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vfiekz [6]
3 years ago
10

mrs.fowler suggested contacting a travel agent to see whether they could get a lower price. the agent found a price that was 30%

lower than $2075. what was the price the travil agent found
Mathematics
2 answers:
CaHeK987 [17]3 years ago
8 0

Answer:

New price found by travel agent was $1452.5

Step-by-step explanation:

We had earlier price as $2075

In search of lower price deals, agent found a price that was 30% lower than $2075.

First we find 30% of $2075.

i.e (30 /100)  * 2075

or  (3/10) * 2075

or $622.5

So the new price was $622.5 less than the earlier price.

Hence , The new price travel agent found was= $2075 - $622.5 = $1452.5

Therefore new price was $ 1452.5


Luden [163]3 years ago
5 0
0.3 x 2075. Use a calculator.
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9

Step-by-step explanation:

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kvv77 [185]
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3 years ago
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Any one Know the Answer To This Algebra 1
alexdok [17]

Answer:

The correct answer is B. -3.

Step-by-step explanation:

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f(-3) = 2(-3) + 3

Now, we can just simplify the expression found on the right side of the equation using PEMDAS.  First, we should perform the multiplication.

f(-3) = -6 + 3

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Hope this helps!

3 0
4 years ago
A cube is built with inside dimensions of 10 inches. The material is 0.2 inches thick. Use a Taylor series approximation to find
larisa86 [58]

Answer:

V(x,y,z) ≈ 61.2 in

Step-by-step explanation:

for the function f

f(X)=x³

then the volume will be

V(x,y,z)= f(X+h) -  f(X) , where h= 0.2 (thickness)

doing a Taylor series approximation to f(x+h) from f(x)

f(X+h) - f(X) = ∑fⁿ(X)*(X-h)ⁿ/n!

that can be approximated through the first term and second

f(X+h) - f(X) ≈ f'(x)*(-h)+f''(x)*(-h)²/2 = 3*x²*(-h)+6*x*(-h)²/2

since x=L=10 in (cube)

f(X+h) - f(X) ≈ 3*x²*(-h)+6*x*(-h)²/2 = 3*L²*h+6*L*h²/2 = 3*L*h*(h+L)

then

f(X+h) - f(X) ≈  3*L*h*(h+L) = 3* 10 in * 0.2 in * ( 0.2 in + 10 in ) = 61.2 in

then

V(x,y,z) ≈   61.2 in

V real = (10.2 in)³-(10 in)³ = 61 in

8 0
4 years ago
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grigory [225]

Answer:

Step-by-step explanation:

Given 3x-4y=65

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So value of X is 27

7 0
3 years ago
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