Can you have me pl4s

The ingredients for making chocolate cookies are 3/4 cup of brown sugar, 1/4 cup of white sugar, 3/2 cup of butter, 11/8 cup of

Naya [18.7K]

Answer:

$\frac{51}{8}$ cups.

Step-by-step explanation:

It is given that the ingredients for making chocolate cookies are $\frac{3}{4}$ cup of brown sugar, $\frac{1}{4}$ cup of white sugar, $\frac{3}{2}$ cup of butter, $\frac{11}{8}$ cup of flour, and $\frac{5}{2}$ cup of chocolate chips.

Now, we have to calculate the total number of cups of ingredients needed in making chocolate cookies.

So, it will be given by

$(\frac{3}{4} + \frac{1}{4} + \frac{3}{2} + \frac{11}{8} + \frac{5}{2})$ cups

= $\frac{3 \times 2 + 1 \times 2 + 3 \times 4 + 11 + 5 \times 4}{8}$ cups

= $\frac{51}{8}$ cups. (Answer)

3 0

3 years ago

Find the critical point for f and then use the second derivative test to decide whether the critical point is a relative maximum

anygoal [31]

Answer:the answer is 9

Step-by-step explanation:

7 0

3 years ago

Rectangle EFGH is graphed on a coordinate plane with vertices at E(-3,5), F(6,2), G(4,-4), and H(-5,-1). Find the slopes of each

Shalnov [3]

Slope of Side EF- 3
Slope of Side FG- 2/-6
Slope of Side GH- 3
Slope of Side HE- 2/6

3 0

3 years ago

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre

Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

(c) Inflection point: (0.56,-0.06)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = $\frac{d}{dx}$ [ $x^{4}ln(x)$ ]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = $4x^{3}ln(x) + x_{4}.\frac{1}{x}$

f'(x) = $4x^{3}ln(x) + x^{3}$

f'(x) = $x^{3}[4ln(x) + 1]$

Now, find the critical points: f'(x) = 0

$x^{3}[4ln(x) + 1]$ = 0

$x^{3} = 0$

x = 0

and

$4ln(x) + 1 = 0$

$ln(x) = \frac{-1}{4}$

x = $e^{\frac{-1}{4} }$

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval x-value f'(x) result

0<x<0.78 0.5 f'(0.5) = -0.22 decreasing

x>0.78 1 f'(1) = 1 increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = $x^{4}ln(x)$

f(0.78) = $0.78^{4}ln(0.78)$