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Basile [38]
3 years ago
15

Geometry help please

Mathematics
1 answer:
kolezko [41]3 years ago
5 0
The side that corresponds to QR is FD or DF
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This k12 question i need help
Yuri [45]

Answer: It's 72.

Step-by-step explanation: They deleted my answer <:(

4 0
3 years ago
Read 2 more answers
Find the coordinate that divides the directed line segment from A(-2,-4) to B(8,1) in the ratio of 2 to 3
zalisa [80]
We have that
A(-2,-4)  B(8,1) <span>

let
M-------> </span><span>the coordinate that divides the directed line segment from A to B in the ratio of 2 to 3

we know that

A--------------M----------------------B
        2                     3
distance AM is equal to (2/5) AB
</span>distance MB is equal to (3/5) AB
<span>so

step 1
find the x coordinate of point M
Mx=Ax+(2/5)*dABx
where
Mx is the x coordinate of point M
Ax is the x coordinate of point A
dABx is the distance AB in the x coordinate
Ax=-2
dABx=(8+2)=10
</span>Mx=-2+(2/5)*10-----> Mx=2

step 2
find the y coordinate of point M
My=Ay+(2/5)*dABy
where
My is the y coordinate of point M
Ay is the y coordinate of point A
dABy is the distance AB in the y coordinate
Ay=-4
dABy=(1+4)=5
Mx=-4+(2/5)*5-----> My=-2

the coordinates of point M is (2,-2)

see the attached figure

7 0
3 years ago
Assume the readings on thermometers are normally distributed with a mean of 0 degrees C and a standard deviation of 1.00 degrees
lozanna [386]

Answer: 0.0035

Step-by-step explanation:

Given : The readings on thermometers are normally distributed with a mean of 0 degrees C and a standard deviation of 1.00 degrees C.

i.e.  \mu=0 and \sigma= 1  

Let x denotes the readings on thermometers.

Then, the probability that a randomly selected thermometer reads greater than 2.17 will be :_

P(X>2.7)=1-P(\xleq2.7)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{2.7-0}{1})\\\\=1-P(z\leq2.7)\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.9965\ \ [\text{By z-table}]\ \\\\=0.0035

Hence, the probability that a randomly selected thermometer reads greater than 2.17 = 0.0035

The required region is attached below .

8 0
3 years ago
This was given to me during a summative test and the teacher didn't bother giving me the correction. I just cannot figure it out
Dmitry [639]
  • Base be y

ATQ

\\ \sf\longmapsto xy=6050\dots 1

\\ \sf\longmapsto 2(x+y)=220\implies x+y=110\dots 2

Now

\\ \sf\longmapsto (x+y)^2=x^2+y^2+2xy

\\ \sf\longmapsto 110^2-2(6050)=x^2+y^2

\\ \sf\longmapsto 12100-12100=x^2+y^2

\\ \sf\longmapsto x^2+y^2=0\dots(3)

From all equations

\\ \sf\longmapsto (x-y)^2=x^2+y^2-2xy

\\ \sf\longmapsto (x-y)^2=0-2(6050)

\\ \sf\longmapsto (x-y)^2=-12100

\\ \sf\longmapsto (x-y)=110\dots(4)

Now

Adding 3 and 4

\\ \sf\longmapsto 2x=220

\\ \sf\longmapsto x=110m

One side won't be covered hence

  • y=2(110)=220m
7 0
3 years ago
Read 2 more answers
The average distance between the Earth and the Moon is 384 400 km.<br> Express it in standard form.
ycow [4]

Answer:

3.84\times10^8\ m

Step-by-step explanation:

It is given that,

The average distance between the Earth and the Moon is 384 400 km.

We need to express in in standard form.

1 km = 1000 m

It means,

384400 km = 384400000 km

or

= 3.84\times10^8\ m

Hence, the average distance between the Earth and the Moon is 3.84\times10^8\ m.

3 0
3 years ago
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