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Dennis_Churaev [7]
4 years ago
15

What is the answer to -9(x-3)-6(3x-2)=93

Mathematics
2 answers:
Anni [7]4 years ago
8 0

Answer:

Step-by-step explanation:

x=-2

WINSTONCH [101]4 years ago
6 0
-9x+27-18x+12=93
-27x=54
x=-2
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4. A fruit vendor packed 100 apples each in 64 baskets. Later, he found some apples were left unpacked. So he packed 30 more app
topjm [15]

Answer:

Here is the answer

Step-by-step explanation:

Apples in 1 baskets=100

Apples in 64 baskets=100*64-6400

apples (before adding 30 apples in each

basket).... ..(1)

Some apples left unpacked

So he added 30 apples to each

Total apples added by him- 30*64(in each basket*no of baskets)

= 1920 apples

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4 0
2 years ago
A thermometer is taken from a room where the temperature is 20◦C to the
lord [1]

Answer:

a) T(2) = 8.265: at time t = 2 minutes the temperature will be 8.265 degress

b) 6 = T(3.55): the temperature will be 6 degrees at time t = 3.55 minutes.

Step-by-step explanation:

When dealing with temperature changes, it's best to work with Newton's Law of Cooling.

T(t) = T_s + Ce^{kt}

here:

T(t) : the temperature in the room.

T_s : ambient (or outdoor) temperature (that always remains constant, in our case: T_s = 5 )

C\,\text{and}\,k: are constants

Our conditions are provided:

1) T(0) = 20

2) T(1) = 12

using the first condition

T(0) = 5 + Ce^{k(0)}\\20 = 5 + C(1)\\C = 15

using the second condition:

T(1) = 5 + Ce^{k(1)}\\12 = 5 + Ce^{k}\\e^k = \dfrac{7}{C}

we can use our calculated value of C to find k

e^k = \dfrac{7}{15}\\k = \ln{(\dfrac{7}{15})}\\k = -0.7621

Finally we can put these constants back in the main equation:

T(t) = T_s + Ce^{kt}

T(t) = 5 + 15e^{-0.7621t} or T(t) = 5 + 15e^{\ln{(\frac{7}{15})t}

a) Reading after one more minute?

so it's asking:

T(2) = ?

T(2) = 5 + 15e^{\ln{(\frac{7}{15})(2)}}\\T(2) = \dfrac{124}{15} \approx 8.267

Hence, after one more minute the temperature of the room will be 8.267 degrees

b) When will it be 6 degrees?

T(t) = 6?

6 = 5 + 15e^{-0.7621t}\\\text{and solve for $t$}\\\\\dfrac{6-5}{15}=e^{\ln{(\frac{7}{15})}t}\\\ln{\left(\dfrac{1}{15}\right)} = \ln{\left(\dfrac{7}{15}\right)t}\\\ln{\left(\dfrac{1}{15}\right)} \div \ln{\left(\dfrac{7}{15}\right)} = t \approx 3.55\\

Hence at t = 3.55 minutes the temperature of the room will be 6 degrees.

8 0
3 years ago
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