What is the midpoint segment with endpoints j(-3,-2),K(4,-6)?

Help plz,BIG POINTS,also plz show me the steps as best as possible

Sidana [21]

Answer:

The correct answer is option 3.

Step-by-step explanation:

Given : ΔPQR, QM is altitude of the triangle

PM = 8

MR = 18

To find = QM

Solution :

PR = 8 + 18 = 26

Let, PQ = x , QR = y, QM = z

Applying Pythagoras Theorem in ΔPQR

$PQ^2+QR^2=PR^2$

$x^2+y^2=(26)^2$ ..[1]

Applying Pythagoras Theorem in ΔPQM

$PM^2+QM^2=PQ^2$

$8^2+z^2=(x)^2$ ..[2]

Applying Pythagoras Theorem in ΔQMR

$MR^2+QM^2=QR^2$

$18^2+z^2=(y)^2$ ..[3]

Putting values of $x^2$ and $y^2$ from [2] and [3 in [1].

$8^2+z^2+18^2+z^2=(26)^2$

$2z^2=(26)^2-(8)^2-(18)^2$

$2z^2=288$

$z^2=144$

z = ±12

z = 12 = QM ( ignoring negative value)

The length of QM is 12.

6 0

2 years ago

Help asap , apx algebra 1

Troyanec [42]

D. 3
F(x) is referring to y. The question is asking what x is when y is 4.

5 0

3 years ago

Help me please ;-;!!!!! Use a coordinate grid to create a map of a town with at least five different locations, such as a house,

Sholpan [36]

A. Our diagram have six location: my house (2,1), a movie theater (5,3), the school (6,6), the police station (3,-2), the airport (-3,3), and the mall (-2,-3)

B. Distance between my house and the movie theater:
To calculate the distance we are going to use the points (2,1) and (5,3) to create a right triangle with tow legs of measures 3 and 2; the hypotenuse of the right triangle will be the distance between my house and the movie theater.
Using the Pythagorean theorem:
$d^2=3^2+2^2$
$d^2=9+4$
$d^2=13$
$d= \sqrt{13}$
We can conclude that the distance between my house and the movie theater is $\sqrt{13}$

Our second distance is the distance between the police station and the airport:
This time we are using the points (3,-2) and (-3,3) to create a right triangle with legs of measures 6 and 5; the hypotenuse of our triangle will be the the distance between our points:
$d^2=6^2+5^2$
$d^2=36+25$
$d^2=61$
$d= \sqrt{61}$
We can conclude that the distance between the police station and the airport is $\sqrt{61}$

C. Distance form the park to the Fire Station:
$d^2=3^2+1^2$
$d^2=9+1$
$d^2=10$
$d= \sqrt{10}$
We can conclude that the distance from the park to the Fire Station is $\sqrt{10}$

Distance from the stadium to the mall:
$d^2=2^2+1^2$
$d^2=4+1$
$d^2=5$
$d= \sqrt{5}$
We can conclude that the distance from the stadium to the mall is $\sqrt{5}$

D. What is the distance between your house and the mall?
Answer:
$d^2=4^2+4^2$
$d^2=16+16$
$d^2=32$
$d= \sqrt{32}$
$d=4 \sqrt{2}$
We can conclude that the distance between my house and the mall is $4 \sqrt{2}$

What is the distance between the movie theater and the school?
answer:
$d^2=3^2+1^2$
$d^2=9+1$
$d^2=10$
$d= \sqrt{10}$
We can conclude that the distance between the movie theater and the school is $\sqrt{10}$

7 0

2 years ago

The midpoint between 94.68 and 49.67

Brilliant_brown [7]

Answer:

72.175

Step-by-step explanation:

The first step is subtract 94.68 by 49.67 which would be 45.01

The next step would be to use 45.01 and divide by two

The last step would be subtract 94.68 by 22.505 and you would get 72.175

Therefore your answer would be 72.175

8 0

3 years ago

Help FAST PLZ!!! marking brainliest!

ss7ja [257]

Answer:

After finding the prime factorization of $2010=2\cdot3\cdot5\cdot67$, divide $5300$ by $67$ and add $5300$ divided by $67^2$ in order to find the total number of multiples of $67$ between $2$ and $5300$. $\lfloor\frac{5300}{67}\rfloor+\lfloor\frac{5300}{67^2}\rfloor=80$ Since $71$,$73$, and $79$ are prime numbers greater than $67$ and less than or equal to $80$, subtract $3$ from $80$ to get the answer $80-3=\boxed{77}\Rightarrow\boxed{D}$.

Step-by-step explanation:

hope this helps

5 0

2 years ago