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professor190 [17]
3 years ago
7

Help? Because my teacher doesn't want too

Mathematics
1 answer:
snow_lady [41]3 years ago
7 0
M<ABC = m<EBD = 36 (vertical angles)

78 - x + 36 + 3x - 10 = 180 (straight angle)

Now solve for x

78 - x + 36 + 3x - 10 = 180
2x + 68 = 180
     -  68     -68 (subtract 68 on both sides)
2x        =  58
  x        =  29
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Ivan purchased n notebooks. They were 4 dollars each. Write an equation to represent the total cost c that Ivan paid.
xenn [34]

Answer:

4n=c

Step-by-step explanation:

your trying to figure out how how much he paid and how much money it cost total so you have to multiply the cost of one by the number of notebooks you have n and by doing that you would be able to get your total cost c

3 0
3 years ago
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How do you solve this?
zavuch27 [327]

Answer:

f(- 10) = 74

Step-by-step explanation:

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3 years ago
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3 years ago
Help. I need help with these questions ( see image).<br> Please show workings.
Andrew [12]

9514 1404 393

Answer:

  4)  6x

  5)  2x +3

Step-by-step explanation:

We can work both these problems at once by finding an applicable rule.

  \text{For $f(x)=ax^n$}\\\\\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h\to 0}\dfrac{a(x+h)^n-ax^n}{h}\\\\=\lim\limits_{h\to 0}\dfrac{ax^n+anx^{n-1}h+O(h^2)-ax^n}{h}=\boxed{anx^{n-1}}

where O(h²) is the series of terms involving h² and higher powers. When divided by h, each term has h as a multiplier, so the series sums to zero when h approaches zero. Of course, if n < 2, there are no O(h²) terms in the expansion, so that can be ignored.

This can be referred to as the <em>power rule</em>.

Note that for the quadratic f(x) = ax^2 +bx +c, the limit of the sum is the sum of the limits, so this applies to the terms individually:

  lim[h→0](f(x+h)-f(x))/h = 2ax +b

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4. The gradient of 3x^2 is 3(2)x^(2-1) = 6x.

5. The gradient of x^2 +3x +1 is 2x +3.

__

If you need to "show work" for these problems individually, use the appropriate values for 'a' and 'n' in the above derivation of the power rule.

3 0
3 years ago
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