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Cerrena [4.2K]
3 years ago
12

Sqrt 2x-1=1 (in picture)

Mathematics
2 answers:
mariarad [96]3 years ago
4 0
(sqrt2x - 1)^2 = (1)^2

2x - 1 = 1

2x = 2

x = 1
Vilka [71]3 years ago
4 0

Answer:

x=1

Step-by-step explanation:

sqrt(2x-1) =1

We will square both sides

sqrt(2x-1)^2 =1^2

2x-1 =1

Add 1 to each side

2x-1+1 = 1+1

2x=2

Divide by 2

2x/2 =2/2

x=1

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Joseph would have 81 quarts of fertilizer left
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Find the slope and use the point to write an equation of the line in point slope form.
Ainat [17]

Answer:

y - 3 = - 1/4(x + 2)

Step-by-step explanation:

m = - 1/4

Point = (-2,3)

y-intercept = 3 - (- 1/4) (-2)

= 3 - 1/2 = 5/2

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4 years ago
Read 2 more answers
A commercial jet can fly 868 miles in 2 hours with a tailwind but only 792 miles in 2 hours into a headwind. Find the speed of t
erik [133]

Answer:

The speed of the jet in still air is 415 mph and the speed of the wind is 19 mph

Step-by-step explanation:

we know that

The speed is equal to divide the distance by the time

Let

x -----> the speed of the wind in miles per hour

y ----> the speed of the jet in still air in miles per hour

we know that

<em>With a tailwind</em>

y+x=\frac{868}{2}

y+x=434 ----> equation A

<em>With a headwind</em>

y-x=\frac{792}{2}

y-x=396 ----> equation B

solve the system of equations A and B by elimination

Adds equation A and equation B

y+x=434\\y-x=396\\------\\y+y=434+396\\2y=830\\y=415

<em>Find the value of x</em>

y+x=434

415+x=434

x=434-415

x=19

therefore

The speed of the jet in still air is 415 mph and the speed of the wind is 19 mph

6 0
3 years ago
Read 2 more answers
Can anyone help me do this please
allochka39001 [22]

Answer:

1) ∆BIG= ∆FAJ (i can't read it perfectly but i think these are the correct letters)

2)TNY

3) QWE(maybe F, cant read it perfectly)

4) VOR

5) AYD

6) GNH

7) MSE

8) TCA

9) ATE

Step-by-step explanation:

you need to look at the hashes, each segment that had one hash, is equivalent to the other segments with one hash, and so on. then you have to look and see which segments match on each shape. so let's say that if H is the same as A, and I is the same as B, and J is the same as C, then in ∆ HIJ, it would be equivalent to ∆ABC

5 0
2 years ago
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