Step-by-step explanation:
-2/3f =9.2
cross multiply
27,6= -2f
divide both sides by 2
f = - 13.8
-1/8n = -2/7
cross multiply
-16 = -7n
divide
n = 2.29
12/20n = -36/44
cross multiply
528 = - 720n
divide
n= -0.733
-64.5g = -25.8
divide both sides by -64.5
g = -0.4
the last one has no variable
1) gradient of line = Δ y ÷ Δ x
= (5 -2) ÷ (3 - (-6))
= ¹/₃
using the point-slope form (y-y₁) = m(x-x₁)
using (3,5)
(y - 5) = ¹/₃ (x -3)
y - 5 = ¹/₃x - 1
⇒ <span> y = ¹/₃ x + 4 [OPTION D]
</span>2) y = 2x + 5 .... (1)
<span> </span>y = ¹/₂ x + 6 .... (2)
by substituting y in (1) for y in (2)
2x + 5 = ¹/₂ x + 6
³/₂ x = 1
x = ²/₃
by substituting found x (2)
y = ¹/₂ (²/₃) + 6
y = ¹⁹/₃
∴ common point is (²/₃ , ¹⁹/₃) thus answer is FALSE [OPTION B]
3) Yes [OPTION A]
This is because the both have a gradient of 5 and if they have the same gradient then that means that the two lines are parallel to each other.
4) No [OPTION B]
Two lines are perpendicular if their gradients multiply to give - 1 and as such one is the negative reciprocal of the other. Since both gradients are ¹/₂ then they are actually parallel and not perpendicular.
Answer:
$2560
Step-by-step explanation:
if 10%=256
then 10 times the amount 100% is 2560
Answer:
Perimeter = 6x² + 8x
Step-by-step explanation:
Perimeter = 2(length + width)
perimeter = 2((x²+x)+(2x²+3x))
perimeter = 2(x²+2x² + x+3x)
perimeter = 2(3x² + 4x)
perimeter = 2*3x² + 2*4x
perimeter = 6x² + 8x
Answer:
8 and 6
Step-by-step explanation:
If you use x for the first number, and y for the second, you get equations of 3x+5y=54 and x-2=y.
You can substitute the second equation into the first one to solve it. This gives 3x+5(x-2)=54.
The brackets can be expanded to 3x+5x-10=54. Collecting like terms makes it 8x-10=54.
Next, we add 10 to both sides. This gives 8x=64.
From there, we isolate the x by dividing both sides by 8, giving x=8.
To work out the second number, we can sub 8 for x in either equation (I'm using the second one as it's simpler).
This comes to y=8-2, therefore y=6.
**This content is simultaneous equations, which you may wish to revise. I'm always happy to help!
