The <em>correct answers</em> are:
5x²+70x+245 ≥ 1050; and
Yes.
Explanation:
Let x be the width of the tablet. Since the width of the TV is 7 inches more than the tablet, the width of the TV would be x+7.
The length of the TV is 5 times the width; this makes the length 5(x+7) = 5x+35.
The area of the TV would be given by
(x+7)(5x+35).
Since Andrew wants the area to be at least 1050, we set the expression greater than or equal to 1050:
(x+7)(5x+35) ≥ 1050
Multiplying this, we have:
x*5x+x*35+7*5x+7*35 ≥ 1050
5x²+35x+35x+245 ≥ 1050
Combining like terms,
5x²+70x+245 ≥ 1050
To see if 8 is a reasonable width for the tablet, we substitute 8 for x:
5(8²)+70(8)+245 ≥ 1050
5(64)+560+245 ≥ 1050
320+560+245 ≥ 1050
1125 ≥ 1050
Since this inequality is true, 8 is a reasonable width.
Answer:
Step-by-step explanation:
if i had to guess i would say a
The value is x= -5/4
<h3>What is LCD?</h3>
The least common denominator is the smallest number of all the common multiples of the denominators when 2 or more fractions are given.
Given:
1/x = 1/5+ 5/4x
1/x = 4x + 25 / 20x
20= 4x +25
4x= -5
x= -5/4
Learn more about this concept here:
brainly.com/question/2605216
#SPJ1
Answer: x = 1/16
Step-by-step explanation:
Since the inverse of a Logarithm is an exponential function, we know that the final solution has to involve an exponential function somewhere in it.
1. log B(2) {x} = -4 || given
2. x = 2 ^ -4 || Logarithm rule that allows you to move the base of the logarithm to the base of the exponent on the other side. For example, if you had log B(5) {x} = 3, the base of 5 would move over to the other side and it would be raised to 3; x = 5^3.
3. x = (1) / (2^4) || Simplify. Use the negative exponent rule. This rule always leaves a numerator of 1, and a denominator of your exponent. In this case, it will be 2 ^ -4, so you will do 2^4 which is 16 and you will put that over 1. Resulting in your final answer of x = 1/16
y = 9ln(x)
<span>y' = 9x^-1 =9/x</span>
y'' = -9x^-2 =-9/x^2
curvature k = |y''| / (1 + (y')^2)^(3/2)
<span>= |-9/x^2| / (1 + (9/x)^2)^(3/2)
= (9/x^2) / (1 + 81/x^2)^(3/2)
= (9/x^2) / [(1/x^3) (x^2 + 81)^(3/2)]
= 9x(x^2 + 81)^(-3/2).
To maximize the curvature, </span>
we find where k' = 0. <span>
k' = 9 * (x^2 + 81)^(-3/2) + 9x * -3x(x^2 + 81)^(-5/2)
...= 9(x^2 + 81)^(-5/2) [(x^2 + 81) - 3x^2]
...= 9(81 - 2x^2)/(x^2 + 81)^(5/2)
Setting k' = 0 yields x = ±9/√2.
Since k' < 0 for x < -9/√2 and k' > 0 for x >
-9/√2 (and less than 9/√2),
we have a minimum at x = -9/√2.
Since k' > 0 for x < 9/√2 (and greater than 9/√2) and
k' < 0 for x > 9/√2,
we have a maximum at x = 9/√2. </span>
x=9/√2=6.36
<span>y=9 ln(x)=9ln(6.36)=16.66</span>
the
answer is
(x,y)=(6.36,16.66)
