3 years ago

Pls help i really need help asap

Mathematics

1 answer:

kompoz [17]3 years ago

7 0

Answer to 8:

February may April march

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Where would you insert parentheses to make the statement true?

9966 [12]

Just around the 15 and 3. So it would be

(15-3)x4+9=57

This is because (15-3)= 12
12x4=48
48+9= 57

8 0

3 years ago

Read 2 more answers

Help i made it worth a lot of points

raketka [301]

Answer:

36x^6/y^10

Step-by-step explanation:

First evaluate the parentheses, 6x^3 x 6x^3 = 36x^6

y^5 x y^5 = y^10 ; remember m^k x m^r = m^k+r

final answer should be 36x^6/y^10

hope this helps :')

4 0

2 years ago

Read 2 more answers

. 4.5w = 5.1W - 30 please help me with this question pls show the work

inn [45]

Answer: w = 50

Step-by-step explanation:

4.5w - 5.1w = -30

-0.6w = -30

0.6w = 30

W = 50

5 0

3 years ago

At what point do the lines y = 6x – 18 and y=8x – 32 intersect?

anzhelika [568]

Answer:

(7, 24)

Step-by-step explanation:

Solve the system:

6x - 18 = 8x - 32

2x = 14

x = 7

y = 6(7) - 18

y = 24

4 0

3 years ago

What is the solution of the equation over the complex numbers

denis-greek [22]

Answer:

$2i\sqrt{3}$

$-2i\sqrt{3}$

Step-by-step explanation:

$x^2 + 12$

First set the equation to 0

$x^2 + 12$

$x^2 + 12 = 0$

Second, get the 12 on the right side of the equal sign by adding a -12 to each side

$x^2 + 12 = 0$

$x^2 + 12 + (-12) = 0 -12$

$x^2 = -12$

Square root both sides of the equal sign.

$x^2 = -12$

$\sqrt{x^2} = \sqrt{-12}$

Take the square root on left sides of the equal sign.

$\sqrt{x^2} = \sqrt{-12}$

$x = \sqrt{-12}$

Take the square root on the right side of the equal sign. Remember $\sqrt{-1} = i$

$x = \sqrt{-12}$

$x = \sqrt{4}*\sqrt{-1}\sqrt{3}}$

$x = 2i\sqrt{3}$

AND

$x = -2i\sqrt{3}$

8 0

4 years ago