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Marina CMI [18]
2 years ago
6

By first calculating the size of angle LMN, calculate the area of triangle MNL. You must show all your working.

Mathematics
1 answer:
RSB [31]2 years ago
3 0

Answer:

m(∠LMN) = 67.44°

Area of ΔMNL = 16.66 square units

Step-by-step explanation:

By applying Sine rule in the ΔLMN,

\frac{\text{Sin}M}{\text{NL}}=\frac{\text{Sin}N}{\text{ML}}

\frac{\text{Sin}M}{7.2}}=\frac{\text{Sin}38}{\text{4.8}}

SinM = \frac{\text{Sin}38\times 7.2}{4.8}

SinM = 0.9235

M = \text{Sin^{-1}}(0.9235)\text{Sin}^{-1} (0.9235)

M = 67.44°

m(∠M) + m(∠N) + m(∠L) = 180°

67.44 + 38 + m(∠L) = 180°

m(∠L) = 180 - 105.44

m(∠L) = 74.56°

Area of ΔMNL = \frac{1}{2}(\text{ML})(\text{NL})\text{Sin}(74.56)

                        = \frac{1}{2}(4.8)(7.2)(0.96391)

                        = 16.66 square units

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2 years ago
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BartSMP [9]

Answer:

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Step-by-step explanation:

We are asked to find the GCF of 44j^5k^4\text{ and }121j^2k^6.

Since we know that GCF of two numbers is the greatest number that is a factor of both of them.

First of all we will GCF of 44 and 121.

Factors of 44 are: 1, 2, 4, 11, 22, 44.

Factors of 121 are: 1, 11, 11, 121.

We can see that greatest common factor of 44 and 121 is 11.

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We can see that greatest common factor of j^5\text{ and }j^2 is j*j=j^2.

Now let us find GCF of k^4\text{ and }k^6.

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Factors of k^6 are:k*k*k*k*k*k

We can see that greatest common factor of  k^4\text{ and }k^6 is k*k*k*k=k^4.

Upon combining our all GCFs we will get,

11j^2k^4  

Therefore, GCF of 44j^5k^4\text{ and }121j^2k^6 is 11j^2k^4 and option D is the correct choice.

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