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write the product of the expression 5^4 * 5^-7 using a positive exponent, then write the product using a negative exponent

pav-90 [236]

Answer with positive exponent = $\left(\frac{1}{5}\right)^3$ or $\frac{1}{5^3}$

Answer with negative exponent = $5^{-3}$

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Explanation:

The rule we use is

$a^b*a^c = a^{b+c}$

If we multiply two exponential expressions with the same base, then we add the exponents.

The base for each is 5. The exponents 4 and -7 add to -3.

This means

$5^4*5^{-7} = 5^{4+(-7)} = 5^{-3}$

To convert to a positive exponent, we apply the reciprocal to the base. We go from 5, aka 5/1, to 1/5.

So, $5^{-3} = \left(\frac{1}{5}\right)^3 = \frac{1^3}{5^3} = \frac{1}{5^3}$

3 0

3 years ago

Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)

masha68 [24]

Given:

The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

To find:

The equation of the quadratic function.

Solution:

A quadratic function is defined as:

$y=ax^2+bx+c$ ...(i)

It is passes through the point (0,11). So, substitute $x=0,y=11$ in (i).

$11=a(0)^2+b(0)+c$

$11=c$

Putting $c=11$ in (i), we get

$y=ax^2+bx+11$ ...(ii)

The quadratic function passes through the point (5,31). So, substitute $x=5,y=31$ in (ii).

$31=a(5)^2+b(5)+11$

$31-11=a(25)+5b$

$20=25a+5b$

Divide both sides by 5.

$4=5a+b$ ...(iii)

The quadratic function passes through the point (3,11). So, substitute $x=3,y=11$ in (ii).

$11=a(3)^2+b(3)+11$

$11-11=a(9)+3b$

$0=9a+3b$

Divide both sides by 3.

$0=3a+b$ ...(iv)

Subtracting (iv) from (iii), we get

$4-0=5a+b-3a-b$

$4=2a$

$\dfrac{4}{2}=a$

$2=a$

Putting $a=2$ in (iv), we get

$0=3(2)+b$

$0=6+b$

$-6=b$

Putting $a=2,b=-6$ in (ii), we get

$y=(2)x^2+(-6)x+11$

$y=2x^2-6x+11$

Therefore, the required quadratic equation is $y=2x^2-6x+11$ .

7 0

3 years ago

Find the value of x. If necessary, round your answer to the nearest tenth. The figure is not drawn to scale. (1 point) 9.7 9.2 9

Vilka [71]

X= 9.2

From the figure, the chord (12) is drawn with a line (from the center of the circle) perpendicular to it. It can be said that the chord 12 is bisected by the 7 long line from the center which gives us with 6 each of the two parts of the chord. Since they are forming (the 6 and 7) a right angled triangle, we can then use Pythagorean theorem to find X, which can be seen as the radius at the same time hypotenuse of the right angled triangle.

4 0

3 years ago

I need help with 20.

irinina [24]

Step-by-step explanation:

why ? you did it right there already.

the numerator is the top part of the fraction, the "counting" part (how many units of a certain type or size are there).

the denominator (the bottom part) is defining the type or size of units described by the fraction.

so, we are adding 4 to the numerator of 2/5, making it (2+4)/5 = 6/5

what do we need to add to the denominator (the bottom part) to make the fraction have the value of 1/2 ?

so, what x would make 6/x = 1/2 ?

6 = x/2

x = 6×2 = 12

so, we need to get 12 in the denominator, where we have currently 5.

therefore, we need to add 12-5 = 7 to it.

so that

6/(5+7) = 6/12 = 1/2

3 0

2 years ago

What is x= 60 × 15 hdvshdvsj

castortr0y [4]

Answer:

X= 9

Step-by-step explanation:

Look up answer on google

8 0

3 years ago