All categories

2 years ago

Please help me . . . . . .

Mathematics

1 answer:

NNADVOKAT [17]2 years ago

3 0

Answer:

Step-by-step explanation:

The figure is a parallelogram. That means the y value of g is the same as the y value of H

G( ? , b)

The x value of G is a little harder. You have to be careful what you do to c.

A is at - c which means you need to add c to get to (0,0)

In addition, you go out d units to get to F. So the x value of G = c + d

the complete answer is G(c + d,b)

You might be interested in

What’s 1/3 • 5/4 and 1/5•2/3

Debora [2.8K]

Answer:

Step-by-step explanation:

1/2 + 2/3 + 5/4 = 29/

= 2 5/

≅ 2.4166667

4 0

2 years ago

Melody is solving the equation x +12=31. Find her mistake and correct it.

nalin [4]

The answer is x = 19

7 0

3 years ago

Which of the following sets are not well-defined?

Julli [10]

Set of all counting number

3 0

2 years ago

HELP!!!!!!!!!!!!!!!!!!!!!!! :)

Anestetic [448]

Hey there!

The answers are as follows:

7.)

m∠1 = 85

m∠2 = 95

m∠3 = 85

m∠4 = 85

8.)

m∠1 = 60

m∠2 = 120

m∠3 = 60

m∠4 = 120

9.)

they all equal 90

10.) No. Parallel lines do not cross each other, but these, if extended, would cross. You could use a protractor by measuring corresponding angles, and if they aren't equal, they aren't parallel.

Have a terrifcly amazing day!

3 0

2 years ago

The following describes a sample. The information given includes the five number summary, the sample size, and the largest and s

lys-0071 [83]

Answer:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

Step-by-step explanation:

For this case we have the 5 number summary

(160,210,220,250,270)

So then we have:

minimum = 160 , Q1 = 210, Q2= Median=220, Q3 = 250, Max=270

If we find the interquartile range we got:

$IQR = Q_3 -Q_1 = 250-210 =40$

For this case we need to find the lower and upper limit with the following formulas:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

4 0

3 years ago