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IRISSAK [1]
3 years ago
7

Pls help me with these 3 problems. * click the picture *​

Mathematics
1 answer:
storchak [24]3 years ago
4 0

Answer:

6. 360 inches

Step-by-step explanation:

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5 ounces of potatoes produce about 20 calories. What is the unit rate in <br> calories per ounce?
NikAS [45]

Answer:

4 calories per ounce.

Step-by-step explanation:

20/5 = 4

7 0
2 years ago
Read 2 more answers
What are the answers to these i put them in the picture
tekilochka [14]
The problem to number 1 is a inequality, so
x-2<-7
x<-5
since it is < it will have open dot.
Answer:  C

2.  x/3>-3
      x>-9

Answer: D

3.  5p+26<72
                5p<46
                 p<9.2

Answer: C

4.  -3w+3>=18
      -3w>=15
       x<=-5

5. 15x-21>= 12x+18
        x>=13
Answer A

6:  |3x|=15
Absolute values positive so {-5,5}
Answer: B
7: |2x-3|=5
    x=4,-1
Answer:  C
8 0
3 years ago
A math textbook has 6 pages of instruction for every 2 pages of practice problems. How many pages of instruction does it have fo
gayaneshka [121]

Answer:

Unit rate is found simply by dividing.

6 pages of instruction/2 pages of practice problems = 3 pages of instruction per page of practice problems

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Cho tập hợp A=(2;7)và tập hợp B=[7;10]tìm tập hợp a∩B
ahrayia [7]

sorry po di ko ma intindihan

Step-by-step explanation:

sorry po

6 0
2 years ago
Find the roots of the equation<br> x ^ 2 + 3x-8 ^ -14 = 0 with three precision digits
scoray [572]

Answer:

Step-by-step explanation:

Given quadratic equation:

x^{2} + 3x - 8^{- 14} = 0

The solution of the given quadratic eqn is given by using Sri Dharacharya formula:

x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}

The above solution is for the quadratic equation of the form:

ax^{2} + bx + c = 0  

x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}

From the given eqn

a = 1

b = 3

c = - 8^{- 14}

Now, using the above values in the formula mentioned above:

x_{1, 1'} = \frac{- 3 \pm \sqrt{3^{2} - 4(1)(- 8^{- 14})}}{2(1)}

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})})

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})} - 3)

Now, Rationalizing the above eqn:

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(- 8^{- 14})} - 3)\times (\frac{\sqrt{9 - 4(- 8^{- 14})} + 3}{\sqrt{9 - 4(- 8^{- 14})} + 3}

x_{1, 1'} = \frac{1}{2}.\frac{(\pm {9 - 4(- 8^{- 14})^{2}} - 3^{2})}{\sqrt{9 - 4(- 8^{- 14})} + 3}

Solving the above eqn:

x_{1, 1'} = \frac{2\times 8^{- 14}}{\sqrt{9 + 4\times 8^{-14}} + 3}

Solving with the help of caculator:

x_{1, 1'} = \frac{2\times 2.27\times 10^{- 14}}{\sqrt{9 + 42.27\times 10^{- 14}} + 3}

The precise value upto three decimal places comes out to be:

x_{1, 1'} = 0.758\times 10^{- 14}

5 0
3 years ago
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