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Marizza181 [45]

3 years ago

5

Chang buys candy that costs $6 per round. He will spen at least $30 on candy. What are the possible pounds he will buy? Use p fo

r the number of pounds Chang will buy. Write your answer as inequality solved for p.
Please please please answer this someone!!!!

1 answer:

exis [7]3 years ago

5 0

Answer:

6p _< 30

Step-by-step explanation:

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Tom put 18 gallons of mid grade gas in his truck and filled up his empty five gallon gas tank with regular gas for his lawn mowe

LuckyWell [14K]

Answer:

Mid grade gas per gallon = $2.67

Regular gas per gallon = $2.37

Step-by-step explanation:

Let, x= mid grade gas, y=regular gas.

So, for 18 gallons of mid grade gas and 5 gallon of regular gas at $59.91, it can be expressed as,

$18x+5y=59.91$ ------------------(equation 1)

for 14 gallon of mid grade gas and 1 gallon of regular gas at 39.75, it can be expressed as,

$14x+y=39.75$ -------------------(equation 2)

$y=39.75-14x$ -------------------(equation 3)

Now substituting the value of y from (equation 3) in (equation 1) we get,

$18x+5y=59.91$

$18x+5(39.75-14x)=59.91$

$18x+198.75-70x=59.91$

$52x=198.75-59.91$

$52x=138.84$

$x=\frac{138.84}{52}$

$x=2.67$ -------------------------(equation 4)

Now substituting value of x from (equation 4) in (equation 3) we get,

$y=39.75-14x$

$y=39.75-(14\times2.67)$

$y=39.75-37.38$

$y=2.37$

Therefore price per gallon of mid grade gas is x = $2.67, and price per gallon of regular gas y = $2.37.

4 0

3 years ago

Whoever helps me with this I'll give brainliest and extra points! :)

Musya8 [376]

What are the options

6 0

2 years ago

I need help with this,I will mark braily to the right answer!

pentagon [3]

Answer:

where the qutsion

Step-by-step explanation:

3 0

2 years ago

Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0<×&lt

brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

$f'(x) = \dfrac{(-x + 2)^4}{dx} = -4 \cdot (-x + 2)^3 = 0$

∴ (-x + 2)³ = 0

x = 2

$f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx} = -12 \cdot (-x + 2)^2$

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0

3 years ago

The algebraic form of an arithmetic sequence is 5n-2<br>a) What is its common difference?

Alexxandr [17]

Answer:

5

Step-by-step explanation:

[math]

U_{n} = 5n - 2 \\

Common Difference = U_{n+1} - U_{n} \\

= [5(n + 1) - 2] - [5n - 2] = [5n + 5 - 2] - [5n - 2]\\

= 5n + 5 - 2 - 5n + 2\\

= 5

[/math]

5 0

2 years ago