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sp2606 [1]
3 years ago
15

I need help please!?!!!

Mathematics
2 answers:
11111nata11111 [884]3 years ago
7 0
8(p-2) = 32
8p - 16 = 32
+16 on each side
8p= 48
Divide by 8 on each side
P= 6
RoseWind [281]3 years ago
4 0

Answer:

6

Step-by-step explanation:

Because 8 x 4 equals 32, so 6 - 2 equals 4.

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PLZ HELP WILL GIVE BRAINLY IF CORRECT
pochemuha

Answer:

1000 MM is the greatest

Step-by-step explanation:

1000 MM to CM = 100

9.7 CM and 4.7 CM are too small, therefore, we ignore them..

450 MM to CM is 45, therefore 1000 MM is the greater length

3 0
3 years ago
Read 2 more answers
Find the constant of variation k for the direct variation ​
Alla [95]

Answer:

The constant of variation is k = -2 ⇒ (B)

Step-by-step explanation:

The equation of the direct variation is y = k x, where

  • k is the constant of variation
  • The constant of variation k = \frac{y}{x}

The given table has 4 points (-1, 2), (0, 0), (2, -4), (5, -10)

We can use one of the points <em>[except point (0, 0)]</em> to find the value of k

∵ (-1, 2) is a given point

∴ x = -1 and y = 2

∵ k = \frac{y}{x}

→ Substitute the values of x and y in the relation above

∴ k = \frac{2}{-1}

∴ k = -2

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7 0
3 years ago
The cost of 12 kg sugar is $240 what will be the cost of 3 kg sugar​
Alexus [3.1K]

Question:

The cost of 12 kg sugar is $240 what will be the cost of 3 kg sugar.

Given:

  • Cost of 12 kg sugar = $240

To find?

  • cost of 3 kg sugar

Answer :

<u>To find </u><u>c</u><u>o</u><u>s</u><u>t</u><u> </u><u>of</u><u> </u><u>3kg sugar first we have to find cost of 1 kg sugar</u><u>.</u>

  • Cost of 1 kg sugar = Cost of total no. of sugar ÷Total sugar
  • Cost of 1 kg sugar = $240/12
  • Cost of 1 kg sugar = $20

<u>Now Let's find cost of 3 kg sugar</u>

Cost of 3kg sugar = total no of sugar ×Cost of 1 kg sugar

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3 0
3 years ago
I'm in 7th grade and need help have a final tomorrow
inna [77]
I can maybe help, what's the question
7 0
3 years ago
Select the correct answer.
marishachu [46]

Answer:

M = \log(10000)

Step-by-step explanation:

Given

M = \log(\frac{I}{I_o})

I = 10000I_o ---- intensity is 10000 times reference earthquake

Required

The resulting equation

We have:

M = \log(\frac{I}{I_o})

Substitute the right values

M = \log(\frac{10000I_o}{I_o})

M = \log(10000)

5 0
3 years ago
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