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disa [49]
3 years ago
13

Can you help me with this problem, please

Mathematics
1 answer:
DanielleElmas [232]3 years ago
5 0

Answer:

  \dfrac{2x}{3y}\sqrt{3y}

Step-by-step explanation:

When we simplify a radical, we want to remove any squares from under the radical. Often, we also want to eliminate any fractions under the radical.

Here, that means we need to multiply by the denominator so as to make it a perfect square, then remove the square from under the radical.

  \sqrt{\dfrac{4x^2}{3y}}=\sqrt{\dfrac{(2x)^2(3y)}{(3y)(3y)}}=\sqrt{\dfrac{(2x)^2(3y)}{(3y)^2}}\\\\=\dfrac{2x}{3y}\sqrt{3y}

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Using the Taylor rule, if inflation is 1 percent, desired inflation is 2 percent, and output is 2 percentage points below potent
antoniya [11.8K]

Answer:

1.5%

Step-by-step explanation:

According to the Taylor rule, the federal funds rate targeted by the Fed is:

FFR = 2+AI+0.5*(AI-DI)+0.5PD

Where AI is the actual inflation (1%), DI is the desired inflation (2%) and PD is the deviation from potential (2%):

FFR = 2+1+0.5*(1-2)+0.5*(-2)\\FFR=1.5\%

The Fed should target a federal funds rate of 1.5%.

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3 years ago
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Answer:

Starting from the left.

4 × 3

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Step-by-step explanation:

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3 years ago
Find the value of x <br> 5x = 20
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Read 2 more answers
Calculate the sum of the multiples of 4 from 0 to 1000
allochka39001 [22]

Answer:

sum is 125,500

sum in summation notation is \sum\limits_{n=0}^n a+nd= (2a+(n-1)d)n/2

Step-by-step explanation:

This problem can be solved using concept of arithmetic progression.

The sum of n term terms in arithmetic progression is given by

sum = (2a+(n-1)d)n/2

where

a is the first term

d is the common difference of arithmetic progression

_____________________________________________________

in the problem

series is multiple of 4 starting from 4 ending at 1000

so series will look like

series: 0,4,8,12,16..................1000

a is first term so

here a is 0

lets find d the common difference

common difference is given by nth term - (n-1)th term

lets take nth term as 8

so (n-1)th term = 4

Thus,

d = 8-4 = 4

d  can also be seen 4 intuitively as series is multiple of four.

_____________________________________________

let calculate value of n

we have last term as 1000

Nth term can be described

Nth term = 0+(n-1)d

1000 =   (n-1)4

=> 1000 = 4n -4

=> 1000 + 4= 4n

=> n = 1004/4 = 251

_____________________________________

now we have

n = 1000

a = 0

d = 4

so we can calculate sum of the series by using formula given above

sum = (2a+(n-1)d)n/2

       = (2*0 + (251-1)4)251/2

       = (250*4)251/2

     = 1000*251/2 = 500*251 = 125,500

Thus, sum is 125,500

sum in summation notation is \sum\limits_{n=0}^n a+nd= (2a+(n-1)d)n/2

3 0
3 years ago
On a science exam, Jackson scored 85 out of 100 possible points. There were 20 questions on the exam. Each question was worth th
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Do 85÷20 and then add it to 100
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