HELP ME PLEASE I REALLY NEED HELP!? :(

Mathematics

1 answer:

Aleksandr [31]2 years ago

5 0

Just answers since long and many simplifiction
it's like doing those multiplication memory things that had like 100 questions but now with algebra fractions

5. fourth or $\frac{5x-15}{x^{2}-3x-28}$
6. third or $\frac{4x+25}{x^{2}+5x+4}$
7.the answe ris x+3 since it can be factored out of the 5x+15 tingie (second )
8. fourth or $\frac{x^{2}+4x}{4}$
9. fourth or $\frac{2y+6}{9y-7}$ times $\frac{3}{2}$
10. third or $\frac{(x+3)(x+8)}{(x-8)(x-8)}$
11. second or $\frac{x-5}{x-4}$
12. third or 11(x+7)=60

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The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.

Mama L [17]

Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

21 m + 40 m + c = 120 m

61 m + c = 120 m

c = 120 m - 61 m

c = 59 m

↓ ↓

A → → → → → → → → → → → C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

$a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0$

Angle B is calculated as follows;

$Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0$

Angle C is calculated as follows;

$Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0$