2x)+cot^2x(1+cot^2x))/((1+cot^2x)(1+tan^2x))=(tan^2x+tan^4x+cot^2x+cot^4x)/(1+tan^2x+cot^2x+tanxcotx)\n\n=(tan^2x+cot^2x+tan^4x+cot^4x)/(1+tan^2x+cot^2x+1)=(tan^2x+2+cot^2x+tan^4x-2+cot^4x)/(tan^2x+cot^2x+2)$

$=((tanx+cotx)^2+(tan^2x-cot^2x)^2)/((tanx+cotx)^2)=((tanx+cotx)^2)/((tanx+cotx)^2)+((tan^2x-cot^2x)^2)/((tanx+cotx)^2)\n\n=1+((tanx-cotx)^2(tanx+cotx)^2)/((tanx+cotx)^2)=1+(tanx-cotx)^2\n\n=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\n\n=\left((sinx)/(cosx)\right)^2+\left((cosx)/(sinx)\right)^2-1=(sin^2x)/(cos^2x)+(cos^2x)/(sin^2x)-1=(sin^4x+cos^4x)/(sin^2x\ cos^2x)-1$

$=((sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x)/(sin^2x\ cos^2x)-1\n\n=((sin^2x+cos^2x)^2-2sin^2x\ cos^2x)/(sin^2x\ cos^2x)-1=(1^2-2sin^2x\ cos^2x)/(sin^2x\ cos^2x)-1\n\n=(1)/(sin^2x\ cos^2x)-(2sin^2x\ cos^2x)/(sin^2x\ cos^2x)-1=(1)/(sin^2x)\cdot(1)/(cos^2x)-2-1\n\n=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

The sum of two numbers is 92. One fifth of the first number added to one half of the second number is 34. Find the two numbers.

Answers

x + y = 92

x/5 + y/2 = 34 / both sides *10
10.x/5 + 10.y/2 = 340
2x + 5y = 340
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x + y = 92 /both sides *-2
2x + 5y = 340

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-2x -2y = -184
+ 2x +5y = 340
3y = 156
y = 32

x + y = 92
x + 32 = 92
x = 60

$x+y=92\n (1)/(5)x+(1)/(2)y=34\n\n x=92-y\n 2x+5y=340\n\n 2(92-y)+5y=340\n\ 184-2y+5y=340\n 3y=156\n y=52\n\n x+52=92\n x=40$

AB and BA name the same ray.
Always
Never
Sometimes

Answers

Answer:

AB in terms of rays, can never be BA.

Step-by-step explanation:

AB and BA name the same ray.

No, this is never true. A ray starts from a point and move in a single direction to infinity.

So, if starting point is A and ray is moving towards B, so it will continue moving towards the points further from B.

Therefore, AB as in terms of rays, can never be BA.

They are never the same ray. ray AB begins at point A and goes through point B to infinity and ray BA begins at point B and goes through point B to infinity

PLEASE HELP! TRYING TO FINISH MATH EARLY! PLEASE ACTUALLY ANSWER, AND NOT JUST TAKE THE POINTS! Thank You! ;3(05.06 MC)
Graph the system of inequalities presented here on your own paper, then use your graph to answer the following questions:

y > −4x − 1
y is less than 3 over 2 times x minus 1

Part A: Describe the graph of the system, including shading and the types of lines graphed. Provide a description of the solution area.

Part B: Is the point (−1, −1) included in the solution area for the system? Justify your answer mathematically.