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3 years ago

Point m is the midpoint of line segment ab, b is at (20,4), m is at (18, -2) and a is at (p,q). what is p+q?

1 answer:

8 0

The x-value of a midpoint is the average of the x-values of the ends.
The y-value of a midpoint is the average of the y-values of the ends.

The x-value of the end at b is 20 and it's 18 at the middle.
x has decreased by 2 in the course of 1/2 of the line.
So it will decrease another 2 during the other half of the line.
The x-value at the other end, at a, then, is 16.

The y-value of the end at b is 4 and it's -2 at the middle.
y has decreased by 6 in the course of 1/2 of the line.
So it will decrease another 6 during the other half of the line.
The y-value at the other end, at a, then, is -8.

The other end of the line, at a, is at (16, -8) .

The sum of the coordinates at a (p+q) is 8 .

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52<br> base six <br><br> Write the normal number

saveliy_v [14]

19770609664 is simple power of 52⁶ .

What is exponent and power?

A base number raised to an exponent is referred to as a power in mathematics, where base number is the factor that is multiplied by itself and exponent is the number of times the same base number is multiplied.
Exponents and powers are two techniques for simplification of extremely big or extremely small numbers.
As an illustration, we can write 3 x 3 x 3 x 3 as 34, where 4 is the exponent and 3 is the base, if we need to represent it simply.
It is claimed that the entire statement 34 has power.

= 52⁶

= 52 × 52 × 52 × 52 × 52 × 52

= 19770609664

Learn more about Exponent

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4 0

10 months ago

Please help I will mark brainliest

Rasek [7]

A) Correct

20-3=17

B) Incorrect

6(16) =96

C) Incorrect

11 does not equal 12

8 0

3 years ago

Read 2 more answers

Find the volume of the triangular prism.

Korvikt [17]

This is the triangular prism

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2 years ago

Find a particular solution to the differential equation y′′+2y′+y=2t2−t+3e−2t. y′′+2y′+y=2t2−t+3e−2t.

Jlenok [28]

$y''+2y'+y=2t^2-t+3e^{-2t}$

The characteristic equation is

$r^2+2r+1=(r+1)^2=0\implies r=-1$

so the characteristic solution is

$y_c=C_1e^{-t}+C_2te^{-t}$

For the particular solution, we can try looking for a solution of the form

$y_p=at^2+bt+c+de^{-2t}$
$\implies{y_p}'=2at+b-2de^{-2t}$
$\implies{y_p}''=2a+4de^{-2t}$

Substituting into the ODE, we have

$(2a+4de^{-2t})+2(2at+b-2de^{-2t})+(at^2+bt+c+de^{-2t})=2t^2-t+3e^{-2t}$
$at^2+(4a+b)t+(2a+2b+c)+de^{-2t}=2t^2-t+3e^{-2t}$
$\implies\begin{cases}a=2\\4a+b=-1\\2a+2b+c=0\\d=3\end{cases}\implies a=2,b=-9,c=14,d=3$

So the general solution to the ODE is

$y=y_c+y_p$
$y=C_1e^{-t}+C_2te^{-t}+2t^2-9t+14+3e^{-2t}$

4 0

2 years ago

7^2 times (15 divided by 5)

butalik [34]

Answer:

147 because 7^2 is 49

15 divided by 5 is 3 so

49x3 is 147

8 0

2 years ago