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4 years ago

1) 2m + 3m2 - 4m

Mathematics

1 answer:

Lilit [14]4 years ago

5 0

Answer:

1) 2m + 3m2 - 4m=7

7) 3m2 – 2m + 4m= 11

8) 20 + 109 + 39 - 4= 164

3) 2m + 4m - 3m2= -3

9) 4xy + x + 2xy= 0

4) 2y + 14x - 7x + 9y= 18

10) 6m2 - 6m - 9m2= -51

Step-by-step explanation:

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There are six households in a rural community. Each household earns $40,000 per year. Suppose that a new resident builds a mansi

dusya [7]

Answer:

Median income has not changed

Mean income has increased

Step-by-step explanation:

Initial number of houses = 6

Earning per household = $40,000

The initial mean :

$(40,000 * 6) / 6 = $40000

Initial median :

40000, 40000, 40000, 40000, 40000, 40000

= 1/2(n+1)th term,

Median = 40000

A 7th household earns 4,000,000

Mean = ((40000*6) + 4,000,000) / 7 = $605,714

Median = 1/2(n+1)th term

40000, 40000, 40000, 40000, 40000, 40000, 4000000

Median = 4th term = 40000

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3 years ago

The graph of f(x) = x2 has been shifted into the form f(x) = (x − h)2 + k: a parabola with a vertex of 4, negative 2 What is the

katrin [286]

Answer:

k = -2

Step-by-step explanation:

f(x) = (x − h)² + k

(h, k) is the (x, y) coordinate of the vertex

For a vertex of (4, -2)

h = 4

k = -2

7 0

3 years ago

Match each spherical volume to the largest cross sectional area of that sphere

zlopas [31]

Answer:

Part 1) $324\pi\ units^{2}$ ------> $7,776\pi\ units^{3}$

Part 2) $36\pi\ units^{2}$ ------> $288\pi\ units^{3}$

Part 3) $81\pi\ units^{2}$ ------> $972\pi\ units^{3}$

Part 4) $144\pi\ units^{2}$ ------> $2,304\pi\ units^{3}$

Step-by-step explanation:

we know that

The largest cross sectional area of that sphere is equal to the area of a circle with the same radius of the sphere

Part 1) we have

$A=324\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

$324\pi=\pi r^{2}$

Solve for r

$r^{2}=324$

$r=18\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=18\ units$

substitute

$V=\frac{4}{3}\pi (18)^{3}$

$V=7,776\pi\ units^{3}$

Part 2) we have

$A=36\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

$36\pi=\pi r^{2}$

Solve for r

$r^{2}=36$

$r=6\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=6\ units$

substitute

$V=\frac{4}{3}\pi (6)^{3}$

$V=288\pi\ units^{3}$

Part 3) we have

$A=81\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

$81\pi=\pi r^{2}$

Solve for r

$r^{2}=81$

$r=9\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=9\ units$

substitute

$V=\frac{4}{3}\pi (9)^{3}$

$V=972\pi\ units^{3}$

Part 4) we have

$A=144\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

$144\pi=\pi r^{2}$

Solve for r

$r^{2}=144$

$r=12\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=12\ units$

substitute

$V=\frac{4}{3}\pi (12)^{3}$

$V=2,304\pi\ units^{3}$

5 0

3 years ago

Read 2 more answers

Think about the sets of rational numbers, integers, and whole numbers.

Rus_ich [418]

Answer:

Step-by-step explanation:

i think it is, sorry if im wrong

7 0

3 years ago

Read 2 more answers

In 1970, a farmer using a two-row combine with a corn head could easily harvest 0.75 ha (1.85 acres) per hour and a typical corn

Anastaziya [24]

Answer:

The value is $w = 6.7 \%$

Step-by-step explanation:

Generally the productivity rate is mathematically represented as

$P_r = K * Z$

Here K is the ha the farmer could harvest in an hour and Z is a typical yield of corn in kg/ha

Now considering 1970

The productivity rate is

$P_r = 0.75 *5000$

=> $P_r = 3750 \ kg/h$

Now considering 1940

The productivity rate is given as $P_R = 250 \ kg/h$

Generally the productivity ratio is mathematically ration is mathematically represented as

$w = \frac{P_R}{P_r} * 100$

=> $w = \frac{250}{3750} * 100$

=> $w = 6.7 \%$

5 0

4 years ago