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lbvjy [14]
3 years ago
6

A $290 suit is marked down by 10%. Find the sale price.

Mathematics
1 answer:
mariarad [96]3 years ago
5 0

Answer:

261 ! Round to the nearest dollar

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Im looking for some help on these questions. The answer that best answers the questions and helps me the most will receive brain
s344n2d4d5 [400]

<em>Detailed</em><em> </em><em>solution</em><em> </em><em>is</em><em> </em><em>attached</em><em>!</em><em>!</em><em>~</em>

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7 0
2 years ago
Which is bigger 500 ft or 40 ft?
Sav [38]

Answer: 500 ft is bigger

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Could someone please help me with the question please
VladimirAG [237]

Answer:

parallel

Step-by-step explanation:

parallel lines are straight lines and never meet if you carried them on. Perpendicular lines meet at a right angle, so those lines are parallel. :)

5 0
3 years ago
15. The average speed that a tsunami (a large tidal wave) travels is represented
irina [24]

Correct question :

The average speed that a tsunami (a large tidal wave) travels is represented

by the functions = (200d)^1^/^2 where s is the speed (in miles per hour) that the tsunami is traveling and d is the average depth (in feet) of the wave.

a. Find the inverse of the function.

b. Find the average depth of the tsunami when the recorded speed of the wave is 250 miles per hour.

Answer:

a) d = \frac{s^2}{200}

b) 312.5 ft

Step-by-step explanation:

Given:

Average speeds, s = (200d)^1^/^2

a) To find the inverse of the function.

s = (200d)^1^/^2

s² = 200d

200d = s²

d = \frac{s^2}{200}

Therefore, inverse of the function =

d = \frac{s^2}{200}

b) average depth when speed is 250 miles per hour.

Average depth = d

Therefore, let's use the formula :

d = \frac{s^2}{200}

= \frac{250^2}{200}

= \frac{62500}{200}

d = 312.5 feet

The average depth when speed is 250 miles per hour is 312.5 ft

3 0
3 years ago
Read 2 more answers
If f(x) = 3x to the 2nd+ 1 and g(x) = 1-x, what is the value of (f-g)(2)?
melisa1 [442]

 

\displaystyle\\f(x)=3x^2+1\\g(x)=1-x\\\\(f-g)(x)=f(x)-g(x)=3x^2+1-(1-x)=3x^2+1-1+x=\boxed{3x^2+x}\\\\(f-g)(2)=f(2)-g(2)=3\cdot2^2+2=3\cdot4+2=12+2=\boxed{14}




3 0
3 years ago
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