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seropon [69]
3 years ago
9

Please help!!!!!!! I’m stuck!

Mathematics
2 answers:
Elina [12.6K]3 years ago
7 0

notice the link above.... that's the relation of the sides in a triangle.


so, we start off by looking at the two smaller sides, nevermind the longest for a second.


on the second one listed there, 12, 5, 17, if we look at 12 and 5 only, how long can the last side be? well, can it be 17? well, 12+5 is 17, but only if 12 and 5 lie down flatly, only then we can put a 3rd side of 17 units long BUT, we're just going to end up with a flat-line, and lose the triangle, so, we can't have a longest leg of 17, because if we do, we don't have a triangle, just a flat-line, so no dice on that.


now checking the third one, 21, 7, 6, looking at 7 and 6, could be have a longest side of 21? wait a second!! 7+6 = 13, how in the dickens can we put a third side of 21 when 7 and 6 lying down flatly only give 13? so no dice there too.


anyhow the four one listed there is 9, 22, 11, again, looking at the small sides of 9 and 11, 9+11 = 20, how can we place a third side joining them that is 22 units long? no dice there either.


now looking at the first one listed, 10, 15, 24, we'll you already know.

vodomira [7]3 years ago
3 0

10 + 15 > 24

10 + 24 > 15

15 + 24 > 10

Answer

10 cm ,15 cm ,24 cm

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Nuetrik [128]

Answer:

-14

Step-by-step explanation:

Range is the difference between the lowest and highest values.

6 0
2 years ago
Xio just took a math test with 25 questions, each worth an equal number of points. The test is worth 100 points total. If a scor
Talja [164]
Each question is worth 4 points (100/25=4)
Therefore, when you divide 83 by 4, you get 20.75, as as he's aiming to get a B, this needs to be rounded up to 21.
Xio would need to get 21 questions right to get a B, hope this helps :)
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2 years ago
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Q # 6 please help to resolve
Rus_ich [418]
The numbers of choices in each category are multiplied together. We assume the order of paint choices matters: using color 1 in area A and color 2 in area B is not the same as using color 2 in area A and color 1 in area B.

P(7,2)*4*3*2 = 42*4*3*2 = 1008 ways

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P(n, k) = n!/(n-k)!
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4 0
2 years ago
Find the minimum and maximum value of the function on the given interval by comparing values at the critical points and endpoint
Kitty [74]

Answer:

maximum: y = 1

minimum: y = 0.

Step-by-step explanation:

Here we have the function:

y = f(x) =  √(1 + x^2 - 2x)

we want to find the minimum and maximum in the segment [0, 1]

First, we evaluate in the endpoints, which are 0 and 1.

f(0)  =√(1 + 0^2 - 2*0) = 1

f(1) = √(1 + 1^2 - 2*1) = 0

Now let's look at the critical points (the zeros of the first derivate)

To derivate our function, we can use the chain rule:

f(x) = h(g(x))

then

f'(x) = h'(g(x))*g(x)

Here we can define:

h(x) = √x

g(x) = 1 + x^2 - 2x

Then:

f(x) = h(g(x))

f'(x)  =  1/2*( 1 + x^2 - 2x)*(2x - 2)

f'(x) = (1 + x^2 - 2x)*(x - 1)

f'(x) = x^3 - 3x^2 + x - 1

this function does not have any zero in the segment [0, 1] (you can look it in the image below)

Thus, the function does not have critical points in the segment.

Then the maximum and minimum are given by the endpoints.

The maximum is 1 (when x = 0)

the minimum is 0 (when x = 1)

7 0
3 years ago
If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle? A) (x
Andrei [34K]

Answer:

\large\boxed{A.\ (x+6)^2+(y+10)^2=20}

Step-by-step explanation:

The standard form of an equation of a circle:

(x-h)^2+(y-k)^2=r^2

(h, k) - center

r - radius

We have the endpoints of the diameter of a circle (-8, -6) and (-4, -14).

The midpoint of a diameter is a center of a circle.

The formula of a midpoint:

\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)

Substitute:

x=\dfrac{-8+(-4)}{2}=\dfrac{-12}{2}=-6\\\\y=\dfrac{-6+(-14)}{2}=\dfrac{-20}{2}=-10

We have h = -6 and k = -10.

The radius is the distance between a center and the point on a circumference of a circle.

The formula of a distance between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Substitute (-6, -10) and (-8, -6):

r=\sqrt{(-8-(-6))^2+(-6-(-10))^2}=\sqrt{(-2)^2+4^2}=\sqrt{4+16}=\sqrt{20}

Finally we have

(x-(-6))^2+(y-(-10))^2=(\sqrt{20})^2\\\\(x+6)^2+(y+10)^2=20

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3 years ago
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