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3 years ago

Find the missing coordinate to complete the ordered-pair solution to the given linear equation: 3x-4y=6

Mathematics

1 answer:

zavuch27 [327]3 years ago

8 0

8 is the answer if u do some equation gymnastics as i call them

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3 years ago

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Goshia [24]

Answer:

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Step-by-step explanation:

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2 years ago

On a normal day, I can walk from my house to school in three quarters of an hour. The return trip takes me the same amount of ti

Likurg_2 [28]

Answer:

208 1/8 hours

Step-by-step explanation:

The amount of time saved on each trip is ...

(3/4 h) - (1/4)(3/4 h) = (3/4)(3/4 h) = 9/16 h

There are 2×5×37 = 370 trips in the school year, so the total savings in time is about ...

370×(9/16 h) = 208 1/8 h

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3 years ago

Write the word sentence as an equation. Then solve the equation.

Nana76 [90]

Answer:

x/-9 = -16 ; 144

Step-by-step explanation:

x is the variable for "a number". It is being divided by -9, so x must be divided by -9. The number divided by -9 equals -16, so -16 must go on the other side of the equation. To solve, multiply both sides by -9 to cancel out the fraction and isolate the variable.

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3 years ago

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Two children own two-way radios that have a maximum range of 3 miles. One leaves a certain point at 1:00 P.M., walking due north

Grace [21]

Answer:

The answer is 21 minutes

Step-by-step explanation:

We use the equation Xf = Xo + vt

1) At 1:00 PM, child one leaves the starting point heading north at a constant velocity of 6 mi/hr or .1 [mi/min] (divide by 60 to convert from [mi/hr] to [mi/min])

2) He walks for 15 minutes before kid 2 starts walking. In 15 minutes he is able to cover 1.5 [mi]

$x_{1f1} =x_{o} +v_{1} t_{1} \\x_{1f1} =0+.1(15)\\x_{1f1} =1.5 [mi]$

3) Now, child 2 starts walking and we know that when the range reaches 3 miles, they won´t be able to communicate. So the sum of the final position of child 1 and child 2 must be 3[mi]

Child 1 final position => $x_{1f} = x_{1f1} +v_{1} t\\x_{1f} =1.5+v_{1} t$
Child 2 final position => $x_{2f} =0+v_{2} t$

4) Sum the equations and equate to 3

$x_{1f} +x_{2f} =3$

5) Substitute the values we already know

$1.5+v_{1} t+v_{2}t=3\\ 1.5+.1t+.15t=3\\1.5+.25t=3\\t=\frac{3-1.5}{.25} \\t=6 [min]$

6) in 15 + 6 minutes they will be 3miles apart

7) In 21 minutes they will still be able to communicate with one another.

7 0

3 years ago