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andreev551 [17]

4 years ago

14

Please help me. I'm desperate

1 answer:

Flauer [41]4 years ago

7 0

Ok so x is the original price and we want to solve if the original price was 48 so x=48. When we substitute this into the equation we get 0.75(48)-0.15(0.75x48). When we simplify this expression we get 30.6. The current price is $30.60. Hope this helps.

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Katie earned $1200 in 8 years on an investment at a 6% annual simple interest rate. How much was Katie’s investment?

Alekssandra [29.7K]

Given
Time = 8 yer a
Rate = 6% = 0.06
Interest earned = $1200
Simple Interest

Let P = the principal (the investment)
Then
P*8*0.06 = 1200
0.48P = 1200
P = 1200/0.48 = $2500

Answer: $2500

8 0

3 years ago

The area of a rectangle a=1(w) is represented by the expression 24x6y15. If the width of the rectangle is 2x3y9, what is the l

natulia [17]

Answer:

Length of rectangle is: $Length=12x^3y^6$

Step-by-step explanation:

The area of rectangle = $24x^6y^{15}$

Width of rectangle = $2x^3y^9$

We need to find length of rectangle

The formula used is: $Area=Length\times Width$

Putting values and finding length

$Area=Length\times Width\\24x^6y^{15}=Length\times 2x^3y^9\\Length=\frac{24x^6y^{15}}{2x^3y^9} \\Length=12x^{6-3}y^{15-9}\\Length=12x^3y^6$

So, Length of rectangle is: $Length=12x^3y^6$

8 0

3 years ago

what is the mass of an object that has a density of 0.7g/cm^3 and a volume of 8cm^3? (The density of an object has the equation

UNO [17]

Mass=5.6
Find this by plugging your values into the density formula and solving for M.

4 0

3 years ago

(b) Find the angle between u and v to the nearest degree.

yulyashka [42]

<u>Question 1 (b)</u>

<u /> $u \cdot v=11\\\\|u|=\sqrt{3^{2}+(-1)^{2}}=\sqrt{10}\\\\|v|=\sqrt{4^{2}+1^{2}}=\sqrt{17}$

So,

$\cos \theta=\frac{11}{\sqrt{10}\sqrt{17}}\\\\\theta=\cos^{-1} \left(\frac{11}{\sqrt{10}\sqrt{17}} \right) \approx \boxed{32^{\circ}}$

<u>Question 2 (b)</u>

<u /> $u \cdot v=22\\\\|u|=\sqrt{5^{2}+2^{2}}=\sqrt{29}\\\\|v|=\sqrt{4^{2}+1^{2}}=\sqrt{17}$

So,

$\cos \theta=\frac{22}{\sqrt{29}\sqrt{17}}\\\\\theta=\cos^{-1} \left(\frac{22}{\sqrt{29}\sqrt{17}} \right) \approx \boxed{8^{\circ}}$

6 0

2 years ago

Find z such that 6% of the standard normal curve lies to the right of z.

yulyashka [42]

P(<em>Z</em> ≥ <em>z</em>) = 1 - P(<em>Z</em> ≤ <em>z</em>) = 0.06

==> P(<em>Z</em> ≤ <em>z</em>) = 0.94

==> <em>z</em> ≈ 1.7507

4 0

3 years ago