All categories

3 years ago

A-b when a=-4.5 b=0.5

Mathematics

2 answers:

Step2247 [10]3 years ago

5 0

Answer:

(-4.5) - (0.5) = -5

Step-by-step explanation:

When are finding things and they give it to you I prefer to use parenthesis sometimes it actually might mean to multiply but since they already give you the what property they want you to do (subtraction) insert the numbers into their appropriate spaces.

Answer by Mikaatkinsbootiful or Mimiwhatsup

Ipatiy [6.2K]3 years ago

4 0

Answer:

The answer is 4.

Step-by-step explanation:

A=4.5

B=0.5

4.5-0.5=4

You might be interested in

Are these fractions equivalent

kykrilka [37]

Answer:

Equivalent

Step-by-step explanation:

Look at 6/18. 18 divided by 6 is 3, which is the denominator. 6 divided by 6 is 1, so the numerator. So 1/3.

6 0

2 years ago

Read 2 more answers

Which of the following are exterior angles? Check all that apply.

Norma-Jean [14]

Answer:

f d a c

Step-by-step explanation:

i took this quix before i think

6 0

2 years ago

Which of the following prefixes means 1,000? Milli Centi Kilo Deca

sesenic [268]

Kilo is the prefix for 1,000, hope i was helpful.

4 0

4 years ago

Read 2 more answers

72 g equals how Many mg?

leva [86]

1 gram = 1000 milligrams

72 * 1000 = 72000

So 72 grams is equal to 72,000 milligrams.

6 0

3 years ago

Read 2 more answers

I really stuck on trying to prove this

Alisiya [41]

$\displaystyle\sum_{r=1}^nr(r+1)\cdots(r+p-1)$

When $n=1$ ,

$\displaystyle\sum_{r=1}^1r(r+1)\cdots(r+p-1)=1(1+1)(1+2)\cdots(1+p-2)(1+p-1)=p!$

Meanwhile, you have on the right

$\dfrac{(1)(1+1)(1+2)\cdots(1+p-2)(1+p-1)(1+p)}{p+1}=(1)(1+1)(1+2)\cdots(p-1)(p)=p!$

so the equality holds for $n=1$ .

Assume it holds for $n=k$ , i.e. that

$\displaystyle\sum_{r=1}^kr(r+1)\cdots(r+p-1)=\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}$

Now for $n=k+1$ , you have

$\displaystyle\sum_{r=1}^{k+1}r(r+1)\cdots(r+p-1)=\sum_{r=1}^kr(r+1)\cdots(r+p-1)+(k+1)(k+2)\cdots(k+1+p-1)$
$=\displaystyle\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}+(k+1)(k+2)\cdots(k+1+p-2)(k+1+p-1)$
$=\displaystyle\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}+(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\left(\dfrac k{p+1}+1\right)(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\dfrac{k+p+1}{p+1}(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\dfrac{(k+1)(k+2)\cdots(k+p-1)(k+p)(k+p+1)}{p+1}$

as required.

3 0

3 years ago