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ehidna [41]
4 years ago
14

Find the result when 5m + 2 is subtracted from 9m

Mathematics
1 answer:
PtichkaEL [24]4 years ago
6 0
9m - (5m + 2)

9m - 5m = 4m

4m + 2 is your answer

hope this helps
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Given the infirmation in the diagram, which theorem best justifies why lines j and k must be parallel
Karo-lina-s [1.5K]

Answer:

Alternate Exterior Angles theorem

6 0
3 years ago
Read 2 more answers
Recall that m(t) = (1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500g sample of phosphorus-32 decays
katrin2010 [14]

The question is incomplete, here is the complete question:

Recall that m(t) = m.(1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500 g sample of phosphorus-32 decays to 356 g over 7 days. Calculate the half life of the sample.

<u>Answer:</u> The half life of the sample of phosphorus-32 is 14.28days^{-1}

<u>Step-by-step explanation:</u>

The equation used to calculate the half life of the sample is given as:

m(t)=m_o(1/2)^{t/h}

where,

m(t) =  amount of sample after time 't' = 356 g

m_o = initial amount of the sample = 500 g

t = time period = 7 days

h = half life of the sample = ?

Putting values in above equation, we get:

356=500\times (\frac{1}{2})^{7/h}\\\\h=14.28days^{-1}

Hence, the half life of the sample of phosphorus-32 is 14.28days^{-1}

7 0
4 years ago
A rectangle is transformed according to the rule r0, 90º. the image of the rectangle has vertices located at r'(–4, 4), s'(–4, 1
Korvikt [17]

Answer:

Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

Counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

Step-by-step explanation:

Given  : rectangle has vertices located at r'(–4, 4), s'(–4, 1), p'(–3, 1), and q'(–3, 4)

To find :  transformed according to the rule 90º , what is the location of q?

Solution : we have given that

vertices located at r'(–4, 4), s'(–4, 1), p'(–3, 1), and q'(–3, 4).

By the rule of 90º rotation clock wise rule : (x ,y ) →→ ( y , -x )

90º rotation counter clock wise rule : (x ,y ) →→ ( -y , x ).

Then   Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

Therefore, Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

4 0
3 years ago
Read 2 more answers
The midpoint between 94.68 and 49.67
Brilliant_brown [7]

Answer:

72.175

Step-by-step explanation:

The first step is subtract 94.68 by 49.67 which would be 45.01

The next step would be to use 45.01 and divide by two

The last step would be subtract 94.68 by 22.505 and you would get 72.175

Therefore your answer would be 72.175

8 0
4 years ago
Help me ASAP giving brianliest!!!!!<br><br>Find the X
ioda
Hey! So I’m pretty sure you multiply 56 and 2, then you swap the sides of the equation. Answer: X=112
6 0
3 years ago
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