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tatuchka [14]
3 years ago
14

Please help me !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Sphinxa [80]3 years ago
5 0

Answer:

33

Step-by-step explanation:

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alina1380 [7]
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7 0
3 years ago
Rewrite the equation as y-2=-2/3(x-(-1)) to identify a point on the graph
Anvisha [2.4K]

Answer:

  point (-1, 2)

Step-by-step explanation:

Your equation is shown here in point-slope form, so a point on the graph can be read from the given equation. (No further rewrite is necesary.)

The point-slope form is ...

  y -k = m(x -h)

Comparing this to your given equation ...

  y -2 = (-2/3)(x -(-1))

we see that ...

  k = 2, m = -2/3, h = -1

The point on the graph is (h, k) = (-1, 2).

7 0
3 years ago
Can someone help me Please!
enyata [817]

Answer:

1.6

Step-by-step explanation:

Lets count how many points are x and y away from each other since we dont have a graph.

X and Y points:

35-15=20

56-24=32

Use the formula:

\frac{rise}{run}

32/20 = 1.6

6 0
3 years ago
Christian rides his motorcycle 150 miles in 3.5 hours. What is his average speed in miles per hour?
galina1969 [7]

Answer:

Step-by-step explanation:

(150 miles)/(3.5 hours) = (150/3.5 miles)/hour = 42.9 miles/ hour

8 0
3 years ago
Read 2 more answers
Find the equation of the tangent line to the curve y=11xex at the point (0,0). the equation of this tangent line can be written
vichka [17]
y=11xe^x\implies y'=11e^x(x+1)

When x=0, the slope of the tangent line is 11e^0(0+1)=11. The equation of the tangent line in point-slope form is then

y-0=11(x-0)\implies y=11x
5 0
4 years ago
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