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Vadim26 [7]
3 years ago
15

Which angle is congruent to <4 <1<2<5<8

Mathematics
1 answer:
Anna007 [38]3 years ago
8 0

Answer:

The angle that will be congruent to angle 4 is :

                             Angle 1

Step-by-step explanation:

It is given that:

angle 4 and angle 5 are complements.

Also, angle 1 and angle 5 are complements.

Congruent complementary Theorem--

It states that if two angles are complementary to the same angle, then the two angles are congruent to each other.

Here both angle 1 and angle 4 are complementary to the same angle i.e. angle 5.

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Answer:

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3 years ago
Write 45/30 in simplest form​
RideAnS [48]

Answer:

1 1/2

Step-by-step explanation:

SO first convert into mixed

1 15/30

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7 0
3 years ago
The graph 4x^2-4x-1 is shown. Use the grpah to find the estimates for the solutions of 4x^2-4x-1=0 and 4x^2 - 4x-1=2
Darina [25.2K]

Answer:

a) The estimates for the solutions of 4\cdot x^{2}-4\cdot x -1 = 0 are x_{1}\approx -0.25 and x_{2} \approx 1.25.

b) The estimates for the solutions of 4\cdot x^{2}-4\cdot x -1 = 2 are x_{1}\approx -0.5 and x_{2} \approx 1.5

Step-by-step explanation:

From image we get a graphical representation of the second-order polynomial y = 4\cdot x^{2}-4\cdot x -1, where x is related to the horizontal axis of the Cartesian plane, whereas y is related to the vertical axis of this plane. Now we proceed to estimate the solutions for each case:

a) 4\cdot x^{2}-4\cdot x -1 = 0

There are two approximate solutions according to the graph, which are marked by red circles in the image attached below:

x_{1}\approx -0.25, x_{2} \approx 1.25

b) 4\cdot x^{2}-4\cdot x -1 = 2

There are two approximate solutions according to the graph, which are marked by red circles in the image attached below:

x_{1}\approx -0.5, x_{2} \approx 1.5

5 0
3 years ago
I need help with this !!
zubka84 [21]

Answer:

is there more to the problem

5 0
3 years ago
What is the inverse of y = 3* ?
grin007 [14]
It doesn’t have an inverse because it doesn’t pass the horizontal line test
8 0
3 years ago
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