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amm1812
3 years ago
8

The table shows Ava's checking account activity for several weeks. Ava's beginning balance was $685. What was her ending balance

? Three column table titled Avas Account. Data in the first column is Date, June 12, June 20, June 29, July 2, and July 15. Data in the second column is Deposit, 175, blank, 180, blank, and blank. Data in the third column is Withdrawal, blank, minus240, blank, minus140, and minus350.
A. –$45
   
B. $310
   
C. $355
   
D. $995
Mathematics
1 answer:
Dafna11 [192]3 years ago
8 0
B. Deposit means she added money to her account so you would add 175 and 180 to 685 (giving you 1040). Withdrawal means she took money out of her account so you would subtract 240, 140, and 350 from 1040 giving you your answer of $310 :)
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You stand a known distance from the base of the tree, measure the angle of elevation the top of the tree to be 15â—¦ , and then
gogolik [260]

Answer:

The maximum possible error of in measurement of the angle is  d\theta_1  =(14.36p)^o

Step-by-step explanation:

From the question we are told that

    The angle of elevation  is  \theta_1  =  15 ^o =  \frac{\pi}{12}

     The height of the tree is  h

      The distance from the base is  D

h is mathematically represented as

            h  = D tan \theta       Note : this evaluated using SOHCAHTOA i,e

                                               tan\theta  =  \frac{h}{D}

Generally for small angles the series approximation of  tan \theta \  is

          tan \theta  =  \theta  + \frac{\theta ^3 }{3}

So given that \theta =  15 \ which \ is \ small

       h = D (\theta + \frac{\theta^3}{3} )

       dh = D (1 + \theta^2) d\theta

=>        \frac{dh}{h} =  \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta

Now from the question the relative error of height should be at  most

        \pm  p%

=>    \frac{dh}{h} =   \pm p

=>    \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta  = \pm p

=>      d\theta  =  \pm  \frac{\theta +  \frac{\theta^3}{3} }{1+ \theta ^2} *    \ p

 So  for   \theta_1

            d\theta_1  =  \pm  \frac{\theta_1 +  \frac{\theta^3_1 }{3} }{1+ \theta_1 ^2} *    \ p

substituting values  

          d [\frac{\pi}{12} ]  =  \pm  \frac{[\frac{\pi}{12} ] +  \frac{[\frac{\pi}{12} ]^3 }{3} }{1+ [\frac{\pi}{12} ] ^2} *    \ p

 =>       d\theta_1  = 0.25 p

Converting to degree

           d\theta_1  = (0.25* 57.29) p

            d\theta_1  =(14.36p)^o

4 0
3 years ago
Just number 7 would be fine but if you could also number 8 would help a lot​
BlackZzzverrR [31]

Answer:

7. r = -5

8. x = -1

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

Step-by-step explanation:

<u>Step 1: Define</u>

r + 2 - 8r = -3 - 8r

<u>Step 2: Solve for </u><em><u>r</u></em>

  1. Combine like terms:                    -7r + 2 = -3 - 8r
  2. Add 8r to both sides:                   r + 2 = -3
  3. Subtract 2 on both sides:            r = -5

<u>Step 3: Check</u>

<em>Plug in r into the original equation to verify it's a solution.</em>

  1. Substitute in <em>r</em>:                    -5 + 2 - 8(-5) = -3 - 8(-5)
  2. Multiply:                              -5 + 2 + 40 = -3 + 40
  3. Add:                                    -3 + 40 = -3 + 40
  4. Add:                                    37 = 37

Here we see that 37 does indeed equal 37.

∴ r = -5 is a solution of the equation.

<u>Step 4: Define equation</u>

-4x = x + 5

<u>Step 5: Solve for </u><em><u>x</u></em>

  1. Subtract <em>x</em> on both sides:                    -5x = 5
  2. Divide -5 on both sides:                      x = -1

<u>Step 6: Check</u>

<em>Plug in x into the original equation to verify it's a solution.</em>

  1. Substitute in <em>x</em>:                    -4(-1) = -1 + 5
  2. Multiply:                               4 = -1 + 5
  3. Add:                                     4 = 4

Here we see that 4 does indeed equal 4.

∴ x = -1 is a solution of the equation.

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