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expeople1 [14]
3 years ago
10

The sum of one-half t and one third s

Mathematics
1 answer:
Anna71 [15]3 years ago
3 0

Answer:

5/6

Step-by-step explanation:

Add the fractions by finding the common denominator.

1/2 + 1/3

3/6 + 2/6

5/6

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What is the a<br> nswer to<br> 5(r-11)=2(r-4)-6
Deffense [45]
5 (r - 11) = 2 (r - 4) - 6   Use the Distributive Property on both sides
 5r - 55 = 2r - 8 - 6      Combine like terms (-8 and - 6)
 5r - 55 = 2r - 14          Subtract 2r from both sides
 3r - 55 = -14                Add 55 to both sides
        3r = 41                  Divide both sides by 3
          r = 14\frac{2}{3}
6 0
4 years ago
PLEASES HELP the screenshots are at the bottom
riadik2000 [5.3K]

Answer:

  1.  D. The water will be about 50°C. (54.8 (rounded up) is 50.)
  2.  C. 23; the amount the temperature increases each minute.

Step-by-step explanation:

x = 6 (6 minutes)

5.3 x 6 + 23 = 54.8

5.3 x 6 = 31.8

31.8 + 23 = 54.48

2.  C.  23: the amount of the temperature increased each minute.

Let me know if this helps!

8 0
3 years ago
Find a quadratic equation which has double root = 9. Write the quadratic form in the simplest standard form? x^2 +bx+c.
yarga [219]

The quadratic equation for an equation with double roots 9 is : x^{2}-18x+81 = 0.

what is a Quadratic equation?

A quadratic equation is an Algebraic equation with the highest power of its variable as 2.

It can be solved by several methods : Factorization, completing the square and graphical methods.

Analysis

if 9 is the double root of the equation, it means x = 9 or  x = 9

So (x-9)(x-9) are factors

Therefore,

(x-9)(x-9) = 0

x^{2} - 9x-9x + 81 = 0

x^{2} -18x+81 = 0

in conclusion, the quadratic equation of the double root 9  is:  x^{2} -18x+81 = 0

Learn more about Quadratic equations : brainly.com/question/1214333

#SPJ1

7 0
2 years ago
1. Find the derivative with respect to x of x +1/x from first principle. <br>​
Murrr4er [49]

If you mean f(x)=x+\frac1x, then the derivative is

\displaystyle f'(x) = \lim_{h\to0} \frac{\left(x+h+\frac1{x+h}\right) - \left(x+\frac1x\right)}h \\\\ = \lim_{h\to0} \frac{(x+h)-x}h + \lim_{h\to0} \frac{\frac1{x+h} - \frac1x}h \\\\ = \lim_{h\to0} \frac hh + \lim_{h\to0} \frac{x-(x+h)}{hx(x+h)} \\\\ = \lim_{h\to0} 1 - \lim_{h\to0} \frac h{hx(x+h)} \\\\ = 1 - \lim_{h\to0} \frac1{x(x+h)} \\\\ = \boxed{1 - \frac1{x^2}}

If you mean f(x) = \frac{x+1}x = 1 + \frac1x, we know from above that

\displaystyle \left(\frac1x\right)' = \lim_{h\to0} \frac{\frac1{x+h}-\frac1x}h = -\frac1{x^2}

which leaves the constant term, whose derivative is

\displaystyle (1)' = \lim_{h\to0}\frac{1 - 1}h = 0

and so

f'(x) = -\dfrac1{x^2}

6 0
2 years ago
How do I create a radical equation with an extraneous solution quickly? How do I go the same for a radical equation with a non-e
Step2247 [10]

Answer:

why r u wafflin

Step-by-step explanation:

5 0
3 years ago
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