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Digiron [165]
2 years ago
5

Which angles are remote interior angles? Triangle L M N. Angle L is 1, angle M is 2, angle N is 3. Side M N extends to form angl

e 4.
Mathematics
1 answer:
Nuetrik [128]2 years ago
6 0

Answer:

Angle L (1) and Angle M (2)

Step-by-step explanation:

Angle N is not a remote angle because it has an exterior angle, angle 4

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A 50-foot flagpole is at the entrance of a building that is 300 feet tall. If the length of the flagpole's shadow is 30 feet at
Soloha48 [4]

The equation you can use to solve the problem is;

50/30 = 300/x

Next Cross Multiply

50x = 9000

x = 9000/50

x = 180

The building's shadow is 180 feet tall

7 0
2 years ago
Please help me, I’m brain dead right now:)
Harlamova29_29 [7]

Answer:

A 40%

Step-by-step explanation:

2/5 = 40/100 (or 40%)

So 40 blocks should be filled in :)

Hope this helps!

3 0
2 years ago
Read 2 more answers
A squirrel moved 200 meters in 7,200 seconds. What was the speed of the squirrel in meters per hour? *
kati45 [8]

Answer:

100 meters per hour.

Step-by-step explanation:

7,200 seconds / 60 = 120 minutes

120 minutes / 60 = 2 hours

200 meters / 2 hours = 100 meters

Since the squirrel moved 200 meters in 2 hours, the squirrel moved 100 metres per hour.

7 0
3 years ago
Which graph represents the function f(x)=(x-5)^2+3
hammer [34]

Answer:

Parabola

Step-by-step explanation:

We are given that a function

f(x)=(x-5)^2+3

The given function is an equation of parabola along y- axis.

General equation of parabola along y- axis  with vertex (h,k) is given by

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

y=(x-h)^2+k

Compare it with given equation then we get

h=5, k=3

Vertex of given parabola =(5,3)

Substitute x=0 then we get

f(0)=(0-5)^2+3=28

y-intercept of parabola is at (0,28).

6 0
3 years ago
Read 2 more answers
For which intervals the graphs of the functions f(x) = x^3 + x^2 - 4x - 4 is positive
Nookie1986 [14]

Step-by-step explanation:

Consider a function

f

(

x

)

which is twice differentiable. The graph of such a function will be concave upwards in the intervals where the second derivative is positive and the graph will be concave downwards in the intervals where the second derivative is negative. To find these intervals we need to find the inflection points i.e. the x-values where the second derivative is 0.

5 0
2 years ago
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