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trasher [3.6K]
3 years ago
7

Pls help this is for math ill give a brainly​

Mathematics
1 answer:
olasank [31]3 years ago
6 0

Answer: The last one - 5 1/3 divided by 1/6

Step-by-step explanation:

since you are trying to figure out how much she runs a day, u divide

Hope this helps :)

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What is the equation in point-slope form for the line parallel to y=-8x - 5 that contains j(-2, 9)
Julli [10]

Answer:

y - 9 = -8(x + 2)

Step-by-step explanation:

What is the equation in point-slope form for the line parallel to y=-8x - 5 that contains j(-2, 9)

Step 1

y intercept =

y = mx + b

y=-8x - 5

m = -8, b = -5

Point slope form

y - y1 =m( x - x1)

We are given coordinates:

j(-2, 9) = (x1, y1) where ,

x1 = -2

y1 = 9

Hence, our point slope form equation is:

y - 9 = -8(x -(-2))

y - 9 = -8(x + 2)

5 0
3 years ago
Anybody know the answer to number 2 !!!
MrRa [10]

Answer:

Well I can't put it on a number line but the order they should go in is: 2 \frac{1}{4}, 2.3 (repeating), 2.5,\sqrt{8}

Step-by-step explanation:

5 0
2 years ago
Print out the graph paper from the "Let's Review" section. Sketch the graph of y = 2tanx for -pi/2 ≤ x ≤ pi/2
snow_lady [41]

Answer:

<h2>The graph is attached below.</h2><h2>Same asymptote as similarity and the slopes are different.</h2>

Step-by-step explanation:

We need to find the graph of y = 2 \times tanx.......... \frac{-\pi }{2} \leq  x \leq \frac{\pi }{2}

When x = 0; y = 0\\x = \frac{\pi }{4} ; y = 2\\x = - \frac{\pi }{4} ; y = -2

and \lim_{x \to \frac{\pi }{2} } y =  \infty\\ \lim_{x \to - \frac{\pi }{2} } y = - \infty

The asymptotes are x = \frac{\pi }{2} \\ x = - \frac{\pi }{2}

If we compare the graphs of y = 2 \times tanx.....(1)\\y = tanx......(2)

then the similarities are, they have same asymptote.

Difference between them is the rate of change that is the slope of the equations.

8 0
4 years ago
How many times does 28 go into 7010000
Zinaida [17]

Answer:

250357.1428

Step-by-step explanation:

all you have to do is divide them

7 0
3 years ago
Read 2 more answers
Find each x-value at which f is discontinuous and for each x-value, determine whether f is continuous from the right, or from th
fredd [130]

Using continuity concepts, the answers are given by:

  • The function is right-continuous at x = 0.
  • The function is left-continuous at x = 1.

------------------------------------

A function f(x) is continuous at x = a if:

\lim_{x \rightarrow a^{-}} f(x) = \lim_{x \rightarrow a^{+}} f(x) = f(a)

  • If \lim_{x \rightarrow a^{-}} f(x) = f(a), the function is left-continuous.
  • If \lim_{x \rightarrow a^{+}} f(x) = f(a), the function is right-continuous.
  • For the given function, the continuity is tested at the points in which the definitions are changed, x = 0 and x = 1.

------------------------------------

At x = 0.

  • Approaching from the left, it is less than 0, thus:

\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0} x + 9 = 0 + 9 = 9

  • Approaching from the right, it is more than 0, thus:

\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0} e^x = e^0 = 1

  • The numeric value is:

f(x) = e^0 = 1

  • Since [tex]\lim_{x \rightarrow 0^+} f(x) = f(0), the function is right-continuous at x = 0.

------------------------------------

At x = 1.

\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1} e^x = e^1 = e

\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1} 4 - x = 4 - x = 3

f(1) = e^1 = e

  • Since [tex]\lim_{x \rightarrow 1^-} f(x) = f(1), the function is left-continuous at x = 0.

A similar problem is given at brainly.com/question/21447009

7 0
3 years ago
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