<span>In this case, the value of the first 3 (the ten-thousands) has a value of 30,000. The 3 next to it, in the hundred-thousands place, has a value of 300,000. To compare the two, the 3 on the right has a value one-tenth as much as that on the left.</span>
Hey there!
4/7 + 5/6
≈ 24/42 + 35/42
≈ 24 + 36/ 42 - 0
≈ 59/42
≈ 1 17/42
Therefore, your answer is:
59/42 or 1 17/42
EITHER OF THOSE SHOULD WORK BECAUSE THEY ARE EQUIVALENT.
Good luck on your assignment & enjoy your day!
~Amphitrite1040:)
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 :)
Given function is
now we need to find the value of k such that function f(x) continuous everywhere.
We know that any function f(x) is continuous at point x=a if left hand limit and right hand limits at the point x=a are equal.
So we just need to find both left and right hand limits then set equal to each other to find the value of k
To find the left hand limit (LHD) we plug x=-4 into 3x+k
so LHD= 3(-4)+k
To find the Right hand limit (RHD) we plug x=-4 into
so RHD= 
Now set both equal
k=-0.47
<u>Hence final answer is -0.47.</u>
Can what perform algebra?
