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3 years ago

-0.5 • 4 to the 18 exponent

Mathematics

2 answers:

worty [1.4K]3 years ago

8 0

Answer:

262,144

Step-by-step explanation:

(-0.5 * 4)^18

(-2)^18

262,144

Answer: 262,144

Ivahew [28]3 years ago

8 0

Answer:

262144

Step-by-step explanation:

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If the supply of computer engineers increases at the same time that the demand for these workers decreases, what would be the MO

Katen [24]

The <em><u>correct answers</u></em> are:

C) Wages would decline as the competition for jobs increases; and C) a monopoly in the production of that medication.

Explanation:

If we have a surplus of employees for a decrease in the amount of jobs, it is likely that employers will lower wages, since there is no difficulty in finding employees.

If one company produces a new drug, and no other companies are producing it, there could potentially be a monopoly for that drug.

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4 years ago

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If PR = 4x – 2 and RS = 3x – 5, which expression represents PS?

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Answer:

The expression that represents the PS is 7x – 7 .

if I am right mark the answer as brainliest

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2 years ago

2. Solve x2 + 2x + 7 = 0 using the Quadratic Formula

pantera1 [17]

Answer:

x=7/4

Step-by-step explanation:

4x=-7=4x/-4=-7/-4= x=7/4

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3 years ago

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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

Why the order of the coordinates does not matter when calculating length

Rudiy27

The formula for calculating length is:
$\sqrt{ (x_b-x_a)^{2} + (y_b-y_a)^{2} }$

We can also write $x_a - x_b$ or $y_a - y_b$
Why it does not matter?
Let's assume we have 2 numbers, a and b.
When we perform a subtraction:
$a-b$ , we get another number $c$
When we perform another subtraction:
$b-a$ , we get a number $-c$

When we raise $c$ or $-c$ to the power of 2, the result is the same, $c^{2}$ .

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4 years ago