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3 years ago

How much sugar should he use ?

Mathematics

1 answer:

Sonja [21]3 years ago

6 0

The answer is A. 1 1/4

You have to subtract 1 3/4 - 1/2 and get 1 1/4

Hope this helps!

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If you try to solve a linear system by the substitution method and get the result 0 = 8, what does this mean? The system has

Andre45 [30]

Answer:

The system has no solution because if 0=8, this means this is an empty set. If you get 2 numbers that don't equal each other for an answer then you know there's no solution

Step-by-step explanation:

5 0

3 years ago

Solve:<br><br> If Sally can walk 1/3 of a mile in 1/4 of an hour. How far will she walk 2 1/2 hours?

Daniel [21]

The answer is 3 and 1/3 miles

8 0

3 years ago

Read 2 more answers

On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or

almond37 [142]

Answer:

a) $\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$ , b) $\theta = \frac{\pi}{4}$ , c) $y_{max} = 84.375\,m$ , $t = 3.75\,s$ .

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

$\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j$

The particular expression for the cannonball is:

$\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$

b) The components of the position of the cannonball before hitting the ground is:

$x = (90\cdot \cos \theta)\cdot t$

$0 = 90\cdot \sin \theta - 6\cdot t$

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

$0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)$

$0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x$

$0 = 4050\cdot \sin 2\theta - 6\cdot x$

$6\cdot x = 4050\cdot \sin 2\theta$

$x = 675\cdot \sin 2\theta$

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

$\frac{dx}{d\theta} = 1350\cdot \cos 2\theta$

$1350\cdot \cos 2\theta = 0$

$\cos 2\theta = 0$

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

$\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta$

$\frac{d^{2}x}{d\theta^{2}} = -2700$

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

$y = 45\cdot t - 6\cdot t^{2}$

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

$\frac{dy}{dt} = 45 - 12\cdot t$

$45-12\cdot t = 0$

$t = \frac{45}{12}$

$t = 3.75\,s$

Now, the second derivative is used to check if such solution leads to a maximum:

$\frac{d^{2}y}{dt^{2}} = -12$

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

$y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}$

$y_{max} = 84.375\,m$

7 0

3 years ago

There was 5/8 of a pie left in the fridge. Daniel are 1/4 of the left over pie. How much of a pie did he eat

oee [108]

Answer:

vvvv

Step-by-step explanation:

1. Make it so 1/4 is an eighth so you can subtract it from 5/8.

1/4 x 2 = 2/8

2. Subtract 2/8 from 5/8.

5/8 - 2/8 = 3/8

Daniel ate 2/8s of the left over pie, and there is 3/8s remaining.

7 0

3 years ago

Read 2 more answers

An observer at the top of a 532 foot cliff measures the angle of depression from the top of the cliff to a point on the ground t

Over [174]

Answer:

Distance from the base of the cliff to the point on the ground = 7608 feet

Step-by-step explanation:

Given: Height of the cliff is 532 feet, angle of depression from the top of the cliff to a point on the ground is equal to 4 degrees.

To find: distance from the base of the cliff to the point on the ground

Solution:

In ΔABC,

$\angle ACB=4^{\circ}$ (Alternate interior angles)

For any angle $\theta$ , $\tan \theta =$ side opposite to angle/side adjacent to angle

$\tan C=\frac{AB}{BC}$

Put $AB=532\,,\,\angle C=4^{\circ}$

$\tan 4^{\circ}=\frac{532}{BC}\\\\BC=\frac{532}{\tan 4^{\circ}}\\\\=7607.95\\\\\approx 7608\,\,feet$

Distance from the base of the cliff to the point on the ground = 7608 feet

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5 0

3 years ago