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Molodets [167]
3 years ago
12

#10 solve the equation 1/3m - 7 = 5

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
5 0
(1/3) m - 7 = 5 
(1/3) m - 7 + 7= 5 + 7 ⇒⇒⇒ edit 7 to both sides
(1/3) m = 12
3 * (1/3) m = 12 * 3     ⇒⇒⇒ multiply both sides by 3
∴ m = 36
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Solve for P. <br><br> S=A+1/2PL
Dovator [93]

9514 1404 393

Answer:

  (b)  P = 2(S -A) ÷ L

Step-by-step explanation:

Undo what is done to P.

  S=A+\dfrac{1}{2}PL\qquad\text{given}\\\\S-A = \dfrac{1}{2}PL\qquad\text{subtract $A$}\\\\\dfrac{2}{L}(S-A)=P \qquad\text{multiply by $\dfrac{2}{L}$}\\\\\boxed{P=\dfrac{2(S-A)}{L}}

5 0
3 years ago
Please help me create a function,and if you can,please explain how
qaws [65]
Current volume = 10 glns
Miles he can drive currently = 28miles/ gln * 10= 280miles
Cost of gas= 1.40/gln
Function= s /1.40 *28 +280= 28s/1.40 +280=20s + 280
Final answer would be 20s + 280
6 0
3 years ago
7s + 8s - 6h please help
Anit [1.1K]

Answer:-6h+15s

Step-by-step explanation:

7s+8s−6h

=7s+8s+−6h

Combine Like Terms:

=7s+8s+−6h

=(−6h)+(7s+8s)

=−6h+15s

Answer:

=−6h+15s

6 0
3 years ago
Which best describes the complement of spinning any number less than 3?
tatuchka [14]

Answer:

  • <u>The complement of spinning any number less than 3, is spinning a number equal to or greater than 3.</u>

Explanation:

The complement of a subset is the subset of elements that are not in the given subset.

You must know which numbers the spinner has.

Assuming the spinner has the numbers 1, 2, 3, 4, the complement of spinning any number less than 3, is spinning a number that is not less than 3.

Then, that is spinning a number that is equal to or greater than 3.

The numbers that are equal to  or greater than four, for a spinner that has the numbers 1, 2, 3, and 4 are 3 and 4.

Thus, the complement of spinning any number less than 3 is spinning a three or a four.

8 0
3 years ago
I forgot to attach the photo before.... please help
SVEN [57.7K]

Part (i)

I'll use H in place of T to represent the heat of the object. That way there isn't a clash of variables lowercase t vs uppercase T.

The equation we're working with is updated to:

H(t) = 22 + a*2^(bt)

Plugging in t = 0 as the initial time value should lead to the temperature being H = 86 degrees Celsius.

So,

H(t) = 22 + a*2^(bt)

86 = 22 + a*2^(b*0)

86 = 22 + a*2^0

86 = 22 + a*1

86 = 22 + a

a+22 = 86

a = 86-22

a = 64

<h3>Answer:  64</h3>

=====================================================

Part (ii)

We'll use the value of 'a' we found earlier. Plus we'll use the fact that H = 28 when t = 0.5 (since 30 min = 30/60 = 0.5 hr).

H(t) = 22 + a*2^(bt)

28 = 22 + 64*2^(b*0.5)

28-22 = 64*2^(0.5b)

64*2^(0.5b) = 6

2^6*2^(0.5b) = 6

2^(6+0.5b) = 6

log(  2^(6+0.5b)   ) = log(6)

(6+0.5b)*log(2) = log(6)

6+0.5b = log(6)/log(2)

6+0.5b = 2.5849625

0.5b = 2.5849625-6

0.5b = -3.4150375

b = -3.4150375/(0.5)

b = -6.830075

<h3>Answer: Approximately   -6.830075</h3>
7 0
3 years ago
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