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3 years ago

Solve the triangle. A = 51°, b = 14, c = 6, I'll give 20 points

Mathematics

1 answer:

ankoles [38]3 years ago

7 0

Answer:

Step-by-step explanation:

We have that

A = 51°, b = 14, c = 6

step 1

find the value of a

Applying the law of cosines

a²=c²+b²-2*c*b*cos A

a²=6²+14²-2*6*14*cos 51-------> 126.27

a=11.2

we know that

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

we have

a=11.2

b=14

c=6

(a+b) > c-------------> (11.2+14)=25.2

25.2 > 6-----> is not correct

therefore

the answer is the option

a. No triangles possible

I don't know if i am right sorry if this is wrong

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Find the hypotenuse of each isosceles right triangle when the legs are of the given measure. 6 sqrt 2

Gemiola [76]

Isosceles right triangles have two equal sides (a and b) that are not the hypotenuse (c). And when two sides are equal, so are their opposite angles. There are only 180° degrees in any triangles, thus the right angle = 90°, so 90 left for the two equal, means that 2x=90,
x = 45°.

There are several ways to go about solving a triangle like this. The best and easiest is simply to memorize that the hypotenuse is exactly root2 times the other sides. Or, each isosceles side is the hypotenuse (c) ÷ root2
$a = b = c \div \sqrt{2} \\ c = a\sqrt{2} \\ c = 6 \sqrt{2} \times \sqrt{2} = 6 \times 2 = 12$
Another way to do it is the longer proof of Pythagorean Theorem:
${c}^{2} = {a}^{2} + {b}^{2}... \: \: c = \sqrt{({a}^{2} + {b}^{2})} \\$
$c= \sqrt{({6 \sqrt{2}) }^{2} + ({6 \sqrt{2})}^{2}} \\ = \sqrt{(2 \times{(6 \sqrt{2} )}^{2} )} = \sqrt{2(36 \times 2)} \\ c = \sqrt{144} = 12$

7 0

3 years ago

Select the correct answer for each statement. (k+2, k+3 , k+4) or (k+10, k+12, k+14) will yield consecutive odd integers. (k+6,

almond37 [142]

Answer:

Step-by-step explanation:

one

The only really certain way of getting an odd integer is to do 2k + 1. Therefore the answer should be something like 2k+1, 2k+3, 2k + 5

The best I can do for the first one is to assume that k is odd and that the answer is k + 10, k+12, k+14, but I would ask your instructor if this is merely the best answer out of a poor lot or if the question really has no answer.

Two things should be noted.

1. k has to be odd.

2. The first choice has to give consecutive integers because what is added on is 2 3 and 4. Those numbers are consecutive.

Two

k+1, k+2, k+3. See point 2 above. k is a constant and any number. 1,2,3 are consecutive so the results are consecutive. Even if k < 0 the numbers will be consecutive.

k= - 10

k+1 = - 9

k+2 = -8

k+3 = - 7

These are consecutive.

Your first choice will give either consecutive odd or even numbers depending on what k is.

If k = even, then k+6, k+ 8, k+10 will all be even.

If k = odd then the givens will be odd.

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3 years ago

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