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2 years ago

Factor c^2 −64. Is this a special product? PLS HELP

1 answer:

8 0

C^2-64 is a square roots factoring one. you take the square root of c^2 which is c and then the square root of 64 which is 8 or -8 so the answer is (c-8)(c+8). and it is a special product

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Mr. randall bought 4 shirts, which were on sale. the shirts were originally priced $20. the salres price of the shirts was $5 do

USPshnik [31]

Answer:

The price Mr. Randall paid = 4 * (20-5)

Step-by-step explanation:

8 0

3 years ago

Solve the system: 5x-7y=-41 -3x-5y=-3

-Dominant- [34]

Answer:

x = -4 , y = 3

Step-by-step explanation:

5x - 7y = -41 ... (i)

-3x - 5y = -3 ... (ii)

Multiplying (i) by -3 and (ii) by 5 ;

-15x + 21y = 123 ... (i)

-15x - 25y = -15 ... (ii)

Subtracting (i) by (ii) ;

0 + 46y = 138

46y = 138

y = 138 ÷ 46 = 3

Returning to equation (ii) ;

-3x - 5(3) = -3

-3x = -3 + 15

-3x = 12

x = -4

8 0

3 years ago

Read 2 more answers

Mr Taylor brought back 180 Euros to the UK. The exchange rate was €1.2 to £1.

solong [7]

The answer is £150.

This could be calculated using proportion. If £1 is €1.2, how many £ is €180:
£1 : €1.2 = x : €180
x = £1 ÷ €1.2 × €180
x = £150

6 0

3 years ago

What are the solutions to the equation x^2-8x=24

Oduvanchick [21]

$x^{2} - 8x = 24$

Transposing the equation to give zero on the RHS:
$x^{2} - 8x - 24 = 0$

Since we can't factorise it directly, we must complete the square to yield any solutions:
$x^{2} - 8x + ( \frac{8}{2})^2 - (\frac{8}{2})^2 - 24 = 0$
$(x^{2} - 8x + 16) - 16 - 24 = 0$
$(x-4)^2 - 40 = 0$
$(x-4)^2 = 40$
$x-4 = +/- \sqrt{40}$ ––> because square rooting yields a positive and negative solution.
x = 4 +/- $\sqrt{40}$
x = 4 +/- $2 \sqrt{10}$

Correct me if I'm wrong anyone.

4 0

3 years ago

N=4; 2i and 3i are zeros f(-1)=50

Nonamiya [84]

Solution:- $\text{Let f(x) be any nth degree polynomial with n=4}\\$

$\text{Given that 2i and 3i are the zeroes of f(x)}$

$\text{so (x-2i) and (x-3i) are factors of f(x)}$

$\text{Since 2i is a zero of f(x) then its conjugate -2i is also a zero of f(x)}$

$\Rightarrow(x+2i) \text{is a factor}\\\text{Similarly, conjugate of 3i is -3i is also a zero of f(x) }\\\Rightarrow(x+3i)\text{is a factor}\\\text{So , }\\f(x)=k(x-2i)(x+2i)(x-3i)(x+3i)\\=k(x^2-(2i)^2)(x^2-(3i)^2)\\=k(x^2-4i^2)(x^2-9i^2)\\=k(x^2+4)(x^2+9)\\=k(x^4+13x^2+36)\\\text{As given}\\f(-1)=50\\\Rightarrow k((-1)^4+13(-1)^2+36)=50\\\Rightarrow k(1+13+36)=50\\\Rightarrow k(50)=50\\\Rightarrow k=1\\\text{So by substituting k=1 in f(x) we get ,}\\f(x)=(x^4+13x^2+36)$

6 0

3 years ago