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3 years ago

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The length of the base of an isosceles triangle is x. The length of a leg is 2x - 2. The perimeter of the triangle is 51. Find x

.

2 answers:

azamat3 years ago

7 0

Answer:

x = 11

Step-by-step explanation:

x + 2(2x-2) = 51

x + 4x - 4 = 51

5x - 4 = 51

5x = 55

x = 11

stealth61 [152]3 years ago

6 0

Answer:

x = 11

Step-by-step explanation:

2x - 2 + 2x - 2 + x = 51

5x - 4 = 51

5x = 55

x = 11

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Suppose we have a right triangle with legs of length a and b and hypotenuse of length c. Suppose b=3 and c=5. Then a= , For the

ANTONII [103]

Answer:

Length of right-angle triangle 'a' = 4

b)

<u><em></em></u> $sin(A) = \frac{opposite side}{Hypotenuse} = \frac{a}{c} = \frac{4}{5}$ <u><em></em></u>

<u><em></em></u> $cos(A) = \frac{Adjacent side}{Hypotenuse} = \frac{b}{c} = \frac{3}{5}$ <u><em></em></u>

<u><em></em></u> $tan(A) = \frac{opposite side}{Adjacent side} = \frac{a}{b} = \frac{4}{3}$ <u><em></em></u>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given b = 3 and hypotenuse c = 5

Given ΔABC is a right angle triangle

By using pythagoras theorem

c² = a² + b²

⇒ a² = c² - b²

⇒ a² = 5²-3²

=25 - 9

a² = 16

⇒ a = √16 = 4

The sides of right angle triangle a = 4 ,b = 3 and c = 5

<u><em>Step(ii):-</em></u>

<u><em></em></u> $sin(A) = \frac{opposite side}{Hypotenuse} = \frac{a}{c} = \frac{4}{5}$ <u><em></em></u>

<u><em></em></u> $cos(A) = \frac{Adjacent side}{Hypotenuse} = \frac{b}{c} = \frac{3}{5}$ <u><em></em></u>

<u><em></em></u> $tan(A) = \frac{opposite side}{Adjacent side} = \frac{a}{b} = \frac{4}{3}$ <u><em></em></u>

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3 years ago

A sports apparel company surveyed the market about the size of T-shirts people most frequently buy: small, medium, large, or ext

goblinko [34]

First off we need to find the denomonator:
11+37+83+19=150

Okay so our probability of getting a small size of a t-shirt will be 11/150
They sold 45,000
That's where we need to change our denominator to get how many shirts were small

45,000÷150=300
300×11=3,300
I multiplied 11 because of the numerator

So our answer is B. 3300

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If n lines are drawn in a plane, how many regions do they separate the plane into? induction

grandymaker [24]

Hello,

Let's assume i the number of lines,
u(i) the number of regions:

$u_0=1\\\\
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u_2=4=2+u_1\\\\
u_3=7=3+u_2\\\\
u_4=11=4+u_3\\\\
u_5=16=5+u_4\\\\
...\\\\
\boxed{u_{n+1}=(n+1)+u_{n}}\\\\

u_{n+1}=(n+1)+n+(n-1)+...+1+u_0\\\\

\boxed{u_{n+1}= \dfrac{(n+1)*(n+2)}{2}+1 }$

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adelina 88 [10]

Answer:

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Step-by-step explanation:

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