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pickupchik [31]

3 years ago

8

How are negative exponents and positive exponents related

1 answer:

Dennis_Churaev [7]3 years ago

7 0

They are both exponents

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Order 3, -4,7,-1 and -6 from least to greatest

katrin [286]

-6, -4, -1, 3, 7

Hope this helps!

5 0

3 years ago

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If 5 + 20 times 2 Superscript 2 minus 3 x Baseline = 10 times 2 Superscript negative 2 x Baseline + 5, what is the value of x? –

sp2606 [1]

Answer:

<h3>Option D) 3 is correct</h3><h3>Therefore the value of x is 3</h3>

Step-by-step explanation:

Given equation is $5+20\times (2)^{2-3x}=10\times (2)^{-2x}+5$

<h3>To find the value of x :</h3>

First solving the given equation we have,

$5+20\times (2)^{2-3x}=10\times (2)^{-2x}+5$

$5+20\times (2)^{2-3x}-5=10\times (2)^{-2x}+5-5$

$20\times (2)^{2-3x}=10\times (2)^{-2x}$

$20\times (2)^2.(2)^{-3x}=10\times (2)^{-2x}$

$20\times 4.(2)^{-3x}=10\times (2)^{-2x}$

$80(2)^{-3x}=10\times (2)^{-2x}$

$\frac{80}{10}(2)^{-3x}=\frac{10\times (2)^{-2x}}{10}$

$8(2)^{-3x}=(2)^{-2x}$

$\frac{(2)^{-3x}}{(2)^{-2x}}=\frac{1}{8}$

$(2)^{-3x}.(2)^{2x}=\frac{1}{2^3}$ ( by using the property $\frac{1}{a^{-m}}=a^m$ )

$2^{-3x+2x}=\frac{1}{2^3}$ ( by using the property $a^m.a^n=a^{m+n}$ )

$2^{-x}=\frac{1}{2^3}$

$\frac{1}{2^x}=\frac{1}{2^3}$ ( by using the property $a^{-m}=\frac{1}{a^m}$ )

Since bases are same so powers are same

Therefore we can equate the powers we get x=3

<h3>Therefore the value of x is 3</h3><h3>Option D) 3 is correct</h3>

5 0

3 years ago

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Cameron rent is 2,308 per month and wants to pay rent in $100 dollar bills and $1 dollar bills. How many $100 dollar bills will

Alinara [238K]

(2308*10)/100=230.3 so he would have to pay 230 hundos (and 30 singles)

6 0

3 years ago

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A recipe calls for 2 1/5 cups of chopped tomatoes and 3 2/5 cups of diced turnips how many more cups of turnips did the recipe c

Zarrin [17]

The answer would be 1 1/5 more turnips.

6 0

3 years ago

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A ball is thrown upward from the top of a building. The function below shows the height of the ball in relation to sea level, f(

Marina86 [1]

$f(t) = -16t^{2} +48t + 100$
Average rate of change from t = 3 seconds to t = 5 seconds = $\frac{f(5)-f(3)}{5-3} = \frac{(-16( 5^{2})+48(5)+100)-(-16( 3^{2})+48(3)+100 }{2}=$ $\frac{(-16(25)+240+100)-(-16(9)+144+100)}{2} = \frac{(-400+240+100)-(-144+144+100)}{2}=$ $\frac{-60-100}{2} = \frac{-160}{2} = -80$

5 0

3 years ago