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3 years ago

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Please answer. I think its B because negative × negative = positive. So -5 × -5 = 25. BUT...since the exponent is negative, mayb

e you divide something. I'm not sure.

1 answer:

yanalaym [24]3 years ago

7 0

Evaluate a^2 where a = -5:
a^2 = (-5)^2
Hint: | Evaluate (-5)^2.
(-5)^2 = 25:
Answer: 25

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5.1% < 0.51 please help been working on this all day I really need help

WITCHER [35]

You can write any decimal as a precent by moving the decimal 2 places to the right so
0.51 = 51%
5.1% < 51%
5.1% < 0.51

or, any precent can be written as a decimal by moving the decimal points two point to the left so
5.1% = 0.051
0.051 < 0.51
5.1% < 0.51

7 0

2 years ago

Vasco needs to know the area of the four-pointed star

Rama09 [41]

Answer:

Step-by-step explanation:

Find the area of the base

The base is a square.

The formula is Area = s^2

s = 1.5

Area = 1.5^2

Area = 2.25 sq cm.

Area of 1 triangle

h = 2.4

b = 1.5

Area = 1/2 * b * h

Area = 1/2 * 1.5 * 2.4

Area = 1.8 cm^2

Area of 4 triangles

Area = 4 * area of 1 triangle

Area = 4 * 1.8

Area = 7.2

Total Area = 2.25 + 7.2 = 9.45 cm^2

7 0

2 years ago

Points of discontinuity.<br> y = cot x, [0, 2 \[\pi \] ]

Triss [41]

<span>cotx = (cosx)/(sinx)
Where is this function not defined?
i.e. where is sin(x)=0?
Because at this point, cotx will be undefined because you cannot divide by zero. So find all points from 0 to 2 pi where sinx is 0.</span>

5 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%20%5Crm%20%20%5Clim_%7Bk%20%5Cto%20%5Cinfty%20%7D%20%5Csqrt%5B%20%20k%5D%7B%20%5CGamma%20%

Naddik [55]

We have

$\sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)\right)}k\right) \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right)\right)+\ln\left( \Gamma\left(\dfrac2k\right)\right)+ \cdots +\ln\left(\Gamma\left(\dfrac kk\right)\right)}k\right)$

and as k goes to ∞, the exponent converges to a definite integral. So the limit is

$\displaystyle \lim_{k\to\infty} \sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\lim_{k\to\infty} \frac1k \sum_{i=1}^k \ln\left(\Gamma\left(\frac ik\right)\right)\right) \\\\ = \exp\left(\int_0^1 \ln\left(\Gamma(x)\right)\, dx\right) \\\\ = \exp\left(\dfrac{\ln(2\pi)}2}\right) = \boxed{\sqrt{2\pi}}$

6 0

1 year ago

The graph shows the function f(x). which value is closest to the average rate of change from x=1 and x=4? a. -0.5 b. -1.5 c. -0.

Lemur [1.5K]

Answer:

c. -0.8

Step-by-step explanation:

From this graph, I estimated the values for the following ordered pairs (1, -2.5) and (4, -4.75)

We can set up the equation for the average rate of change like this

$\frac{-4.75-(-2.5)}{4-1}$

This simplifies to

$\frac{-2.25}{3}$

Which simplifies further to

$-0.75$

This is closes to c. -0.8

4 0

3 years ago