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lapo4ka [179]
3 years ago
11

How many 3/4 cup servings of cereal can be made from the box of cereal that has 11 1/4 cereal?

Mathematics
2 answers:
Kipish [7]3 years ago
8 0
15 cereal servings that measure 3/4 cup 
Daniel [21]3 years ago
7 0
11 1/4 = 45/4
 then we divide this bu 3/4
45/4 / 3/4 = 45/4 x 4/3 = 15
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Need answers to the ones that i haven’t written anything for
Trava [24]

Answer:

domain - (-infinity,infinity)

range - (-infinity, 2]

axis of sym - x=-1

y int - (0,-1)

at what value (row 1) - (-1,2)

Step-by-step explanation:

domain is side to side, it never ends bc of the arrows

range is top to bottom, there is no bottom bc of the arrow, but it has a top at 2 (the value on the y axis)

the axis of sym is basically the line that runs through the vertex, it's where you could fold it in half and it would still match up

8 0
3 years ago
Determine whether each expression can be used to find the length of side AB. Match Yes or No for each
tankabanditka [31]

Answer:

(a)\ AB = \frac{7}{\sin (B)}  \to Yes

(b)\ AB = \frac{24}{\cos (B)} \to Yes

(c)\ AB = \frac{24}{\cos (A)} \to No

(d)\ AB = \frac{7}{\cos (A)}  \to Yes

Step-by-step explanation:

Given

BC =24

AC = 7

Required

Select Yes or No for the given options

(a)\ AB = \frac{7}{\sin (B)}  \to Yes

Considering the sine of angle B, we have:

\sin(B) = \frac{Opposite}{Hypotenuse}

\sin(B) = \frac{7}{AB}

Make AB, the subject

AB = \frac{7}{\sin(B)}

(b)\ AB = \frac{24}{\cos (B)} \to Yes

Considering the cosine of angle B, we have:

\cos(B) = \frac{Adjacent}{Hypotenuse}

\cos(B) = \frac{24}{AB}

Make AB the subject

AB = \frac{24}{\cos(B)}

(c)\ AB = \frac{24}{\cos (A)} \to No

Considering the cosine of angle B, we have:

\cos(A) = \frac{Adjacent}{Hypotenuse}

\cos(A) = \frac{7}{AB}

Make AB the subject

AB = \frac{7}{\cos(A)}

(d)\ AB = \frac{7}{\cos (A)}  \to Yes

<em>This has been shown in (c) above</em>

3 0
3 years ago
A rock thrown vertically upward from the surface of the moon at a velocity of 36​m/sec reaches a height of s = 36t - 0.8 t^2 met
Verdich [7]

Answer:

a. The rock's velocity is v(t)=36-1.6t \:{(m/s)}  and the acceleration is a(t)=-1.6  \:{(m/s^2)}

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

  • Velocity is defined as the rate of change of position or the rate of displacement. v(t)=\frac{ds}{dt}
  • Acceleration is defined as the rate of change of velocity. a(t)=\frac{dv}{dt}

a.

The rock's velocity is the derivative of the height function s(t) = 36t - 0.8 t^2

v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t

The rock's acceleration is the derivative of the velocity function v(t)=36-1.6t

a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6

b. The rock will reach its highest point when the velocity becomes zero.

v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = \frac{405}{2} =202.5 \:m

To find the time it reach half its maximum height, we need to solve

36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45

The rock is aloft for 45 seconds.

5 0
3 years ago
How many times larger is the value 0.75 than the value 0.0075?
azamat
0.75 is greater the 0.0075
4 0
3 years ago
Read 2 more answers
A candle is 9 inches long. Ronnie lights the candle and records the height of the candle, y inches, for x hours.​
DochEvi [55]

In the future, please post the full problem with all included instructions. After doing a quick internet search, I found your problem listed somewhere else. It mentions two parts (a) and (b)

Part (a) asked for the equation of the line in y = mx+b form

That would be y = -2x+9

This is because each time y goes down by 2, x goes up by 1. We have slope = rise/run = -2/1 = -2. This indicates that the height of the candle decreases by 2 inches per hour. The slope represents the rate of change.

The initial height of the candle is the y intercept b value. So we have m = -2 and b = 9 lead us from y = mx+b to y = -2x+9

----------------------------------------------------------------

Part (b) then asks you to graph the equation. Because this is a linear equation, it produces a straight line. We only need 2 points at minimum to graph any line. Let's plot (0,9) and (1,7) on the same xy grid. These two points are the first two rows of the table. Plot those two points and draw a straight line through them. The graph is below

7 0
2 years ago
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