7) - 1
tan <BAD = BD / AD
BC = √(64 + 36)
BC = √100
BC = 10
now you can solve BD
BD/6 = 6/10
BD = 6*6 / 10
BD = 36/10
BD = 3.6
AD = √36 - 12.96
AD = √23.04
AD = 4.8
tan <BAD = BD / AD
tan <BAD = 3.6 / 4.8
tan <BAD = 0.75
7) - 2
cos <DAC + cos <DAB
as you know cos = Adj / Hypo
cos <DAC = AD/AC
cos <DAC = 4.8 / 8
cos <DAC = 0.6
cos <DAB = AD / AB
cos <DAB = 4.8/6
cos <DAB = 0.8
so
cos <DAC + cos <DAB = 0.6 + 0.8
cos <DAC + cos <DAB = 1.4
hope it helps
The multiplication and divison equation involves using the '×' and '÷' operator on numbers. Few equations using the divison and multiplication operator are given below.
Given the numbers 6, 2, 3
Their opposite values are :
Divison equation possible include :
- 6 ÷ 2 = 3
- 6 ÷ 3 = 2
- -6 ÷ 2 = - 3
- -6 ÷ 3 = - 2
- -6 ÷ - 3 = 2
- -6 ÷ - 2 = 3
Multiplication equations :
- 6 × 2 = 12
- 6 × 3 = 18
- -6 × 2 = - 12
- -6 × 3 = - 18
- -6 × - 3 = 18
- -6 × - 2 = 12
There are other multiplication and division equations which can be extracted from the given values, above are just a few.
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F=ma so 160N=20kg*8m/s
160 N is the answer
Answer:
if i am readin this correctly, it is subtraction. 3-8 is -5.
Step-by-step explanation:
Answer:
Rectangular area as a function of x : A(x) = 200*x + 2*x²
A(max) = 5000 m²
Dimensions:
x = 50 m
l = 100 m
Step-by-step explanation:
"x" is the length of the perpendicular side to the wall of the rectangular area to be fenced, and we call "l" the other side (parallel to the wall of the barn) then:
A(r) = x* l and the perimeter of the rectangular shape is
P = 2*x + 2*l but we won´t use any fencing material along the wll of the barn therefore
P = 2*x + l ⇒ 200 = 2*x + l ⇒ l = 200 - 2*x (1)
And the rectangular area as a function of x is:
A(x) = x * ( 200 - 2*x) ⇒ A(x) = 200*x + 2*x²
Taking derivatives on both sides of the equation we get:
A´(x) = 200 - 4*x ⇒ A´= 0
Then 200 - 4*x = 0 ⇒ 4*x = 200 ⇒ x = 50 m
We find the l value, plugging the value of x in equation (1)
l = 200 - 2*x ⇒ l = 200 - 2*50 ⇒ l = 100 m
A(max) = 100*50
A(max) = 5000 m²