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melomori [17]
4 years ago
9

Find the result when 90 is decreased by 85%.

Mathematics
1 answer:
alexira [117]4 years ago
8 0

Answer:

13.5

Step-by-step explanation:

100 - 85 = 15

90 x 0.15 = 13.5

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Is 12 degrees below zero a negative quantity​
Natalka [10]

Answer:

Yes

Step-by-step explanation:

Because it is 12 <u><em>below</em></u> 0, it is a negative number.(-12)

4 0
3 years ago
Read 2 more answers
The length of a rectangle is 3 cm greater than its width. The perimeter is 24 cm. Find the dimensions of the rectangle.
Shalnov [3]

Answer:

The width is 4.5, and the length is 7.5.

Step-by-step explanation:

We know that the perimeter is 24 cm, and the length is 3 cm greater than the width. We know that the perimeter = l*w. This info gives us an equation:

24 = (x + 3)*2 + x * 2. The 24 is the perimeter, and x is the width, so l is x + 3. Now we just have to solve for x.

24 = (x + 3)*2 + x * 2 | Simplify the Parenthesis

24 = 2x + 6 + x * 2 | Subtract 6 from both sides

18 = 2x + x * 2 | x * 2 is 2x

18 = 2x + 2x | 2x + 2x = 4x

18 = 4x | Now we divide by 4

4.5 = x. This mean the width is 4.5, and the length is 4.5 + 3, or 7.5. We can plug these into the original formula to confirm.

24 = 7.5*2 + 4.5*2, 24 = 15 + 9, 24 = 24.

6 0
3 years ago
An oval track is made by erecting semicircles on each end of a 44 m by 88 m rectangle. Find the length of the track and the area
Dmitry_Shevchenko [17]

Answer:

The length of the track is 247 m.

The area enclosed by the track is  5391 . 76 sq m.

Step-by-step explanation:

The length of the rectangle  = 88 m

The width of the rectangle  =  44 m

Now, as we know the semicircles are at the end of rectangular track.

So, the diameter of semicircles = Width of the rectangle

⇒ The diameter of the semi circle  = 44 m

⇒Radius of the semi circle  = 44/2 = 22 m

TOTAL LENGTH OF TRACK  

= 2 ( Length of rectangle)  + 2 (Circumference of 1 semicircle)

= 2 x (88 m)  + 2 (\frac{2\pi r}{2})

= 176 m + 2 (3.14)(22 m)  = 176 + 71.08 =  247 m

Hence, the length of the track is 247 m

TOTAL AREA OF TRACK  

= (Area of rectangle)  + 2 (Area of 1 semicircle)

= (Length x Width)  +    2 (\frac{\pi r^2}{2})

=  (44  m x 88 m) +  (3.14)(22 m)(22 m)   = (3872  + 1519.76) sq m

=  5391 . 76 sq m

Hence, the area enclosed by the track is  5391 . 76 sq m.

7 0
3 years ago
Can someone please help me with these problems?
HACTEHA [7]

Answer:

number 1 is 5 inches

Step-by-step explanation:

it fits in the 15 inches 3 times.

5*3=15

6 0
3 years ago
ALGEBRA 1, SEMESTER 2- please help, im failing my class already
Maru [420]

Velocity is given as 80 feet per second, so replace V with 80.

The rocket is launched from the ground so the initial height is 0, so replace h with 0


The formula now becomes" h(t) = -16t^2 + 80t +0


To find the maximum we need to find the coordinates of the vertex by finding the axis of symmetry:

-b/2a = -80/(2*-16) = -80/-32 = 2.5

This represents the amount of time for the rocket to reach maximum height.

So we now replace t in the equation with 2.5 and solve for the height:

h(2.5) = -16(2.5)^2 + 80(2.5)

h(2.5) = 100


The maximum height is 100 feet

4 0
3 years ago
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