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Snowcat [4.5K]
2 years ago
12

Original price $29.99, markdown is 33 1/3%

Mathematics
1 answer:
Allushta [10]2 years ago
7 0
Subtract 33.333% from 100%; you'll get 66.667%.

Mult the original price by 0.66667:  ($29.99)(0.667) = $20.00 (answer)

This answer reflects a 33.333% discount from the regular price of the item.

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Solve the equation 4x^2+36=0
raketka [301]

Answer:

it have no answer in real number(but it have in complex number if you know)

Step-by-step explanation:

4x^2+36=0

4x^2=-36

(if you know complex numbers see this: x^2=-9 then

x =  \frac{ + }{} 3i

)

7 0
2 years ago
Please solve this question withvfull process i will wark as braineast a) After allowing 5 % discount on the marked price of a ra
schepotkina [342]

Answer:

Let the marked price be x

Discount % = 10%

Discount = 10\% \times x = 0.1x10%×x=0.1x

Cost after discount = x-0.1x = 0.9x

10%vat is charged on it

Cost including VAT = 0.9x+0.1(0.9x)=0.99x0.9x+0.1(0.9x)=0.99x

We are given that  it's price became rs 1672

So, 0.99x = 1672

x=\frac{1672}{0.99}x=0.991672

x=1688.88

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Hence The discount

7 0
3 years ago
find the average rate of change from t=2 to t=5 for the velocity function v(t)=t^2-t+10. WILL GIVE THE BRAINLIEST ANSWER--EXPLAI
Elina [12.6K]
Answer: The average rate of change is 6.
First, plug in each value of <em>t</em> into the function, v(t) to find there coordinate pairs.

v(2) = (2)^2 - (2) + 10
v(2) = 4 + 8
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v(5) = (5)^2 - (5) + 10
v(5) = 25 + 5
v(5) = 30

You can write these values as coordinate pairs, like so: (2, 12) and (5, 30).
The formula for the average rate of change is A(x) =  \frac{f(x)-(f(a)}{x-a}. When you plug in the values from this particular case, the average rate of change formula becomes A(t) = \frac{30-12}{5-2}, or A(t)= \frac{18}{3}.

Looking at the equation, you can solve for the average rate of change between t = 2 and t = 5, which equals 6.
8 0
3 years ago
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STatiana [176]

Answer:

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Step-by-step explanation:

9^2 = 81 bc 9x9=81

3 0
3 years ago
Finding the area of A=1/2bh
DiKsa [7]

Yes that is correct.

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