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sleet_krkn [62]
3 years ago
11

Please help with these 2 questions for 15 points I think

Mathematics
1 answer:
Yuki888 [10]3 years ago
3 0

Answer:

$13.50

$750

Step-by-step explanation:

You might be interested in
15/55 in simplest form
poizon [28]
The answer you are looking for is 3/11 because you can divide both numbers by five without having to further simplify it
3 0
4 years ago
Please help me to prove this!​
Ymorist [56]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π              → A + B = π - C

                                              → B + C = π - A

                                              → C + A = π - B

                                              → C = π - (B +  C)

Use Sum to Product Identity:  cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use the Sum/Difference Identity: cos (A - B) = cos A · cos B + sin A · sin B

Use the Double Angle Identity: sin 2A = 2 sin A · cos A

Use the Cofunction Identity: cos (π/2 - A) = sin A

<u>Proof LHS → Middle:</u>

\text{LHS:}\qquad \qquad \cos \bigg(\dfrac{A}{2}\bigg)+\cos \bigg(\dfrac{B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)

\text{Sum to Product:}\qquad 2\cos \bigg(\dfrac{\frac{A}{2}+\frac{B}{2}}{2}\bigg)\cdot \cos \bigg(\dfrac{\frac{A}{2}-\frac{B}{2}}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)

\text{Given:}\qquad \quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)

\text{Sum/Difference:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)

\text{Double Angle:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{2(A+B)}{2(2)}\bigg)\\\\\\.\qquad \qquad  \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+2\sin \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)

\text{Factor:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg]

\text{Cofunction:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi}{2}-\dfrac{A+B}{4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{2\pi-(A+B)}{4}\bigg)\bigg]

\text{Sum to Product:}\ 2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[2 \cos \bigg(\dfrac{2\pi-2B}{2\cdot 4}\bigg)\cdot \cos \bigg(\dfrac{2A-2\pi}{2\cdot 4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)

\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{\pi -C}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)

LHS = Middle \checkmark

<u>Proof Middle → RHS:</u>

\text{Middle:}\qquad 4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)\\\\\\\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)

Middle = RHS \checkmark

3 0
3 years ago
With food prices becoming a great issue in the world; wheat yields are even more important. Some of the highest yielding dry lan
Rzqust [24]

Answer:

Option E) 61.6

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100 bushels per acre

Standard Deviation, σ = 30 bushels per acre

We assume that the distribution of yield is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(X>x) = 0.90

We have to find the value of x such that the probability is 0.90

P(X > x)  

P( X > x) = P( z > \displaystyle\frac{x - 100}{30})=0.90  

= 1 -P( z \leq \displaystyle\frac{x - 100}{30})=0.90  

=P( z \leq \displaystyle\frac{x - 100}{30})=0.10  

Calculation the value from standard normal table, we have,  

P(z

\displaystyle\frac{x - 100}{30} = -1.282\\x = 61.55 \approx 61.6  

Hence, the yield of 61.6 bushels per acre or more would save the seed.

7 0
3 years ago
Which General won the Battle of Gettysburg?
lesya692 [45]

Answer: The Cautious Meade

Battle of Gettysburg: Aftermath and Impact

The Union had won the Battle of Gettysburg. Though the cautious Meade would be criticized for not pursuing the enemy after Gettysburg, the battle was a crushing defeat for the Confederacy.

Hope this helped<3.

5 0
3 years ago
5. A rectangular pool ha a length that is 13 meters longer than its width. perimeter is74 meters. What is the width Its of this
gregori [183]
Good evening Brian,

For this problem, let's look at what we're given.  So we have a pool with a length that is 13 m (meters) longer than its width, and we're given the perimeter, or distance around the entire pool, which is 74 m.

We know that the pool has a rectangular shape and that the perimeter of a rectangle is <span>width + length + width + length ⇒ 2(W) + 2(L)</span>.

Given the information provided, we can rewrite the equation to fit this problem.  Since we're told that the length is equal to 13 m + the width, so we can represent the length as W +13 m.  We can now rewrite the perimeter equation to be:

2(w) + 2 (w + 13 m) ⇒ 2(w) + 2(w) + 2(13 m) ⇒ 4(w) + 26 m = 74 m.

We're down to one variable now so this should be easy.  Subtract 26 m from both sides,

4(w) = 48 m

Now divide each side by 4 in order to find the width.

w = 12 m

-Hope this helps!


6 0
3 years ago
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