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3 years ago

What is the area of the figure? 5m. 2m. 12m.

Mathematics

2 answers:

Iteru [2.4K]3 years ago

8 0

5 x 2 x 12 = 120 m^3

notsponge [240]3 years ago

6 0

5 x 2 x 12 = 120
this is the area of the figure

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Determine the value of s, the arc length (measured in inches) cut off in a circle with a radius of 3.4 inches by an angle with a

AleksandrR [38]

Answer:

Arc length s = 15.3 inches

Step-by-step explanation:

We know that length of an arc that subtends an angle θ radians at the center is given by

$s=r\times \Theta$

where

r is the radius of the circle

θ is the angle subtended in radians

Thus applying values we get $s=3.4\times \ 4.5radians$

thus s = 15.3 inches

4 0

3 years ago

Given that (ax^2 + bx + 3) (x + d) = x^3 + 6x^2 + 11x + 12<br> a + 2b - d = ?

Mariulka [41]

Answer:

Let's solve for a.

(ax2+bx+3)(x+d)=x3+6x2+11x+12a+2b−d

Step 1: Add -12a to both sides.

adx2+ax3+bdx+bx2+3d+3x+−12a=x3+6x2+12a+2b−d+11x+−12a

adx2+ax3+bdx+bx2−12a+3d+3x=x3+6x2+2b−d+11x

Step 2: Add -bdx to both sides.

adx2+ax3+bdx+bx2−12a+3d+3x+−bdx=x3+6x2+2b−d+11x+−bdx

adx2+ax3+bx2−12a+3d+3x=−bdx+x3+6x2+2b−d+11x

Step 3: Add -bx^2 to both sides.

adx2+ax3+bx2−12a+3d+3x+−bx2=−bdx+x3+6x2+2b−d+11x+−bx2

adx2+ax3−12a+3d+3x=−bdx−bx2+x3+6x2+2b−d+11x

Step 4: Add -3d to both sides.

adx2+ax3−12a+3d+3x+−3d=−bdx−bx2+x3+6x2+2b−d+11x+−3d

adx2+ax3−12a+3x=−bdx−bx2+x3+6x2+2b−4d+11x

Step 5: Add -3x to both sides.

adx2+ax3−12a+3x+−3x=−bdx−bx2+x3+6x2+2b−4d+11x+−3x

adx2+ax3−12a=−bdx−bx2+x3+6x2+2b−4d+8x

Step 6: Factor out variable a.

a(dx2+x3−12)=−bdx−bx2+x3+6x2+2b−4d+8x

Step 7: Divide both sides by dx^2+x^3-12.

a(dx2+x3−12)

dx2+x3−12

−bdx−bx2+x3+6x2+2b−4d+8x

dx2+x3−12

−bdx−bx2+x3+6x2+2b−4d+8x

dx2+x3−12

Answer:

−bdx−bx2+x3+6x2+2b−4d+8x/

dx2+x3−12

Step-by-step explanation:

8 0

3 years ago

What is the mathematical notation for absolute value?

alukav5142 [94]

Use vertical pipes around the value.

Example:

abs(-4) = |-4| = 4

4 0

3 years ago

Read 2 more answers

If you have 36 ft of fencing what are the area of the different rectangles you could enclose with the fencing?consider only whol

n200080 [17]

The fence you have must fit the perimeter of the rectangle.

With 36 feet of fence, these are the rectangles that you can enclose:

1-ft x 17-ft . . . Area = 17 ft²

2 x 16 . . . Area = 32 ft²
3 x 15 . . . Area = 45 ft²
4 x 14 . . . Area = 56 ft²
5 x 13 . . . Area = 65 ft²
6 x 12 . . . Area = 72 ft²
7 x 11 . . . Area = 77 ft²
8 x 10 . . . Area = 80 ft²
9 x 9 . . . Area = 81 ft²

6 0

4 years ago

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If 8 and 15 are sides to a right triangle, whats the missing side?

mr Goodwill [35]

Answer:

Step-by-step explanation:

a[2]+b[2]=c[2]

8*8+15*15=hypotenuse.

64+225=289

$\sqrt{x} 289$