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kodGreya [7K]
1 year ago
12

Which lines are parallel if M^4 +M^5? Justify your answer.

Mathematics
1 answer:
Igoryamba1 year ago
4 0
Opposite angles formed by two intersecting lines are equal, so angle 1 is the same as angle 4. That means angle 1 = angle 5 as well. 

<span>When a line intersects two parallel lines, the corresponding angles are equal. That is, if r and s are parallel, then the angles formed when l intersects r are the same s the angles formed when l intersects s. Angle 1 = Angle 5, Angle 2 = Angle 6, and so forth. Since we know angle 1 = angle 5, so from that you can see that r and s are parallel</span>
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What common angle do ACE and ACD share?
SCORPION-xisa [38]
The answer is the second option, <ACE = <ECA.

This is because C is in the middle of both, so it's referring to the same angle. The other options are all comparing different angles, that aren't congruent, so they're incorrect.
6 0
2 years ago
Read 2 more answers
() and () are inverses of one another and drawn on the same graph with the same scale on both the horizontal and vertical axis.
Gelneren [198K]

Answer: Option A.

Step-by-step explanation:

We have the functions f(x) and g(x), that are inverses between them.

This means that if:

f(x) = y

then:

g(y) = x.

now, remember that:

When we have a point (x, y), and we reflect it over the line y = x, our new point will be (y, x).

So before we whe had:

f(x) = y.

and now in that same place, we have:

g(y) = x.

So the old graph of f(x) now coincides with the graph of g(x). (And the old graph of g(x) now coincides with the graph of f(x) )

So A is true.

B) This depends on the function:

if we have f(x) = x  + 1.5

then f(0) = 1.5

now we want that:

g(1.5) = 0, then we can write:

g(x) = x - 1.5

Now f(x) and g(x) are inverses, and we would have that:

f(x) = g(x) + 3.

So f(x) is g(x) translated up by 3 units, but this is a particular case, not a general one, so B is not always true.

C and D) When we do rotations of 90° or 180°, we are effectively changing the quadrant of our point. so rotations will cause not only changes as the reflection over the x = y line, those will also cause changes in the sign of our variables, so, while for some functions f(x) and g(x) we can have that the rotations will map one into the other, this is not the general case.

6 0
1 year ago
Evaluate the triple integral ∭Tx2dV, where T is the solid tetrahedron with vertices (0,0,0), (3,0,0), (0,3,0), and (0,0,3).
bixtya [17]

Answer:

the integral I=81

Step-by-step explanation:

for the integral I

I=\int\limits^{}_{}\int\limits^{}_{}\int\limits^{}_{T} {x^{2} } \, dxdydz

where T is the solid tetrahedron , then

I=\int\limits^{3}_{0}\int\limits^{3}_{0}\int\limits^{3}_{0} {x^{2} } \, dxdydz = \int\limits^{3}_{0}dz\int\limits^{3}_{0}dy\int\limits^{3}_{0} {x^{2} } \, dx = (3-0)*(3-0)*1/3*(3^{3}-0^{3}) = 3^{4} = 81

the integral is equal to 81

8 0
1 year ago
Evaluate 1/4c + 3d when c = 6 and d = 7
Natalka [10]
Replace the C and D with 6 and 7.

1/4 (6) + 3 (7)

6/4 + 21 

6/4 + 84/4 = 90/4

90/4 = 22.5

Hope this helps!
7 0
2 years ago
Read 2 more answers
Show that the Fibonacci numbers satisfy the recurrence relation fn = 5fn−4 + 3fn−5 for n = 5, 6, 7, . . . , together with the in
Sonja [21]

Answer with step-by-step explanation:

We are given that the recurrence relation

f_n=5f_{n-4}+3f_{n-5}

for n=5,6,7,..

Initial condition

f_0=0,f_1=1,f_2=1,f_3=2,f_4=3

We have to show that Fibonacci numbers satisfies the recurrence relation.

The recurrence relation of Fibonacci numbers

f_n=f_{n-1}+f_{n-2},f_0=0,f_1=1

Apply this

f_n=(f_{n-2}+f_{n-3})+f_{n-2}=2f_{n-2}+f_{n-3}

f_n=2(f_{n-3}+f_{n-4})+f_{n-3}=3f_{n-3}+2f_{n-4}

f_n=3(f_{n-4}+f_{n-5})+2f_{n-4}=5f_{n-4}+3f_{n-5}

Substitute n=2

f_2=f_1+f_0=1+0=1

f_3=f_2+f_1=1+1=2

f_4=f_3+f_2=2+1=3

Hence, the Fibonacci numbers satisfied the given recurrence relation .

Now, we have to show that f_{5n} is divisible by 5 for n=1,2,3,..

Now replace n by 5n

f_{5n}=5f_{5n-4}+3f_{5n-5}

Apply induction

Substitute n=1

f_5=5f_1+3f_0=5+0=5

It is true for n=1

Suppose it is true for n=k

f_{5k}=5f_{5k-4}+3f_{5k-5} is divisible 5

Let f_{5k}=5q

Now, we shall prove that for n=k+1 is true

f_{5k+5}=5f_{5k+5-4}+3f_{5k+5-5}=5f_{5k+1}+3f_{5k}=5f_{5k+1}+3(5q)

f_{5k+5}=5(f_{5k+1}+3q)

It is multiple of 5 .Therefore, it is divisible by 5.

It is true for n=k+1

Hence, the f_{5n} is divisible by 5 for n=1,2,3,..

8 0
1 year ago
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