All categories

3 years ago

Eight oranges cost $1.00.How much will 5 dozen oranges cost?

Mathematics

2 answers:

vazorg [7]3 years ago

7 0

5x12=60 60 divided by 8=7.5

Nutka1998 [239]3 years ago

5 0

1 dollar is to 8
1:8
so 5 dozen
1 dozen=12
5 dozen=12 times 5=60
1:8 and x:60
1/8=x/60
multiply both sides by 60
60/8=x
30/4=15/2=7.5=x
$7.50 is how much you pay

You might be interested in

**Will give brainiest

Masteriza [31]

Answer:

707

tan(3)=a/4960

4 0

3 years ago

Read 2 more answers

Given f(x)=1/4(5-x)2 what is the value of f(11)

iren [92.7K]

Answer:

f(x)=1/4(5-x)²

Step-by-step explanation:

f(11)=1/4(5-11)²

1/4(-6)²

1/4(36)

7 0

3 years ago

Read 2 more answers

Is the relationship =~ or 180 set up on equation and solver for x find the measure the gíren

larisa86 [58]

Answer:ññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññ

8 0

3 years ago

I need help

const2013 [10]

Answer:

8 x 8 = 64

Step-by-step explanation:

Just including the second explosion, each of the 8 sparks bursts into 8, that means 8 sparks were produced 8 times, or 8x8

4 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