A = bh = 11 x 8 = 88
answer
<span>88 m²</span>
No it is not a perfect square
Just plug and chug :)
h(-29)=-1/2(-29)+3
h(-29)=29/2+3
h(-29)=-11.5
Problem:
find line perpendicular to 4x+7y+3=0 passing through (-2,1).
Solution:
We will first find the general form of the perpendicular line.
The perpendicular line has the form
7x-4y+k=0 .........................(1)
by switching the coefficients of x and y, and switching the sign of one of the two coefficients. This way, the slope of (1) multiplied by that of the original equation is -1, a condition that the two lines are perpendicular. The value of k is to be determined from the given point (-2,1).
To find k, we substitute x=-2, y=1 into equation (1) and solve for k.
7(-2)-4(1)+k=0
=>
k=14+4=18
Therefore the required line is
7x-4y+18=0
Answer:
The right statement is sin(J) = cos(L) ⇒ answer D
Step-by-step explanation:
* Lets describe the figure
- LKJ is a right triangle, where K is a right angle
∵ m∠K = 90°
∵ LJ is opposite to angle K
∴ LJ is the hypotenuse
∵ LJ = 219
∵ KJ = 178
- By using Pythagoras Theorem
∵ (LJ)² = (LK)² + (KJ)²
∴ (219)² = (LK)² + (178)² ⇒ subtract (178)² from both sides
∴ (LK)² = (219)² - (178)²
∴ (LK)² = 16277
∴ LK = √16277 = 127.58
* Lets revise how to find the trigonometry function
# sin Ф = opposite/hypotenuse
# cos Ф = adjacent/hypotenuse
# tan Ф = opposite/adjacent
∵ LK is the opposite side to angle J
∵ LJ is the hypotenuse
∵ sin(J) = LK/LJ
∵ LK = 127.58 , LJ = 219
∴ sin(J) = 127.58/219 = 0.583
∵ LK is the adjacent side to angle L
∵ LJ is the hypotenuse
∵ cos(L) = LK/LJ
∵ LK = 127.58 , LJ = 219
∴ cos(L) = 127.58/219 = 0.583
∴ sin(J) = cos(L)
* The right statement is sin(J) = cos(L)
