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3 years ago

Plzzz help me i can not find it nowhere

Mathematics

1 answer:

ratelena [41]3 years ago

6 0

Answer: D. 50

Step-by-step explanation:

the sum of opposite angles of a cyclic quadrilateral are always 180 degrees

80 +2x= 180

2x= 180 - 80

x= 100/2

x= 50

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Stephanie drew this triangle how many obtuse triangle’s are the triangle have

Mice21 [21]

The triangle has no obtuse angles.

Obtuse angles are greater than 90 degrees, or are wider than a right angle. These are all acute angles in this diagram because they are all less than 90 degrees.

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3 years ago

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Exponential functions algebra 1

meriva

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5 0

3 years ago

Find the point P on the line yequals=66x that is closest to the point (74 comma 0 )(74,0). What is the least distance between P

tekilochka [14]

Answer:

The point P that is closest to the point (74, 0) is P(7.4, 22.2), and the least distance between P and (74, 0) is 70.2

Step-by-step explanation:

Given

y = 66x

And a point A(74, 0)

We are required to find the point on the line that is closest to A, and then find the distance between these points.

Let the point be P(s, t)

Because it's coordinate satisfies the given equation

y = 3x

Let t = 3s, then we have

p(s, t) = P(s, 3s)

Next, we find the distance between P and A

Let the distance be

D = √[(x1 - y1)² + (x2 - y2)²]

Here x1 = 74, x2 = 0, y1 = s, y2 = 3s.

D = √[(74 - s)² + (0 - 3s)²]

= √(s² - 148s + 5476 + 9s²)

= √(10s² - 148s + 5479)

To maximize the distance, we find the derivative of D with respect to s, and set it to 0.

That is dD/ds = 0

dD/ds =

(20s - 148)/2√(10s² - 148s 5479) = 0

=> 20s - 148 = 0

20s = 148

s = 148/20

s = 7.4

Since t = 3s, we have

t = 22.2

So, the point is P(7.4, 22.2)

Finally, we find the distance D.

D = √[(74 - s)² + (0 - 3s)²]

D = √[(74 - 7.4)² + (0 - 3×7.4)²]

D = √(4435.56 + 492.84)

= √4928.4

D = 70.2

Therefore, the distance is 70.2

4 0

3 years ago

Kim S age is twice that of her sister when you add Kim's age to her sisters age of 36 how old is each sister right in equation t

Damm [24]

equation (s+k=36)
s=18
k=18

6 0

3 years ago

Find a second-degree polynomial (that is, an equation of the form y =a + bx + cx2 d that goes through the three points (0, 1), (

Lostsunrise [7]

Answer:

$y = 1 - x^2$

Step-by-step explanation:

we're given three points (0,1),(1,0) and (-1,0). and and equation of a parabola

$y = a + bx + cx^2$

we can plug in each of the coordinates, and 3 equations

(x,y) = (0,1)

$1 = a + b(0) + c(0)^2$

$a=1\quad\quad\Rightarrow A$

(x,y) = (1,0)

$0 = a + b(1) + c(1)^2$

$a + b + c=0\quad\quad\Rightarrow B$

(x,y) = (-1,0)

$0 = a + b(-1) + c(-1)^2$

$a - b + c=0\quad\quad\Rightarrow C$

These are are three equations, well we can simultaneously solve them to find the values of a, b and c.

we already found that, a = 1. so we plug this value in the rest of the equations. we'll use equation B.

$a + b + c=0$

$1 + b + c=0$

$b=-1-c$

we can substitute this value of b and a = 1, equation C

$a - b + c=0$

$1 -(-1-c) + c=0$

$1 +1+c + c=0$

$2+2c=0$

$c=-1$

we can use this value of c back in b

$b=-1-(-1)$

$b=0$

hence our equation of the 2-degree polynomial will be:

$y = a + bx + cx^2$

$y = 1 + (0)x + (-1)x^2$

$y = 1 - x^2$

and this polynomial indeed passes through all the points (0, 1), (1, 0), and (-1, 0).

And since these are the only solutions to the simultaneous equation we solved (i.e. we have single values of a,b and c). there's no other possibility.

8 0

3 years ago