expand the following 4 (x - 1)

Answer:

x^2 - 2x

Step-by-step explanation:

Distribute the x to every term in the parenthesis.

Answer:

x^2 -2x

Step-by-step explanation:

x(x - 2)

Distribute

x*x - 2*x

x^2 -2x

Answer:

Step-by-step explanation:

8x² + 7x - 1 = 0

Rewrite 7x as a difference

That's

8x² + 8x - x - 1 = 0

Factorize

8x( x + 1) - ( x + 1) = 0

(8x - 1)( x + 1) = 0

8x - 1 = 0 x + 1 =0

8x = 1 x = - 1

x = 1/8

The solutions are

Hope this helps you

Answer:

12.56 [tex]cm^2[/tex] is the error in area of circle.

Step-by-step explanation:

Given that:

Radius of the circle, r = 10 cm

Error in measurement of radius, [tex]\triangle r[/tex] = 0.2 cm

To find:

The error in area of circle = ?

Solution:

First of all, let us have a look at the percentage error in measurement of radius:

[tex]\dfrac{\triangle r}{r}\times 100 = \dfrac{0.2}{10}\times 100 = 2\%[/tex]

Now, we know that Area of a circle is given as:

[tex]A = \pi r^2[/tex]

[tex]\Rightarrow \dfrac{\triangle A}{A} \times 100 = 2 \times \dfrac{\triangle r}{r} \times 100\\\Rightarrow \dfrac{\triangle A}{A} = 4\%[/tex]

Area according to r = 10

[tex]A = 3.14\times 10^2 = 314 cm^2[/tex]

Now, error in area = 4% of 314 [tex]cm^2[/tex]

[tex]\Rightarrow \dfrac{4}{100} \times 314 = 12.56 cm^2[/tex]

So, the answer is:

12.56 [tex]cm^2[/tex] is the error in area of circle.

Answer:

The correct answer is:

[tex]+6x^{2}\\-9y^2[/tex]

Step-by-step explanation:

We are given the term:

[tex]3x^{2} +5y^{2} [\text{ \ }] +3 [\text{ \ }] +4y^{2} +6 = 9x^{2} -y^{2} +9[/tex]

We have to fill in to the empty spaces such that the above equation gets satisfied.

First of all, let us simplify the LHS (Left Hand Side):

[tex]3x^{2} +5y^{2} [\text{ \ }] +3 [\text{ \ }] +4y^{2} +6\\\Rightarrow 3x^{2} +5y^{2} +4y^{2} [\text{ \ }] [\text{ \ }] +6 +3\\\Rightarrow 3x^{2} +9y^{2} [\text{ \ }] [\text{ \ }] +9[/tex]

Now, let us equate the LHS and  RHS (Right Hand Side):

[tex]\Rightarrow 3x^{2} +9y^{2} [\text{ \ }] [\text{ \ }] +9 = 9x^{2} -y^{2} +9[/tex]

Equating the coefficients of [tex]x^{2}\ and\ y^{2}[/tex] in LHS and RHS:

One box will have value = [tex]9x^{2} -3x^{2} =+6x^{2}[/tex]

Other box will have value = [tex]-y^{2} -9y^{2} =-10y^{2}[/tex]

The correct answer is:

[tex]+6x^{2}\\-9y^2[/tex]

So, if we fill the boxes with above values, the expression will be simplified as given.

Answer:

The correct answer is B. Positive 6 x squared and Negative 10 y squared

Step-by-step explanation: