Answer:
all of those are shown inside of the triangle ...
Step-by-step explanation:
Answer:
not enough info, make another question with the problem attached
Step-by-step explanation:
11. You’ve done it correctly
12. Let x^2=y
y^2+13y+40=0
(y+8)(y+5)=0
y=8, 5
Since y=x^2
x^2=8 x^2=5
x=+/-√5 x= +/-2√2
13. x^4-x^2-x^2-8=0
x^4-2x^2-8=0
let x^2=y
Y^2-2y-8=0
(y-4)(y+2)=0
y=4, -2
Since y=x^2
X^2=4 X^2=-2
X= +/- 2 This wouldn’t be a real solution
14. It’s pretty much the same process, just substitute y in for x^2. If you’re confused feel free to ask and I can do it, or you can put it through Photomath
15. You’re on the right track so I’m just going to continue from where you left off
x^2(4x+5)-4(x+5)=0
(x^2-4)(4x+5)=0
x= +/- 2 4x=5
x=5/4 or 1 1/4
Hope this helped :)
Answer: The gradient of the line passing through is 3, thus the equation of the line becomes y=3x-4.
Step-by-step explanation: The parallel lines gradients that have similar gradients, so now we have the gradient of 3 because its similar to the other gradient. To get the equation of the line we use this equation: y=mx+c, we have to substitute the x and y to get the constant c.
5=3(3)+c
c= -4
y=3x+c
