What is the decimal equivalent of 11/50

What is the complete factorization of r2 - 6r - 9s2 + 9? A) (r - 3 + 3s)(r - 3 - 3s) B) (r + 3 + 3s)(r - 3 - 3s) C) (r - 3 + 3s)

nikitadnepr [17]

$r^2-6r-9s^2+9=r^2-6r+9-9s^2=r^2-2\cdot r\cdot3+3^2-9s^2\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=(r-3)^2-9s^2=(r-3)^2-(3s)^2\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=(r-3-3s)(r-3+3s)\to A$

6 0

3 years ago

What is the volume of the composite figure? Round your answer to the nearest hundredth. Use 3.14 for π.

Reil [10]

Answer:

263.76 in^3

VOLUME OF FIGURE=VOL.OF CONE+VOL.OF CYLINDER +VOL.OF HEMISPHERE

=(1/3×3.14×3×3×4)+(3.14×3×3×6)+(2/3×3.14×3×3×3)

=37.68+169.56+56.52

=263.76in^3

7 0

3 years ago

Find the area of the parallelogram. The figure is not drawn to scale.

valina [46]

Answer:

1330 in.

Step-by-step explanation:

38 x 35 = 1330

3 0

3 years ago

Read 2 more answers

Evaluate the function at x = –2. y = 0 y = 3 y = –4 y = 2

dezoksy [38]

When x=-2, y is equal to 3

4 0

3 years ago

A certain forest covers an area of

iVinArrow [24]

Answer:

2598 square kilometers

Step-by-step explanation:

Hello

Step 1

year one

using a rule of three is possible to find how much is 8.75 od 4500 km2

Let

if

4500 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$4500:100\\x:8.75\\\frac{4500}{100}=\frac{x}{8.75}\\x=\frac{4500*8.75}{100} \\x=393.75\\$

at the end of the year one, the area will be

4500-393.75=4106.25

this will be the initial area for the year 2.

Step 2

repite the step 1 with area initial =4106.25 km2

4106.25 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$4106.25:100\\x:8.75\\\frac{4106.25}{100}=\frac{x}{8.75}\\x=\frac{4106.25*8.75}{100} \\x=359.29\\$

at the end of the year 2, the area will be

4106-359.29=3746.70

this will be the initial area for the year 3.

Step 3

repite the step 1 with area initial =4106.25 km2

3746.70 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$3746.70:100\\x:8.75\\\frac{3746.70}{100}=\frac{x}{8.75}\\x=\frac{3746.7*8.75}{100} \\x=327.83\\$

at the end of the year 3, the area will be

3746.70-327.83=3419.09

this will be the initial area for the year 4.

Step 4

year four

repite the step 1 with area initial =3419.09 km2

3419.09 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$3419.09:100\\x:8.75\\\frac{3419.09}{100}=\frac{x}{8.75}\\x=\frac{3419.09*8.75}{100} \\x=299.17\\$

at the end of the year 4, the area will be

3419.09-299.173=3119.82

this will be the initial area for the year 5.

Step 5

year five

repite the step 1 with area initial =3119.82 km2

3119.82 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$3119.82:100\\x:8.75\\\frac{3119.82}{100}=\frac{x}{8.75}\\x=\frac{3119.82*8.75}{100} \\x=272.99\\$

at the end of the year 5, the area will be

3119.82-272.99=2846.92

this will be the initial area for the year 6.

Step 6

year six

repite the step 1 with area initial =2846.92km2

2846.92 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$2846.92:100\\x:8.75\\\frac{2846.92}{100}=\frac{x}{8.75}\\x=\frac{2846.92*8.75}{100} \\x=249.10\\$

at the end of the year six, the area will be

2846.92-249.10=2597.82 square kilometers

Have a great day.

8 0

3 years ago