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aleksklad [387]
3 years ago
6

If two angles are supplement of each other, then one of the angles must be acute

Mathematics
1 answer:
oksian1 [2.3K]3 years ago
7 0

Answer:

this is true

Step-by-step explanation:

a supplementary angle adds up to 180 degrees. two accute angles would equal to around 90 and two obtuse angles would add up to more than 180. two right angles are perpendicular and they could work but since the question asks about acute and obtuse, its true that one angle must be acute and the other must be obtuse for it to add up to 180.

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Please show work and help me pls
Gekata [30.6K]

Answer:

The answer is 207.

Step-by-step explanation:

If you do 1 ÷ 3 you'd get 3. Then move on too the 1 in 3105. 1 ÷ 1 equals 1. You do that with the 0 and the 5 in 3105. The last step too do is is do the same thing with the 5 in 15. Such as, 5 ÷ 3, 5 ÷ 1, 5 ÷ 0, and lastly 5 ÷ 5. Now if you understand just divide the two numbers in 15, which are 1 and 5 to get the quotient in the division problem. Hope this helps, I'm not the greatest at explaining!

7 0
3 years ago
Conner has $3 more than Chi
Tju [1.3M]
I think it’s 20 but I may be wrong
6 0
3 years ago
The sum of two integers is 23 and the positive difference of the same two integers is 13. What is the product of these two integ
Lady_Fox [76]

Answer:

A) 90

Step-by-step explanation:

"The sum of two integers is 23" becomes

a + b = 23

"the positive difference of the same two integers is 13" becomes

a - b = 13        (difference means subtract)

Now solve the system..

a + b = 23

a - b = 13            *use addition or elimination method, since b has opposite coefficients...

2a = 36             ( we add the two equations together)

   a = 18             (divide by 2 on both sides)

Since a = 18,   18 + b = 23, gives us b = 5.     (18 + 5 = 23)

ab = (18)(5) = 90

7 0
3 years ago
Consider the region, R, bounded above by f(x)=x2−6x+9 and g(x)=−3x+27 and bounded below by the x-axis over the interval [3,9]. F
Salsk061 [2.6K]

Answer:

22.5

Step-by-step explanation:

The region R contains every point of the plane with coordinate x between 3 and 9, and with coordinate y positive such that y < f(x) and y < g(x).

We can note that both f and g are positive on [3,9] because g is a decreasing linear function and g(9) = 0, thus g is positive in every other point of the interval, and f(x) = (x-3)^2 is always positive excpept when x = 3, where it reaches the value 0.

The interception of the graphs takes place for a value x such that f(x) = g(x).

We compute x^2-6x+9 = -3x + 27, thus x^2-6x+9-(-3x + 27) = x^2-3x -18 = 0.

The roots of that quadratic function are

r_1, r_2 = \frac{3 ^+_- \sqrt { 9 +72}}{2} = \frac{3^+_-9}{2} , thus r1 = 6, r2 = -3. We dont care about -3 because it is outside the interval, but we know that f and g graphs intersects on x = 6. Thus, we obtain, due to Bolzano Theorem:

  • On the interval [3,6), the function f in smaller because it takes the value 0 on x=3, while g is always positive.
  • On the interval (6,9]. the function g is smaller because it takes the value 0 on x=9, while f is always positive

Hence, the upper bound is f on the interval [3,6) and g on the interval (6,9]. While the lower bound is the 0 function.

We need to calculate the following integral, using Barrow's rule

\int\limits^6_3 {x^2-6x+9} \, dx + \int\limits^9_6 {-3x+27} \, dx = (\frac{x^3}{3} - 3x^2 + 9x) |^6_3 + (\frac{-3x^2}{2} + 27x)|^9_6 = \\  (18 - 9) + (121.5-108) = 22.5

As a result, the area of the region R is 22.5

6 0
3 years ago
I did it twice but i cant get it
Sladkaya [172]

Answer:

i dont no my guess is -4.5 sorry if it does not work

Step-by-step explanation:


3 0
3 years ago
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