Answer:
<h3>Hence required integers are 11 and 13</h3>
Let x an odd positive integer
Then, according to question
x^2 +(x+2)^2=290
2x^2 +4x−286=0
x^2 +2x−143=0
x ^2+13x−11x−143=0
(x+13)(x−11)=0
x=11 as x is positive
Hence required integers are 11 and 13
Step-by-step explanation:
Hope it is helpful....
A' is (-12, 6)
B' is (0,6)
C' is (3, -3)
D' is (-9, -3)
Hope this helps!
Answer:
B. a^2x
Step-by-step explanation:
squaring a^x gives. (a^x)²
by applying the rules of indices, ^2 multiplies x therefore making it a^2x
Given:
Current revenue = $6000
<span>R(f)= -100f^2 + 400f + 6000
where f is a whole number of $5 fee increases
We are told to find f, when R(f) </span><span>< 6000
Since </span>R(f)= -100f^2+400f+6000
R(f) < 6000 ⇒ -100f^2+400f+6000 < 6000
Subtract 6000 from both sides
-100f^2 + 400f + 6000 - 6000 < 6000 - 6000
-100f^2 + 400f < 0
⇒ 400f - 100f^2 < 0
Divide the equation by 100
400f/100 - 100f^2/100 < 0/100
4f - f^2 < 0
Add f^2 to both sides of the equation
4f - f^2 + f^2 < 0 + f^2
4f < f^2
Divide both sides by f
4f/f < (f^2)/f
4 < f
⇒ f > 4
⇒ f ≥ 5
Therefore, <span> for 5 or more numbers of $5 fee increases, the revenue from fees will actually be less than its current value.</span>
Answer:
24yrds is (a) 24.12yrds is (b) 64yrds is (c) And that's all i can do, don't get mad at me please.
Step-by-step explanation:
