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rewona [7]
1 year ago
10

The figure shown represents the pattern of a glass ceiling. Based on the markings shown, which triangles can be proven congruent

? answers: A) ΔQNI ≅ ΔNPK by SSS B) ΔJKL ≅ ΔNKP by SSA C) ΔJKL ≅ ΔNKP by AAS D) ΔPNK ≅ ΔIKN by SAS

Mathematics
1 answer:
umka2103 [35]1 year ago
5 0

Answer:

D) ΔPNK ≅ ΔIKN by SAS

Step-by-step explanation:

The Side Angle Side (SAS) postulate states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then these two triangles are congruent.

In this case, the congruent sides are IK with NP and NK with itself. And the included angle is ∠K in ΔIKN and ∠N in ΔPNK.

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Help me with this math equation.<br>​
Elodia [21]

Answer:

7y+21

Step-by-step explanation:

7(y+4)-7

=7y+28-7

=7y+21

4 0
1 year ago
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I don't know how to do this. I forget plz help
Leno4ka [110]
Ok so 
0.5%=0.005
0.005x=8
x=8/0.005
x=1600
8 is 0.5% of 1600
Hope this helps
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If you continue with the same formula with a different problem then you will get it correct 
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3 0
1 year ago
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the figure shown is a rectangle with a semicircle on each end. what is the area of the figure? use 3.14 for <img src="https://te
nirvana33 [79]

Answer:

\boxed{a.\:\:\:76.3\:in^2}

Step-by-step explanation:

The figure is made up of a rectangle and two semicircles.


We find the area of the rectangle using the formula;

Area = l\times w


\Rightarrow Area = 8\times 6


\Rightarrow Area = 48\:in^2.


We find the area of one of the semicircles and multiply by 2;


The area of the semicircles

=2\times \frac{1}{2}\times \pi \times r^2


We substitute \pi =3.14 and r=3\:in to obtain;


=2\times \frac{1}{2}\times 3.14 \times 3^2


= 3.14 \times 9


=28.26\:in^2


The area of the figure is

=48+28.26


=76.26\:in^2

To the nearest tenth we have;

=76.3\:in^2

The correct answer is A





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1 year ago
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I NEED HELP PLZZZ AN EXPERT ASAP
Svetradugi [14.3K]
A is on the graph. When you plug in 3 in for x you get -4

B is not. When 1 is plugged in you do notnget 4
4 0
1 year ago
The differential equation y′′=0 has one of the following two parameter families as its general solution: yyyy=C1ex+C2e−x=C1cos(x
svet-max [94.6K]

Answer:

y_c = 2 + 10*x

Step-by-step explanation:

Given:

                                                y'' = 0

Find:

- The solution to ODE such that y(0) = 2, y'(0) = 10

Solution:

- Assuming a solution y = Ce^(mt)

So,                                y' = C*me^(mt)

                                    y'' = C*m^2e^(mt)

- Back substitute into given ODE, we get:

                                    y'' = C*m^2e^(mt) = 0

                                    e^(mt) can not be equal to zero

- Hence,                       m^2 = 0

                                     m = 0 , 0 - (repeated roots)

- The complimentary function for repeated roots is:

                                    y_c = (C1 + C2*x)*e^(m*t)

                                    y_c = C1 + C2*x  

- Evaluate @ y(0) = 2

                                    2 = C1 + C2*0

                                    C1 = 2

-Evaluate @ y'(0) = 10

                                    y'(t) = C2 = 10

Hence,                         y_c = 2 + 10*x

5 0
1 year ago
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