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Pavel [41]
3 years ago
11

Simplify the expression (5^-2 * 4^-4)^-2 so there is only one positive power for each base

Mathematics
2 answers:
Nonamiya [84]3 years ago
7 0
I got 5^4 x 4^8....................
lapo4ka [179]3 years ago
6 0
(5^4*4^8) that's is what I got
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The temperature fell from 0 Degrees Fahrenheit to 15 and one-half Degrees Fahrenheit below 0 in 5 and three-fourths hours. Wen t
Kryger [21]

Answer:

The correct answer will be:

-\dfrac{62}{23}

Step-by-step explanation:

It is given that :

Initial temperature, T_1 = 0^\circ F

Final temperature,

T_2 = -15\dfrac{1}{2}^\circ F\\\Rightarrow T_2 = -\dfrac{15\times 2+1}{2} ^\circ F\\\Rightarrow T_2 = -\dfrac{31}{2} ^\circ F

Time taken :

5\dfrac{3}{4}\ hrs = \dfrac{5 \times 4+3}{4}\ hrs = \dfrac{23}{4}\ hrs

Change in temperature per hour:

\dfrac{\text{Difference of temperature}}{\text{Total Time Taken}}\\\Rightarrow \dfrac{T_2-T_1}{\text{Total Time Taken}}

Putting the values of temperatures and time:

\dfrac{\dfrac{-31}{2}-0}{\dfrac{23}{4}}\\\Rightarrow \dfrac{\dfrac{-31}{2}}{\dfrac{23}{4}}\\\Rightarrow \dfrac{-31 \times 4}{2 \times 23}} \text{---- Error done by Wen at this step}\\\Rightarrow \dfrac{-31 \times 2}{23}}\\\Rightarrow \dfrac{-62}{23}}

The error done by Wen was during calculating the values of fraction.

So, the correct answer is :\frac{-62}{23}} instead of \frac{-713}{8}

5 0
4 years ago
Read 2 more answers
Can you all please help me with problems 6 through 10? Please don’t use for points I’m extremely stressed and need help
patriot [66]

Answer:

See below.

Step-by-step explanation:

6.) (5)/6 ≤ 1 (Yes)

7.) 1.4(11) > 16

    15.4 > 16 (No)

8.) 11.1 + 9.8 ≥ 21.01

    20.9 ≥ 21.01 (No)

9.) 2.5 < (90)/30

    2.5 < 3 (Yes)

10.) 1/2 > 3(1/6)

      1/2 > 1/2 (No)

11.) 2.16 ≥ 3(0.6) - 0.5

     2.16 ≥ 1.8 - 0.5

     2.16 ≥ 1.3 (Yes)

12.) x < 2 (x is less than 2.)

13.) x ≥ -1 (x is greater than or equal to -1.)

5 0
3 years ago
Read 2 more answers
Add 77 -10 with its additive inverse.​
Ganezh [65]

Answer: 77 - 10 = 67, and the additive inverse of 77 - 10 is 77 + 10. 77 + 10 = 87.

Hope this helps! This question was kind of confusing for me, because I didn’t understand why you said to add 77 - 10… is this is the incorrect answer sorry your question was worded weird

4 0
2 years ago
-4÷[1/3(7-1/2)]+3<br>(pls show your work) ​
jek_recluse [69]

Step-by-step explanation:

= -4 ÷ ( 1/3 × (7 - 1/2) ) + 3

= -4 ÷ ( 1/3 × 13/2 ) + 3

= -4 ÷ 13/6 + 3

= -4 × 6/13 + 3

= -24/13 + 3

= -24/13 + 39/13

= 15/13

3 0
3 years ago
For the cost function c equals 0.1 q squared plus 2.1 q plus 8​, how fast does c change with respect to q when q equals 11​? Det
Soloha48 [4]

Answer:

Rate of change of c with respect to q is 4.3

Percentage rate of change c with respect to q is  9.95%

Step-by-step explanation:

Cost function is given as,  c=0.1\:q^{2}+2.1\:q+8

Given that c changes with respect to q that is, \dfrac{dc}{dq}. So differentiating given function,  

\dfrac{dc}{dq}=\dfrac{d}{dq}\left (0.1\:q^{2}+2.1\:q+8 \right)

Applying sum rule of derivative,

\dfrac{dc}{dq}=\dfrac{d}{dq}\left(0.1\:q^{2}\right)+\dfrac{d}{dq}\left(2.1\:q\right)+\dfrac{d}{dq}\left(8\right)

Applying power rule and constant rule of derivative,

\dfrac{dc}{dt}=0.1\left(2\:q^{2-1}\right)+2.1\left(1\right)+0

\dfrac{dc}{dt}=0.1\left(2\:q\right)+2.1

\dfrac{dc}{dt}=0.2\left(q\right)+2.1

Substituting the value of q=11,

\dfrac{dc}{dt}=0.2\left(11\right)+21.

\dfrac{dc}{dt}=2.2+2.1

\dfrac{dc}{dt}=4.3

Rate of change of c with respect to q is 4.3

Formula for percentage rate of change is given as,  

Percentage\:rate\:of\:change=\dfrac{Q'\left(x\right)}{Q\left(x\right)}\times 100

Rewriting in terms of cost C,

Percentage\:rate\:of\:change=\dfrac{C'\left(q\right)}{C\left(q\right)}\times 100

Calculating value of C\left(q \right)

C\left(q\right)=0.1\:q^{2}+2.1\:q+8

Substituting the value of q=11,

C\left(q\right)=0.1\left(11\right)^{2}+2.1\left(11\right)+8

C\left(q\right)=0.1\left(121\right)+23.1+8

C\left(q\right)=12.1+23.1+8

C\left(q\right)=43.2

Now using the formula for percentage,  

Percentage\:rate\:of\:change=\dfrac{4.3}{43.2}\times 100

Percentage\:rate\:of\:change=0.0995\times 100

Percentage\:rate\:of\:change=9.95%

Percentage rate of change of c with respect to q is 9.95%

7 0
3 years ago
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