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NeTakaya
3 years ago
6

If the lengths of two sides of a triangle are 5 and 9, respectively, which of the following could be the length of the third sid

e of the triangle? Indicate all such lengths. [A] 3 [B] 5 [G] 8 [D] 15
Mathematics
1 answer:
slamgirl [31]3 years ago
3 0

There are two answers:

B) 5

C) 8

==============================================

Explanation:

If we had a triangle with sides a, b and c, then we can say

b-a < c < b+a

where b is larger than 'a'. This is the triangle inequality theorem

In this case, a = 5 and b = 9 so,

b-a < c < b+a

9-5 < c < 9+5

4 < c < 14

Telling us that c is some number between 4 and 14, not including either endpoint. If c is a whole number, then c could be any value from this set: {5,6,7,8,9,10,11,12,13}

We see that the numbers 5 and 8 are in this set. The values 3 and 15 are not in the set.

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juin [17]

#1:

A,B,C,E

#2:

12 edges, 12 vertices

7 0
3 years ago
Read 2 more answers
Help me<br> On this pleaseee
Evgen [1.6K]
2 rectangle shaped
1) Length = 160 mi ; Width = 40 mi
2) Length = 440 mi - 160 mi ; Width = 240 mi - 70 mi

1 triangle shape
1) base = 70 mi ; height = 440 mi - 160 mi

Area Rectangle 1 = 160 mi * 40 mi = 6,400 mi²
Area Rectangle 2 = 280 mi * 170 mi = 47,600 mi²
Area Triangle 1 = ((440 mi - 160 mi) * 70mi)/2 = (280mi * 70mi)/2 = 9,800 mi²

Total Area = 6,400 mi² + 47,600 mi² + 9,800 mi² = <span>63,800 mi²</span>
4 0
3 years ago
If A = C + 3A=C+3, B = 2CB=2C, and C = 4C=4, which is the correct order from least to greatest?
kotykmax [81]

Answer:

B, C, A.

Step-by-step explanation:

A = C + 3A = C + 3

A = C + 3

A = C + 3A

C + 3A = C + 3

3A = 3

A = 1

A = C + 3

1 = C + 3

C + 3 = 1

C = -2

B = 2C

B = 2(-2)

B = -4.

So, from least to greatest, B = -4, C = -2, and A = 1.

Hope this helps!

6 0
3 years ago
G(1) = 4<br> g(n) = g(n-1)-(-3)<br> g(3)
Licemer1 [7]

Answer:

2

g

n

+

3

+

3

g

Step-by-step explanation:

8 0
3 years ago
Which of the following is a solution to 2cos2x − cos x − 1 = 0?
Pavel [41]

Answer:

Option A is correct.

Solution for the given equation is, x = 0^{\circ}

Step-by-step explanation:

Given that : 2\cos^2x -\cos x -1 =0

Let \cos x =y

then our equation become;

2y^2-y-1= 0           .....[1]

A quadratic equation is of the form:

ax^2+bx+c =0.....[2] where a, b and c are coefficient and the solution is given by;

x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}

Simplify:

y = \frac{ 1 \pm \sqrt{1+8}}{4}

or

y = \frac{ 1 \pm \sqrt{9}}{4}

y = \frac{ 1 \pm 3}{4}

or

y = \frac{1+3}{4} and y = \frac{1 -3}{4}

Simplify:

y = 1 and y = -\frac{1}{2}

Substitute y = cos x we have;

\cos x = 1

⇒x = 0^{\circ}

and

\cos x = -\frac{1}{2}

⇒ x = 120^{\circ} \text{and} x = 240^{\circ}

The solution set:  \{0^{\circ}, 120^{\circ} , 240^{\circ}\}

Therefore, the solution for the given equation  2\cos^2x -\cos x -1 =0 is, 0^{\circ}





8 0
3 years ago
Read 2 more answers
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