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AURORKA [14]
3 years ago
8

Easy// need help please

Mathematics
1 answer:
ASHA 777 [7]3 years ago
6 0

Answer:

41

Step-by-step explanation:

Yes, it is easy.

Square root of (40*40+9*9) = 41

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A ferris wheel goes around once every 20 seconds . How many times will a rider be ag thetop during an 8-minuteride?
allsm [11]

Answer:

The answer is B.

Step-by-step explanation:

There are three 20 second time periods in 1 minute. Three times eight is equal to 24.  

8 0
3 years ago
9. Which transformation will place the trapezoid onto
galben [10]
Answer: B

The rotation will go back to the place where the trapezoid started
8 0
3 years ago
What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0
Tom [10]

Answer:

x=\frac{-7-\sqrt{5}}{2} or x=\frac{-7+ \sqrt{5}}{2}

Step-by-step explanation:

The given equation is

(2x+3)^2+8(2x+3)+11=0

Let us treat this as a quadratic equation in (2x+3).

where a=1,b=8,c=11

The solution is given by the quadratic formula;

(2x+3)=\frac{-b\pm \sqrt{b^2-4ac} }{2a}

We substitute these values into the formula to obtain;

(2x+3)=\frac{-8\pm \sqrt{8^2-4(1)(11)} }{2(1)}

(2x+3)=\frac{-8\pm \sqrt{64-44} }{2}

(2x+3)=\frac{-8\pm \sqrt{20} }{2}

(2x+3)=\frac{-8\pm2\sqrt{5} }{2}

(2x+3)=-4\pm \sqrt{5}

(2x+3)=-4-\sqrt{5} or (2x+3)=-4+ \sqrt{5}

2x=-3-4-\sqrt{5} or 2x=-3-4+ \sqrt{5}

2x=-7-\sqrt{5} or 2x=-7+ \sqrt{5}

x=\frac{-7-\sqrt{5}}{2} or x=\frac{-7+ \sqrt{5}}{2}

5 0
3 years ago
I NEED HELP PLEASE<br> directions:Work and Solution
Volgvan

Answer:

2(x + 4) / 6(x² - 3x - 28)

Step-by-step explanation:

Area of a rectangle = length × width

Length = 2/(x² - 3x - 28)

Width = x² - 16/6x - 24

= (x + 4)(x - 4) / 6(x - 4)

= (x + 4) / 6

Area of a rectangle = length × width

= 2/(x² - 3x - 28) × (x + 4) / 6

= 2(x + 4) / (x² - 3x - 28)6

= 2(x + 4) / 6x² - 18x - 168

= 2(x + 4) / 6(x² - 3x - 28)

Area of a rectangle =

2(x + 4) / 6(x² - 3x - 28)

4 0
2 years ago
Explain the tangent line problem
ale4655 [162]

The Tangent Line Problem  1/3How do you find the slope of the tangent line to a function at a point Q when you only have that one point? This Demonstration shows that a secant line can be used to approximate the tangent line. The secant line PQ connects the point of tangency to another point P on the graph of the function. As the distance between the two points decreases, the secant line becomes closer to the tangent line.
8 0
3 years ago
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