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3 years ago

Someone plz help me :(

Mathematics

1 answer:

kondaur [170]3 years ago

7 0

Answer:

the answer is 9 mark me brainlest

Step-by-step explanation:

A P E X

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Directed line segment PT has endpoints whose

crimeas [40]

Answer:

(2 , 5)

Step-by-step explanation:

point (x,y) that divides the segment PI in the ratio 2 to 1

PD=4 - (-2) = 6 PE/ED=2/1 PE=2ED PE+ED=6 3ED=6 ED=2 PE=4

coordinate of E (2,1) .... 2 is x coordinate of E

DI=7-1=6 DF/FI=2/1 DF=4

coordinate of F (4,5) .... .... 5 is y coordinate of E

CF // PD PC/CI=DF/FI=2/1

C (2,5)

4 0

3 years ago

Please help me with this question

stiks02 [169]

Do you know how to find the area and perimeter

7 0

3 years ago

Read 2 more answers

Find the IQR for the data set 24, 36, 42, 57, 65

Zielflug [23.3K]

Answer:

Step-by-step explanation:

The median of this data set is 42.

The 1st quadrant is the mean of 24 and 36: 30.

The 3rd quadrant is the mean of 57 and 65, or 61.

Thus, the IQR is the difference between 61 and 30: It is 31.

3 0

4 years ago

Read 2 more answers

Given the rectangle abcd shown below has a total area of 72. E is in the midpoint of bc and f is the midpoint of dc. What is the

scoray [572]

Refer to the attached image.

Given the rectangle ABCD of length 'l' and height 'h'.

Therefore, CD=AB = 'l' and BC = AD = 'h'

We have to determine the area of triangle AEF.

Area of triangle AEF = Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE

Area of triangle ADF = $\frac{1}{2}bh$

= $\frac{1}{2}(DF \times AD)$

= $\frac{1}{2}(\frac{l}{2} \times h)$

$=\frac{lh}{4}$

Area of triangle ECF = $\frac{1}{2}bh$

= $\frac{1}{2}(CF \times CE)$

= $\frac{1}{2}(\frac{l}{2} \times \frac{h}{2})$

$=\frac{lh}{8}$

Area of triangle ABE = $\frac{1}{2}bh$

= $\frac{1}{2}(AB \times BE)$

= $\frac{1}{2}(l \times \frac{h}{2})$

$=\frac{lh}{4}$

Now, area of triangle AEF =

Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE

= $72 - (\frac{lh}{4} + \frac{lh}{8} + \frac{lh}{4})$

= $72 - (\frac{2lh+lh+2lh}{8})$

= $72 - (\frac{5lh}{8})$

= $72 - (\frac{5 \times 72}{8})$

$=\frac{72 \times 8 - (5 \times 72)}{8}$

= 27 units

Therefore, the area of triangle AEF is 27 units.

8 0

3 years ago

Find the major axis for the ellipse <br> x² + 16y2-96y + 128 = 0

Softa [21]

The major axis for the ellipse, x² + 16y² - 96y + 128 = 0 is the x-axis

To answer the question, we need to write it in the standard form of the equation of an ellipse

<h3>Equation of an ellipse</h3>

The equation of an ellipse centered at (h,k) is

(x - h)²/a² + (y - k)²/b² (1) where a > b and the major axis is parallel to the x axis

Given x² + 16y² - 96y + 128 = 0, we convert it into the standard equation of an ellipse.

So, x² + 16y² - 96y + 128 = 0

Dividing through by 16, we have

x²/16 + 16y²/16 - 96y/16 + 128/16 = 0/16

x²/16 + y² - 6y + 8 = 0

Completing the square in y by adding and subtracting (-6/2)² = (-3)²

x²/16 + y² - 6y + (-3)² - (-3)² + 8 = 0

x²/16 + (y - 3)² - 9 + 8 = 0

x²/16 + (y - 3)² - 1 = 0

x²/16 + (y - 3)² = 1

x²/4² + (y - 3)²/1² = 1 (2)

Comparing equations (1) and (2), we have that a = 4 and b = 1.

Since a = 4 > b = 1, the major axis for the ellipse is the x-axis

So, the major axis for the ellipse, x² + 16y² - 96y + 128 = 0 is the x-axis

Learn more about ellipse here:

brainly.com/question/26679189

#SPJ1

8 0

2 years ago