4 students equally share 2/3 of a pizza

She wants to save 3/5 of 180.00 and spend the rest hpw much would she have left

Anastasy [175]

3/5 of 180 is 108
180-108=72
she would have 72$ left

6 0

3 years ago

Country Motorbikes Incorporated finds that it costs $400 to produce each motorbike, and that fixed costs are $1600 per day. The

faltersainse [42]

Answer:

30units,2900

Step-by-step explanation:

Given that Country Motorbikes Incorporated finds that it costs $400 to produce each motorbike, and that fixed costs are $1600 per day.

The price function is p(x) = 700 − 5x, where p is the price (in dollars) at which exactly x motorbikes will be sold.

If x units are produced and sold we have

Costs for x units = variable costs *x +Fixed costs $= 400x+1600$

Sales revenue = no of units sold * price = $x(700-5x) = 700x-5x^2$

Profit funciton = P(x) = Sales revenue - Total cost

= $700x-5x^2-400x-1600\\=300x-5x^2-1600$

To get maximum profit we use derivative test I derivative =0 and II derivative =negative

$P'(x) = 300-10x\\p"(x) = -10$

Producing 30 units will maximize the profit.

Max profit

=P(30) = 2900

8 0

3 years ago

Looking at the top of tower A and base of tower B from points C and D, we find that ∠ACD = 60°, ∠ADC = 75° and ∠ADB = 30°. Let t

katrin2010 [14]

Answer:

$\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24$

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of $\triangle ACD$ and the hypotenuse of $\triangle ADB$ .

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

$\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}$

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is $\angle CAD$ . The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

$\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}$

Now use this value in the Law of Sines to find AD:

$\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}$

Recall that $\sin 45^{\circ}=\frac{\sqrt{2}}{2}$ and $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ :

$AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}$

Now that we have the length of AD, we can find the length of AB. The right triangle $\triangle ADB$ is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio $x:x\sqrt{3}:2x$ , where $x$ is the side opposite to the 30 degree angle and $2x$ is the length of the hypotenuse.

Since AD is the hypotenuse, it must represent $2x$ in this ratio and since AB is the side opposite to the 30 degree angle, it must represent $x$ in this ratio (Derive from basic trig for a right triangle and $\sin 30^{\circ}=\frac{1}{2}$ ).