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yanalaym [24]
3 years ago
10

0.78 =? as a simplified fraction?

Mathematics
1 answer:
Murljashka [212]3 years ago
8 0

Answer: 39/50

Step-by-step explanation:

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Write the point-slope form of the equation for a line that passes through (6, –1) with a slope of 2. The value of x1 is . The va
ELEN [110]

Answer:

The value of x1 is:6

The value of y1 is:-1

The point-slope form of the equation is:

y – 1 = 2(x + 6)

y + 1 = 2(x – 6)✔

1 – y = 2(6 + x)

1 + y = 2(6 – x).

9 0
4 years ago
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Juan walks 3.5 miles in 56 minutes. How long will it take him to walk 1 mile?
mestny [16]

distance = time * speed

1 mile = time * 3.5 miles /56 minutes

so time = 56 minutes/3.5 = 16 minutes

5 0
3 years ago
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What is this and how do you do it? I’m confused...
Maksim231197 [3]

this means alternate interior angle are congruent. to solve this, you would equal them together.

you would write out the equation, and then subtract 6x on both sides.

7x-7 = 6x+11

-6x -6x

this will give you

1x-7 = 11

you then add 7 on both sides

1x - 7 = 11

+ 7 +7

1x = 18

divide by 1 to solve for x

1x÷ 1

18÷ 1

x=18

5 0
3 years ago
Given the force field F, find the work required to move an object on the given oriented curve. F = (y, - x) on the path consisti
timofeeve [1]

Answer:

0

Step-by-step explanation:

We want to compute the curve integral (or line integral)

\bf \int_{C}F

where the force field F is defined by

F(x,y) = (y, -x)

and C is the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9).

We can write  

C = \bf C_1+C_2

where  

\bf C_1 =  line segment from (1, 5) to (0, 0)  

\bf C_2 = line segment from (0, 0) to (0, 9)

so,

\bf \int_{C}F=\int_{C_1}F+\int_{C_2}F

Given 2 points P, Q in the plane, we can parameterize the line segment joining P and Q with

<em>r(t) = tQ + (1-t)P for 0 ≤ t ≤ 1 </em>

Hence \bf C_1 can be parameterized as

\bf r_1(t) = (1-t, 5-5t) for 0 ≤ t ≤ 1

and \bf C_2 can be parameterized as

\bf r_2(t) = (0, 9t) for 0 ≤ t ≤ 1

The derivatives are

\bf r_1'(t) = (-1, -5)

\bf r_2'(t) = (0, 9)

and

\bf \int_{C_1}F=\int_{0}^{1}F(r_1(t))\circ r_1'(t)dt=\int_{0}^{1}(5-5t,t-1)\circ (-1,-5)dt=0

\bf \int_{C_2}F=\int_{0}^{1}F(r_2(t))\circ r_2'(t)dt=\int_{0}^{1}(9t,0)\circ (0,-9)dt=0

In consequence,

\bf \int_{C}F=0

6 0
4 years ago
Please ⟟ need help fast
allochka39001 [22]

Answer:

do you have a picture or?

have a good day :)

Step-by-step explanation:

4 0
3 years ago
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