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Scorpion4ik [409]
3 years ago
14

Math question shown below question and graph

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
4 0

Answer:

The graph would look something like this...

Step-by-step explanation:


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What are the coordinates of the point on the directed line segment from (1,-5) to
Black_prince [1.1K]

Answer:

Q (2.5, -4)

Step-by-step explanation:

Let A (1, -5), C = (10, 1), find Q (x, y), so that AQ:QC = 1:5

AQ = (x - 1, y + 5)

QC = (10 - x, 1 - y)

AQ:QC = 1:5

so 5*AQ = QC

5(x - 1) = (10 - x)

5(y + 5) = 1 - y

Solve them:

5x - 5 = 10 - x

6x = 15

x = 15/6 = 2.5

5y + 25 = 1 - y

6y = -24

y = -24/6 = -4

Then answer is Q (2.5, -4)

6 0
2 years ago
The graph of a system of linear equations is shown below.
lara31 [8.8K]

Answer:

wheres the graph lol ?

5 0
2 years ago
Read 2 more answers
Andy has a stamp collection with 343 stamps, Of these 296 are from Germany. Is 40,50 or 60 a more reasonable estimate for how ma
mariarad [96]

Answer:

40 Stamps

Step-by-step explanation:

According to the scenario, given data are as follows,

Total stamp collected = 343

Stamps from Germany = 296

To estimate the stamps from other contact, we first round off the given number of stamps.

So, Total stamp collected = 340 ( Rounded off)

Stamps from Germany = 300 ( Rounded off)

So, Stamps from other contact = 340 - 300 = 40 stamps

3 0
3 years ago
Round 591,753,460 to the hundrend thousand
never [62]

Answer:

591,800,000

Step-by-step explanation:

The 7 in 753,460 is rounded to an 8 because if the number after it is 5 or higher you turn all the numbers after the 7 into a 0 and turn 7 into a 8.

5 0
3 years ago
the value of c guaranteed to exist by the Mean Value Theorem for V(x) = x² in the interval [0, 3] is...? a) 1 b) 2 c) 3/2 d) 1/2
lilavasa [31]

Answer:  c) \dfrac{3}{2} .

Step-by-step explanation:

Mean value theorem : If f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that

\begin{displaymath}f'(c) = \frac{f(b) - f(a)}{b-a} \cdot\end{displaymath}

Given function : f(x) = x^2

Interval : [0,3]

Then, by the mean value theorem, there is at least one number c in the interval (0,3) such that

f'(c)=\dfrac{f(3)-f(0)}{3-0}\\\\=\dfrac{3^2-0^2}{3}=\dfrac{9}{3}\\\\=3

\Rightarrow\ f'(c)=3\ \ \ ...(i)

Since f'(x)=2x

then, at x=c, f'(c)=2c\ \ \ ...(ii)

From (i) and (ii), we have

2c=3\\\\\Rightarrow\ c=\dfrac{3}{2}

Hence, the correct option is c) \dfrac{3}{2} .

4 0
3 years ago
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