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asambeis [7]
2 years ago
12

Please Help!!

Mathematics
1 answer:
Margaret [11]2 years ago
8 0
That’s literally the easy problem ever
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2(x + 4)= 2(-8 - x) - 2x<br> Plz explain or show how u got the answer
Andrews [41]

Answer: x=-4

Step-by-step explanation:

See attachment

5 0
4 years ago
Please help answer :)
Umnica [9.8K]

Answer:

Step-by-step explanation:

1. You just plot the points

2. To plot the line you have to find the slope and y-intercept

Slope = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  } = \frac{30-15}{4-2} = \frac{15}{2}

Slope = \frac{15}{2}

Y-intercept = y-y_{1}=m(x-x_{1} )

Y-intercept: y - 15 = \frac{-15}{2}(x - 2)

y - 15 = \frac{-15}{2}x + 15

Add 15 to both sides

y = \frac{-15}{2}x + 30 this is the equation of the line.

So you start with the starting point and then the slope!

5 0
4 years ago
3 = m/12<br>What is m?<br><br>It is for math
JulsSmile [24]

Answer:

m is equal to 36.

Step-by-step explanation:

To get rid of division, all you need to do is multiply both sides of the equation by the denominator, which is 12 in this case:

3 = \frac{m}{12}\\\\3 \times 12 = \frac{m}{12} \times 12\\\\36 = m

So m is equal to 36

3 0
3 years ago
Solve for x<br> X=In sqrt e
sesenic [268]

Step-by-step explanation:

e = 2.7182818 so you can just plug that in to your calculator and solve like normal

4 0
3 years ago
A 2000-gallon metal tank to store hazardous materials was bought 15 years ago at cost of $100,000. What will a 5,000-gallon tank
harkovskaia [24]

Answer:

The value  is P_o = \$ 561958.9

Step-by-step explanation:

From the question we are told that

  The capacity of the metal  tank is  C =  2000 \  gallon

   The duration usage is  t = 15\ years \ ago

   The cost of 2000-gallon tank 15 years ago is P =  \$100,000

    The capacity of the second tank considered is C_1 = 5,000  

    The power sizing exponent is e = 0.57

     The initial construction cost index is  u_1 = 180

      The new construction after 15 years cost index is  u_2 =600

Equation for the power sizing exponent is mathematically represented as

      \frac{P_n}{P} = [\frac{C_1}{C} ]^{e}

=> Here P_n is the cost of 5,000-gallon tank as at 15 years ago  

So

     P_n  =  [\frac{5000}{2000} ] ^{0.57} * 100000

      P_n  =  \$168587.7

Equation for the cost index exponent is mathematically represented as

      \frac{P_o}{P_n}  =  \frac{u_2}{u_1}

HereP_o is the cost of 5,000-gallon tank today

So

       \frac{P_o}{168587.7}  =  \frac{600}{180}

=>    P_o = \frac{600}{180} * 168587.7

=>      P_o = \$ 561958.9

8 0
4 years ago
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