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3 years ago

Three vertices of parallelogram ABCD are A(−2, −2), B(2, 1), and C(0, 7). What are the coordinates of vertex D?

Mathematics

1 answer:

Alinara [238K]3 years ago

5 0

C
you have the 2 side ones and the top one, you are missing the bottom vertice

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Eastern Meats pays its delivery workers regular time for up to 36 hours per week and

Cloud [144]

Answer:

42-36=6

42+6=48

545.28 / 48=11.36

11.36

8 0

3 years ago

Please help me with these. Thanks

DIA [1.3K]

5.

f(K) = D^3 => f(25) = 125 => 25 * t = 125 ( because K is directly proportional with D^3 )=> t = 125 / 25 => t = 5 => f(25) = 25 * 5 => K * 5 = D^3 ;

6.

f(L) = F^3 => f(2) =3^3 =>f(2) = 27 => 2 / t =27 => t = 2 / 27 => t = 0.074 => f(2) = 2 / 0.074 => K / 0.074 = F^3 ;

6 0

3 years ago

How would the mean change if the number 52 replaced the number 12 in the set?

m_a_m_a [10]

Answer:

increase

Step-by-step explanation:

First, we need to find the mean of the given data set without any change.

10 + 61 + 10 + 44 + 21 + 79 + 27 + 12 = 264

264 ÷ 8 = 33

Now that we have the mean, we can find the mean of the data set with the change of 52.

10 + 61 + 10 + 44 + 21 + 79 + 27 + 52 = 304

304 ÷ 8 = 38

From this, we can see that the mean has increased with the change of 12 to 52. Thus, that's the correct option.

hope this helps!

6 0

2 years ago

Someone please help me answer this

Brrunno [24]

B= 30
a2+b2=c2
72^2+b^2 = 78^2
5184+b^2 = 6084
b2 = 900
b = 30

8 0

2 years ago

Read 2 more answers

An observer (O) spots a plane flying at a 42° angle to his horizontal line of sight. If the plane is flying at an altitude of 15

NeTakaya

Answer:

D. 22,417 feet

Step-by-step explanation:

Fine the diagram in the attachment for proper elucidation. Using the SOH, CAH, TOA trigonometry identity to solve for the distance (x) from the plane (P) to the observer (O), the longest side x is the hypotenuse and the side facing the angle of elevation is the opposite.

Hypotenuse = x and Opposite = 15,000feet

According to SOH;

$sin 42^0 = \frac{Opposite}{Hypotenuse} \\\\Sin42^0 = \frac{15000}{x}\\ \\x = \frac{15000}{sin42^0}\\\\ \\$

$x = \frac{15000}{ 0.6691} \\\\x = 22,417 feet$

Hence the distance (x) from the plane P to the observer O is approximately 22,417 feet

3 0

3 years ago