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erastovalidia [21]
3 years ago
13

What temperature is 20 degrees lower than 6 degree celsius

Mathematics
1 answer:
Hatshy [7]3 years ago
3 0
The answer is -14 degrees centigrade
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Help me please thank you.
OLga [1]

<u>Top row - Left Row :</u>

Order : Left to Right

{11.7 - below(negative)} , {11.6 - below(negative)} , {12 - exactly filled} , {12.2 - above(positive)}

<u>Middle Row </u>:

Order : Left to Right

{11.1 - below(negative)} , {11.2 - below(negative)} , {11.9 - below(negative)} , {12.5 - above(positive)}

<u>Right Row </u>:

Order : Left to Right

{12 - exactly filled} , {11.4 - below(negative)} , {11.5 - below(negative)} , {10.8 - below(negative)}

8 0
3 years ago
Write m=-3 and (3,-1) in point slope form
Art [367]

Answer:

(y+1)= -3(x-3)

Step-by-step explanation:

Use Point-Slope Formula

(y-y_{1} )=m(x-x_{1} )

Substitute

(y+1) =-3(x-3)

7 0
3 years ago
Complete each proof. Remember to mark the diagram as you go. You may not need all the rows.
Illusion [34]

Answer:

did u ever get hte answers for this ?

6 0
3 years ago
find the component equation of the plane which is normal to the vector -2i+5j+k and which contains the point (-10;7;5).​
maksim [4K]

Given:

A plane is normal to the vector = -2i+5j+k

It contains the point (-10,7,5).​

To find:

The component equation of the plane.

Solution:

The equation of plane is

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

Where, (x_0,y_0,z_0) is the point on the plane and \left< a,b,c\right> is normal vector.

Normal vector is -2i+5j+k and plane passes through (-10,7,5). So, the equation of the plane is

-2(x-(-10))+5(y-7)+1(z-5)=0

-2(x+10)+5y-35+z-5=0

-2x-20+5y-35+z-5=0

-2x+5y+z-60=0

-2x+5y+z=60

Therefore, the equation of the plane is -2x+5y+z=60.

4 0
2 years ago
How do I do this.?.?.
viktelen [127]
108 = ( x)(48-x)
108 = 48x - x^2
48x -x^2 -108 = 0

I'm assuming you're doing solve the square right now, so you should be able to do the rest.
3 0
3 years ago
Read 2 more answers
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