1:0,1+2:0,2+3:0,3...+9:0

Question below. Please answer it, its math.

AleksAgata [21]

Answer:

- 0.8

Step-by-step explanation:

The first thing we want to do here is simplify the expression -

$\frac{3}{5}$ ( 2x + 5 ) - 2x, Distribute the " $\frac{3}{5}$ " to elements within the parenthesis

= $\frac{3}{5}$ $*$ 2x + $\frac{3}{5}$ $*$ 5 - 2x, Focus on simplifying the expression " $\frac{3}{5}$ $*$ 2x + $\frac{3}{5}$ $*$ 5 "

= $2\cdot \frac{3}{5}x+5\cdot \frac{3}{5}$ - 2x

= $\frac{6x}{5}+3$ - 2x, Combine fractions

= $-\frac{4x}{5}$ + 3

= $-\frac{4}{5}$ x + 3

So we have our simplified expression " $-\frac{4}{5}$ x + 3, " with $-\frac{4}{5}$ being the coefficient of x. Our requirements are that this fraction should be expressed as a decimal, so we can simply divide the numerator by the denominator to figure that out,

- 4 / 5 = - 0.8,

Solution = - 0.8

4 0

3 years ago

What is another way to write 42 + 63

satela [25.4K]

You could write this by going 63+42 or add 63 and 42
hope that's what you are looking for ;-)<span />

7 0

3 years ago

Read 2 more answers

Micheal was asked to give examples of the identity property of addition and identity multiplication. Below are his answers.

xxMikexx [17]

Answer:

identity property of addition-

a+0=a

identity property of multiplication-

a*1=a

Step-by-step explanation:

i cant give u an exact answer as u didnt give Micheals answers so i just gave some examples about what addition and multiplication identity property should look like. Identity property's concept is to keep the same identity. Basically, "a" shouldnt change. In addition, to keep a the same all u hv to do is add 0 as anything plus 0 is the same. for multiplication, just multiply by 1. Hope this helps!!

5 0

3 years ago

solve the equations below. make sure you say how many solutions each equation has(MCC.8.EE.7a) A. 5p-9=7p-19 b. 5p-9= 5p-19 C. 5

Sedaia [141]

<h3>Answer:</h3>

A) p = 5, one solution
B) no solutions
C) infinite solutions

<h3>Step-by-step explanation:</h3>

A) Add 19-5p to each side of the equation:

... 10 = 2p

... 5 = p . . . . . divide by the coefficient of p

B) Subtract 5p from both sides of the equation:

... -9 = -19 . . . . . there is <em>no value of p</em> that will make this true. (No solution.)

C) Subtract 5p from both sides of the equation:

... -9 = -9 . . . . . this is true for <em>every value of p</em>. (Infinite solutions.)

6 0

3 years ago

5x3 + 14x2 + 9x <br> help

Cerrena [4.2K]

<h3>Answer: x(x+1)(5x+9) </h3>

===================================================

Work Shown:

5x^3 + 14x^2 + 9x

x( 5x^2 + 14x + 9 )

To factor 5x^2 + 14x + 9, we could use the AC method and guess and check our way to getting the correct result.

A better way in my opinion is to solve 5x^2 + 14x + 9 = 0 through the quadratic formula

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(14)\pm\sqrt{(14)^2-4(5)(9)}}{2(5)}\\\\x = \frac{-14\pm\sqrt{16}}{10}\\\\x = \frac{-14\pm4}{10}\\\\x = \frac{-14+4}{10} \ \text{ or } \ x = \frac{-14-4}{10}\\\\x = \frac{-10}{10} \ \text{ or } \ x = \frac{-18}{10}\\\\x = -1 \ \text{ or } \ x = \frac{-9}{5}\\\\$

Then use those two solutions to find the factorization

x = -1 or x = -9/5

x+1 = 0 or 5x = -9

x+1 = 0 or 5x+9 = 0

(x+1)(5x+9) = 0

So we have shown that 5x^2 + 14x + 9 factors to (x+1)(5x+9)

-----------

Overall,

5x^3 + 14x^2 + 9x

factors to

x(x+1)(5x+9)

6 0

3 years ago

1:0,1+2:0,2+3:0,3...+9:0​

1:0,1+2:0,2+3:0,3...+9:0