All categories

3 years ago

How many grams are there in 4.5 kilograms

Mathematics

2 answers:

zavuch27 [327]3 years ago

3 0

Answer: 4500 grams

Step-by-step explanation: 1 kilogram is 1,000 grams, so 4 kilograms is 4,000 grams, and half a kilogram is 500 grams. So, 4000 + 500 = 4,500

Harlamova29_29 [7]3 years ago

3 0

Answer: 4,500 Grams

Step-by-step explanation:

1 Kilogram = 1,000 grams

1,000 x 4.5 = 4,500

- R3KTFORGOOD ☕

You might be interested in

Multiply by hundreds: 545 times 200

fiasKO [112]

Answer:

109000

Step-by-step explanation:

5 0

3 years ago

Evaluate: . C=4x? rind the value of c- given thatx = 16 pour

Butoxors [25]

Answer:

c=4x

c-4(16)

c=64

Step-by-step explanation:

6 0

3 years ago

Can I have somebody answer a few more of the questions that I need please and this one too?

tankabanditka [31]

Answer:

x > 22

Step-by-step explanation:

Hey there!

Well to solve,

52 - 3x < -14

we need to single out x

52 - 3x < -14

-52 to both sides

-3x < -66

Divide both sides by -3

x > 22

The < changes to > because the variable number is a - being divided.

Hope this helps :)

7 0

2 years ago

Read 2 more answers

2xy-(x-(+3) when x =-3 and y=1

RSB [31]

0 should be the answer

6 0

2 years ago

The polynomial of degree 5, P ( x ) has leading coefficient a=1, has roots of multiplicity 2 at x = 3 and x = 0 , and a root of

ser-zykov [4K]

Answer:

$p_{5} (t) = x^{5} - 5\cdot x^{4} + 3\cdot x^{3} +9\cdot x^{2}$ , for $r_{1} = 0$

Step-by-step explanation:

The general form of quintic-order polynomial is:

$p_{5}(t) = a\cdot x^{5} + b\cdot x^{4} + c\cdot x^{3} + d\cdot x^{2} + e \cdot x + f$

According to the statement of the problem, the polynomial has the following roots:

$p_{5} (t) = (x - r_{1})\cdot (x-3)^{2}\cdot x^{2} \cdot (x+1)$

Then, some algebraic handling is done to expand the polynomial:

$p_{5} (t) = (x - r_{1}) \cdot (x^{3}-6\cdot x^{2}+9\cdot x) \cdot (x+1)\\p_{5} (t) = (x - r_{1}) \cdot (x^{4}-5\cdot x^{3} + 3 \cdot x^{2} + 9 \cdot x)$

$p_{5} (t) = x^{5} - (5+r_{1})\cdot x^{4} + (3 + 5\cdot r_{1})\cdot x^{3} +(9-3\cdot r_{1})\cdot x^{2} - 9 \cdot r_{1}\cdot x$

If $r_{1} = 0$ , then:

$p_{5} (t) = x^{5} - 5\cdot x^{4} + 3\cdot x^{3} +9\cdot x^{2}$

5 0

2 years ago