All categories

3 years ago

Determine a and b so that figure ABCD is a parallelogram

Mathematics

1 answer:

ziro4ka [17]3 years ago

4 0

Answer:

C(8,2)

That is the answer for a and b

You might be interested in

A right triangle has a leg measuring 52 m and a hypotenuse measuring 63 m. find the length of the other leg.

tangare [24]

By Pythagoras Theorem,
the length of the other leg= (63^2-52^2)^1/2
hope this helps :)

4 0

3 years ago

Find the midpoint of the line segment whose endpoints are (3, 10) and (7, 8).

Stels [109]

Answer:

Option 2 is correct.

Step-by-step explanation:

Given the coordinates of lines segment (3, 10) and (7, 8). we have to find the mid-point of given line segment.

Mid-point formula states that if $(x_1,y_1)$ and $(x_2,y_2)$ are the coordinates of end points of line segment then the coordinates of mid-point are

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

∴ Coordinates of mid-point of line segment joining the points (3, 10) and (7, 8) are

$(\frac{3+7}{2},\frac{10+8}{2})=(\frac{10}{2},\frac{18}{2})=(5,9)$

Hence, option 2 is correct.

6 0

2 years ago

Read 2 more answers

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

3 years ago

Which expression is equivalent to

lisabon 2012 [21]

Answer:

i think the answer is, A...

8 0

2 years ago

The company has a sample box that is a rectangular prism with a rectangular base with an area of 231⁄3 in2. The height of the pr

vodka [1.7K]

Answer:

$29.17 in^3$

Step-by-step explanation:

We are given that

Base area of sample box,a= $23\frac{1}{3} in^2=\frac{70}{3}in^2$

Height of the sample box,h= $1\frac{1}{4}=\frac{5}{4} in$

We have to find the volume of the sample box.

We know that

Volume of rectangular prism, $V=ah$

Where a=Area of base

Using the formula

Volume of sample box= $\frac{70}{3}\times \frac{5}{4}=29.17 in^3$

Hence, the volume of the sample box= $29.17 in^3$

6 0

2 years ago