There are 2 way to solve this.

one using Pythagoras theorem and 2nd using trigonometry

so lets solve it by both

using Pythagoras theorem we know

base^2 + perpendicular^2 = hypotanes^2

6^2 + x^2 = 12^2

36 + x^2 = 144

x^2 = 144- 36 = 108

x = √(108) = √( 2×2×3×3×3)

= (2×3) √ (3) = 6 √3

so answer is option 2

bow lets use trigonometry

we know
sin theta = perpendicular / hypotanes
sin 60 = x /12
x = 12 × sin 60
we kNow sin 60 = √3/ 2
so
x = 12×√3 /2 = 6√3

7 0

3 years ago

Sam’s business manufactures 5 times as many widgets as Mary does every hour but Sam needs to stop for a few minutes every hour a

natita [175]

Answer:
A) S=5M-3

Explanation:

6 0

3 years ago

A quadratic equation ax^2+bx+c=0 has -11 and 4 as solutions. Find the values of b and c if the value of a is 1 (hint, use zero f

Alina [70]

Answer:

b = 7, c = -44

Step-by-step explanation:

If the quadratic equation has the solutions -11 and 4, the two factors are:

$(x+11)(x-4)=0$

Since when we use the zero factor property we get

x+11=0 ⇒ x= -11

x-4=0 ⇒ x=4

Thus, we have used the zero factor property in reverse to find the factorization of the quadratic equation.

Now we develop the multiplications between parenthesis:

$(x+11)(x-4)=0\\x^2-4x+11x-44=0\\x^2+7x-44=0$

So b is the number that accompanies the x: b = 7

and c is the independent number: c = -44

3 0

3 years ago