There is my question

Simplify the expression by combining like terms<br>-7k-3+8j+4j-3k+18

Nutka1998 [239]

Answer:

-10k+12j+15

Step-by-step explanation:

-7k-3+8j+4j-3k+18

-7k-3k+8j+4j-3+18

-10k+12j+15

If you're satisfied with the answer, then please mark me as Brainliest.

3 0

3 years ago

Is the fraction 1/3 equivalent to a terminating decimal or a decimal that does not terminate

Lapatulllka [165]

Answer:

does not terminate

Step-by-step explanation:

3 0

3 years ago

Write a quadratic function whose zeros are -3 and -4

andreev551 [17]

Answer:

f(x) = (x -(-3))(x -(-4))

Step-by-step explanation:

The function can be written as the product of binomial terms whose values are zero at the given zeros.

(x -(-3)) is one such term

(x -(-4)) is another such term

The product of these is the desired quadratic function. In the form easiest to write, it is ...

f(x) = (x -(-3))(x -(-4))

This can be "simplified" to ...

f(x) = (x +3)(x +4) . . . . simplifying the signs

f(x) = x^2 +7x +12 . . . . multiplying it out

6 0

4 years ago

Please help me thanks

KiRa [710]

Answer:

1,2,4,5,10

1,2,3,5,6

60

Step-by-step explanation:

7 0

3 years ago

If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(

timama [110]

Use the change-of-basis identity,

$\log_x(y) = \dfrac{\ln(y)}{\ln(x)}$

to write

$xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}$

Use the product-to-sum identity,

$\log_x(yz) = \log_x(y) + \log_x(z)$

to write

$xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}$

Redistribute the factors on the left side as

$xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}$

and simplify to

$xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)$

Now expand the right side:

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}$

Simplify and rewrite using the logarithm properties mentioned earlier.

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1$

$xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}$

$xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}$

$xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)$

$\implies \boxed{xyz = x + y + z + 2}$

(C)

6 0

2 years ago