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klemol [59]
3 years ago
5

7x^2-2x+1-(9x^2+2x-1)

Mathematics
2 answers:
likoan [24]3 years ago
6 0

Answer:

-2x^2-4x+2

Step-by-step explanation:

(7x^2-2x+1)-(9x^2+2x-1)=\\(7x^2-9x^2)+(-2x-2x)+(1+1)=\\-2x^2-4x+2

Hope this helps!

zalisa [80]3 years ago
6 0
(7x2-2x-1)+(9x2-4x+5)-(4x2+6x-7)

Final result :

12x2 - 12x + 11
Step by step solution :

Step 1 :

Equation at the end of step 1 :

((((7•(x2))-2x)-1)+(((9•(x2))-4x)+5))-((22x2+6x)-7)
Step 2 :

Equation at the end of step 2 :

((((7•(x2))-2x)-1)+((32x2-4x)+5))-(4x2+6x-7)
Step 3 :

Equation at the end of step 3 :

(((7x2-2x)-1)+(9x2-4x+5))-(4x2+6x-7)
Step 4 :

Trying to factor by splitting the middle term

4.1 Factoring 12x2-12x+11
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The distance from Rodney's house to the park is 100 meters. If he walks to the park and back to his house, how far will he have
konstantin123 [22]

Answer:

c. 20,000cm

Step-by-step explanation:

from is house to the park is 100m

from the park back to his house is another 100m

100m+100m=200m

1m = 100cm

200m = xcm

cross multiply

xcm = 200×100

xcm= 20,000cm

3 0
3 years ago
Write an equation for the line that passes through the given point and is perpendicular to the graph of the given equation.
Nezavi [6.7K]

Answer:

  • y + 2 = - 3/2(x - 4)

Step-by-step explanation:

<u>Given line:</u>

  • y + 4 = 2/3(x - 2)

It has a slope of 2/3.

<u>Perpendicular lines have negative-reciprocal slopes, so the perpendicular slope is:</u>

  • m = - 3/2

<u>Find its equation provided it passes through the point (4, - 2):</u>

  • y - (-2) = - 3/2(x - 4)
  • y + 2 = - 3/2(x - 4)
3 0
2 years ago
If 4 (x-2/3) = -18, what is the value of 2x
BaLLatris [955]

Answer:

-11.5

Step-by-step explanation:

4(x-2/3)=-18

4x-8/3=-18

4x-8=-18×3

4x-8=-54

4x=-54+8

4x=-46

x=-46/4

x=-11.5

4 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
Plsss help i’ll give brainliest if you give a correct answer
Ugo [173]

Answer:

23

Step-by-step explanation:

8 0
2 years ago
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