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2 years ago

Please help me. Multiply using the power of ten. Enter your answer using digits.

Mathematics

1 answer:

aksik [14]2 years ago

7 0

Answer:

14.5 × 10²(100) = 1,450

Move the decimal place r places to the right for every 0, in this case, 2 zeros, so move the decimal 2 places to the right.

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State if the given functions are inverses NO LINKS!!!! Part 1

Dafna11 [192]

Step-by-step explanation:

h(x)=(-3x-15)/5

let y=(-3x-15)/5

interchanging role of x &y

x=(-3y-15)/5

5x+15=-3y

y=-(5x+15)/3

h-1(x)=-(5x+15)/3

not

equal to f(x)=(-3x-6)/4

Given function are not function of each other .

g(x)=2/3x-2/3

let

y=2/3(x-1)

interchanging role of x &y

x=2/3(y-1)

3/2x+1=y

g-1(x)=3/2x+1

not equal to f(x)=½x+1

Given function are not function of each other .

3 0

3 years ago

Read 2 more answers

25m/s to miles/h Can anyone solve it? With explanation if you can

Dimas [21]

25 meters per second x 60 seconds = 1500 meters per minute

1500 meters per minute x 60 minutes per hour = 90,000 meters per hour

1 mile = 1609.34 meters

90000/1609.34 = 55.92 miles per hour

round answer to either 55.9 or 56 mph depending on what you need.

6 0

3 years ago

The power generated by an electrical circuit (in watts) as a function of its current c (in amperes) is modeled by: P(c)= -12c^2

sveta [45]

Answer:

5 amperes will produce the maximum power

Step-by-step explanation:

The power generated by an electrical circuit (in watts) as a function of its current c (in amperes) is modeled by: P(c)= -12c^2 + 120c.

To find out the maximum power we need to find the vertex

formula to find x -coordinate of vertex is $\frac{-b}{2a}$

P(c)= -12c^2 + 120c is given in the form of y=ax^2+bx+c

a= -12, b = 120

$c=\frac{-b}{2a}=\frac{-120}{2(-12)}=5$

c= 5 amperes will produce the maximum power

3 0

3 years ago

Read 2 more answers

Choose the equation that represents a line that passes through points (−1, 2) and (3, 1).

Sholpan [36]

B) X+4y=7

It works with both equations

7 0

3 years ago

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At an effective annual interest rate of i > 0, each of the following two sets of payments has present value K: (i) A payment

IgorLugansk [536]

Answer:

The present value of K is, $K=251.35$

Step-by-step explanation:

First of all, we need to construct an equation system, so

$(1)K=\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}$

$(2)K=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}$

Then we equalize both of them so we can find $i$

$(3)\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}$

To solve it we can multiply $(3)*(1+i)^{4}$ to obtain $(1+i)^{4}*(\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}})$ , then we have $225(1+i)^{2}+225=169(1+i)^{3}+169(1+i)^{2}$ .

This leads to a third-grade polynomial $169i^{3}+451i^{2}+395i-112=0$ , after computing this expression, we find only one real root $i=0.2224$ .

Finally, we replace it in (1) or (2), let's do it in (1) $K=\frac{169}{(1+0.2224)} +\frac{169}{(1+0.2224)^{2}}\\\\K=251.35$

8 0

3 years ago