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Mashutka [201]
3 years ago
15

3x+20=110 true or false

Mathematics
1 answer:
lys-0071 [83]3 years ago
6 0
Falseeeeeeeeee
:) hope this helps

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Which of the following is the solution set of the given equation? 14 + 8m = 14 - 3m - 5m ∅ {0} {all reals}
Yanka [14]

Answer:

{0}

Step-by-step explanation:

14 + 8m = 14 - 3m - 5m

Combine like terms

14 + 8m = 14 - 8m

Add 8m to each side

14 +8m +8m = 14 -8m +8m

14 +16m = 14

Subtract 14 from each side

14-14 +16m = 14-14

16m = 0

Divide by 16

16m/16 = 0/16

m = 0

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What's the divisor for <br> 4/12 =1/3
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I'm gonna go with 1 (:
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3 years ago
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Need help with this question; What is the measure of one leg of a right angle given that the hypotenuse measure 14cm and the oth
melamori03 [73]

Answer:

11.5 cm

Step-by-step explanation:

The dimensions of a right angled triangle can be determined using Pythagoras theorem

The Pythagoras theorem : a² + b² = c²

where a = length

b = base

c = hypotenuse

a² + 8² = 14²

a² + 64 = 196

a² = 196 - 64

a² = 132

find the square root of 132

a = 11.5 cm

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2 years ago
Use the following cell phone airport data speeds​ (Mbps) from a particular network. Find the percentile corresponding to the dat
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In a certain town, the amount of sulfur oxide in the air, S, in tons, is related to the town’s population, P, in people. The rel
Lady bird [3.3K]

Answer:

Change in sulfur oxide in the air = \frac{\textup{110010}}{\textup{S}}

Step-by-step explanation:

Data provided in the question:

Relation between the amount of sulfur oxide in the air and the population as:

S² = 110P² + 20P + 600

Population growth rate, \frac{\textup{dP}}{\textup{dt}}  = 10 people per month

Now,

change in sulfur oxide with time i.e \frac{\textup{dS}}{\textup{dt}}

differentiating the given relation with respect to time 't'

we have

2S\frac{\textup{dS}}{\textup{dt}} =  2\times110P\frac{\textup{dP}}{\textup{dt}}  + 20

at P = 100 and  \frac{\textup{dP}}{\textup{dt}}  = 10 people per month

we have

2S\frac{\textup{dS}}{\textup{dt}} = 2 × 110 × 100 × 10 + 20

or

2S\frac{\textup{dS}}{\textup{dt}} = 220020

or

\frac{\textup{dS}}{\textup{dt}} = \frac{\textup{220020}}{\textup{2S}}

or

Change in sulfur oxide in the air = \frac{\textup{110010}}{\textup{S}}

8 0
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