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3 years ago

Subtract (5x2 3) −(2x2 4x − 12)

Mathematics

2 answers:

timurjin [86]3 years ago

6 0

Distribute invisible-1 to the 2x^2+4x-12
5x^2+3-2x^2-4x+12
group like terms
5x^2-2x^2-4x+3+12
3x^2-4x+15

Airida [17]3 years ago

3 0

Answer:

3x2 − 4x + 15

(just took the quiz)

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Order each set of numbers from least to greatest 0.99 0.89 7/8

kifflom [539]

7/8 = 0.875
1. 7/8
2. 0.89
3. 0.99

3 0

3 years ago

The question is there

Katarina [22]

Answer:

3 cm

Step-by-step explanation:

Set up an equation that solves for the surface area, "112 $cm^{2}$ " :

10(2)+10(2)+2x+2x+10x+10x = 112

20+20+4x+20x = 112

24x+40=112

24x=72

x=3

So, 3 cm.

7 0

3 years ago

PLLLLZ HELP ME!!!!!! What is the area of the parallelogram plotted in the Coordinate Plan in square units?

charle [14.2K]

I believe the answer is B

5 0

3 years ago

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

4 years ago

Find the surface area and volume of the composite figure. Use 3.14 for π and round to the nearest tenth if needed.

Vesnalui [34]

Answer:

Step-by-step explanation:

The shape consists of three parts: a half sphere, a round cylinder and a round cone.

The half sphere volume = 1/2 * 4/3 * π * (4)^3

= 32/3 π

The round cyllinder volume = π * (4)^2 * 10

= 160 π

The round cone volume = 1/3 * π * (4)^2 * 3

= 16 π

Adding them together, the volume of shape

= 32/3 π + 160π + 16π

= (32/3 + 160 + 16)*3.14

= 586.13

Similarly the surface area

= (4π4^2)/2 + 2π4*10 + π*4*5

= (32+80+25)*3.14

= 430.18

3 0

3 years ago

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