F it takes 14 cup of flour to make 16 of a cake, how many cups of flour are needed to make a whole cake?

What is the range of this piecewise function

White raven [17]

Answer as a compound inequality: $-4 \le y < 2$

Answer in interval notation: [-4, 2)

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Explanation:

The range is the set of all possible y outputs of a function. When dealing with a graph like this, we just look at the highest and lowest points to determine which y values are possible.

The lowest point occurs when y = -4. We include this value. So far we have $y \ge -4$ which is the same as $-4 \le y$

The upper ceiling for the y value is y = 2. We can't actually reach this value because of the open hole at (-3,2). So we say that $y < 2$

Combine $-4 \le y$ and $y < 2$ to get the compound inequality $-4 \le y < 2$

This says y is between -4 and 2, including -4 but excluding 2.

To convert this to interval notation, we write [-4, 2) where the square bracket says to include the endpoint and the curved parenthesis says to exclude the endpoint.

6 0

3 years ago

A)A cuboid with a square x cm and height 2xcm². Given total surface area of the cuboid is 129.6cm² and x increased at 0.01cms-¹.

Nutka1998 [239]

Answer: (given assumed typo corrections)

(V ∘ X)'(t) = 0.06(0.01t+3.6)^2 cm^3/sec.

The rate of change of the volume of the cuboid in change of volume per change in seconds, after t seconds. Not a constant, for good reason.

Part B) y'(x+Δx/2)×Δx gives exactly the same as y(x+Δx)-y(x), 0.3808, since y is quadratic in x so y' is linear in x.

Step-by-step explanation:

This problem has typos. Assuming:

Cuboid has square [base with side] X cm and height 2X cm [not cm^2]. Total surface area of cuboid is 129.6 cm^2, and X [is] increas[ing] at rate 0.01 cm/sec.

129.6 cm^2 = 2(base cm^2) + 4(side cm^2)

= 2(X cm)^2 + 4(X cm)(2X cm)

= (2X^2 + 8X^2)cm^2

= 10X^2 cm^2

X^2 cm^2 = 129.6/10 = 12.96 cm^2

X cm = √12.96 cm = 3.6 cm

so X(t) = (0.01cm/sec)(t sec) + 3.6 cm, or, omitting units,

X(t) = 0.01t + 3.6

= the length parameter after t seconds, in cm.

V(X) = 2X^3 cm^3

= the volume when the length parameter is X.

dV(X(t))/dt = (dV(X)/dX)(X(t)) × dX(t)/dt

that is, (V ∘ X)'(t) = V'(X(t)) × X'(t) chain rule

V'(X) = 6X^2 cm^3/cm

= the rate of change of volume per change in length parameter when the length parameter is X, units cm^3/cm. Not a constant (why?).

X'(t) = 0.01 cm/sec

= the rate of change of length parameter per change in time parameter, after t seconds, units cm/sec.

V(X(t)) = (V ∘ X)(t) = 2(0.01t+3.6)^3 cm^3

= the volume after t seconds, in cm^3

V'(X(t)) = 6(0.01t+3.6)^2 cm^2

= the rate of change of volume per change in length parameter, after t seconds, in units cm^3/cm.

(V ∘ X)'(t) = ( 6(0.01t+3.6)^2 cm^3/cm )(0.01 cm/sec) = 0.06(0.01t+3.6)^2 cm^3/sec

= the rate of change of the volume per change in time, in cm^3/sec, after t seconds.

Problem to ponder: why is (V ∘ X)'(t) not a constant? Does the change in volume of a cube per change in side length depend on the side length?

Question part b)

Given y=2x²+3x, use differentiation to find small change in y when x increased from 4 to 4.02.

This is a little ambiguous, but "use differentiation" suggests that we want y'(4.02) yunit per xunit, rather than Δy/Δx = (y(4.02)-y(4))/(0.02).

Neither of those make much sense, so I think we are to estimate Δy given x and Δx, without evaluating y(x) at all.

Then we want y'(x+Δx/2)×Δx

y(x) = 2x^2 + 3x

y'(x) = 4x + 3

y(4) = 44

y(4.02) = 44.3808

Δy = 0.3808

Δy/Δx = (0.3808)/(0.02) = 19.04

y'(4) = 19

y'(4.01) = 19.04

y'(4.02) = 19.08

Estimate Δy = (y(x+Δx)-y(x)/Δx without evaluating y() at all, using only y'(x), given x = 4, Δx = 0.02.

y'(x+Δx/2)×Δx = y'(4.01)×0.02 = 19.04×0.02 = 0.3808.

In this case, where y is quadratic in x, this method gives Δy exactly.

6 0

3 years ago

Find the value for the given variables. Give the variables in simplest radical form.

Svetach [21]

$\huge\boxed{ \boxed{\mathrm{ \underline{Answer}࿐ }}}$

The given triangle is a right angled triangle,

so let's apply trigonometry here :

$\sin(30) = \dfrac{7}{y}$

$\dfrac{1}{2} = \dfrac{7}{y}$

$y = 7 \times 2$

$y = 14 \: \: units$

Now,

$\cos(30) = \dfrac{x}{y}$

$\dfrac{ \sqrt{3} }{2} = \dfrac{x}{14}$

$x = \dfrac{14 \sqrt{3} }{2}$

$x = 7 \sqrt{3}$

$x = 12.12 \: \: units$

_____________________________

$\mathrm{ ☠ \: TeeNForeveR \:☠ }$

3 0

2 years ago

Read 2 more answers

Find the area of a triangle with a base of 12.5 cm and a height of 8 cm

dybincka [34]

Answer:

51.2 i think

Step-by-step explanation:

7 0

2 years ago

Fiona made a score of 76 on her midterm exam for her to get a b in the course the average of her midterm exam and final exam mus

qaws [65]

Answer:

84 <= x <= 102

Step-by-step explanation:

Fiona scored 76 on her 1st midterm. Lets call x the score she will get on her 2nd exam.

We know she needs between 80 and 89 inclusive so get a b. So the question is, how much does she need to get on the 2nd test? Or, what values x can take for her to get a b?

The average she gets between both test is the arithmetic mean, this is, the sum of both tests divided by 2:

(76+x)/2

So, she need this value to be between 80 and 89. Lets use inequalities:

80 <= (76+x)/2 <= 89

We can solve it in a similar way we do for equalities. Lets start by multiplying terms in every side by 2, to eliminate the 2 dividing the central term:

2*80 <= 76+x <= 89*2

160 <= 76+x <= 178

Now, lets subtract 76 in every side so we get x alone in the center:

160 - 76 <= x <= 178 - 76

84 <= x <= 102

This is, for getting a b she need her second score to be equal or greater than 84 and less or equal to 102.

Lets verify for some values. For example, if she gets 88:

(76+84)/2 = 160/2 =80---> verifies!

And any value greater than 84 will give her a better score. Lets see what happens with 102"

(76+102)/2 = 178/2 = 89 ---> verifies

Any score greater than 102 will giver here a total sore greater than 89 so it won't be a b.

So, if she scores 102 she will get 89 (the statement does not say the range of scores, it is not usually common that the pass 100, but as the statement does not restrict it we consider here. We the maximum score is 100 we will use 84 <= x <= 100).

3 0

3 years ago