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3 years ago

Solve the formula for v 1 v1 . Average acceleration of an object: a= v 1 − v 0/ t pls help

Mathematics

1 answer:

ziro4ka [17]3 years ago

3 0

In this question , we have the formula of acceleration given, which is

$a = \frac{v_{1} - v_{0}}{t}$

And we have to solve for $v_{0}$

First we need to get rid of denominator t from right side and for that, we multiply both sides by t, that is

$at = v_{1} - v_{0}$

Now to solve for v1 , we need to get rid of v0, and for that , we add v0 to both sides, that is

$v_{1} = at + v_{0}$

And that's the required answer.

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David made a class banner out of a large rectangular piece of paper. He cut a triangular piece out of one side. what is the area

VikaD [51]

The height would be the same as the rectangular piece of paper as well as the length so what you would do is you do heights times Lanks divide that by two and you would get the Square inches of the banner hope it helps

8 0

2 years ago

Insert brackets to make this calculation right 4x2+5-3=

pentagon [3]

Salutations!

Insert brackets to make this calculation right 4x2+5-3

The brackets should be inserted outside 2+5. If you add brackets outside teh rest of the numbers the calculation would not make sense.

4 × (2 + 5) - 3

Hope I helped (:

have a great day!

8 0

3 years ago

Match each equation with its solution set.

taurus [48]

Answer:

1- The solution of I2x + 5I = 9 is {-7 , 2}

2- The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- The solution of I5 - xI = 6 is {-1 , 11}

4- The solution of I6x - 8I + 7 = 5 is ∅

5- The solution of Ix + 3I = 12 is {-15 , 9}

6- The solution of Ix - 3I = -12 is ∅

Step-by-step explanation:

* At first lets explain the meaning of IxI = a

- If IxI = a ⇒ then x = a or x = -a

- IxI never give a negative answer, because IxI means the

magnitude of x is always positive

Ex: I-2I is 2

* Now lets find the solution of each equation

1- ∵ I2x + 5I = 9

∴ 2x + 5 = 9 ⇒ subtract 5 from both sides

∴ 2x = 4 ⇒ divide both sides by 2

∴ x = 2

∴ 2x + 5 = -9 ⇒ subtract 5 from both sides

∴ 2x = -14 ⇒ divide both sides by 2

∴ x = -7

* The solution of I2x + 5I = 9 is {-7 , 2}

2- ∵ I2x + 7I + 2 = 11 ⇒ Subtract 2 from both sides

∴ I2x + 7I = 9

∴ 2x + 7 = 9 ⇒ subtract 7 from both sides

∴ 2x = 2 ⇒ divide both sides by 2

∴ x = 1

∴ 2x + 7 = -9 ⇒ subtract 7 from both sides

∴ 2x = -16 ⇒ divide both sides by 2

∴ x = -8

* The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- ∵ I5 - xI = 6

∴ 5 -x = 6 ⇒ subtract 5 from both sides

∴ -x = 1 ⇒ divide both sides by -1

∴ x = -1

∴ 5 -x = -6 ⇒ subtract 5 from both sides

∴ -x = -11 ⇒ divide both sides by -1

∴ x = 11

* The solution of I5 - xI = 6 is {-1 , 11}

4- ∵ I6x - 8I + 7 = 5 ⇒ Subtract 7 from both sides

∴ I6x - 8I = -2

- I I never give negative answer

* The solution of I6x - 8I + 7 = 5 is ∅

5- ∵ Ix + 3I = 12

∴ x + 3 = 12 ⇒ subtract 3 from both sides

∴ x = 9

∴ x + 3 = -12 ⇒ subtract 3 from both sides

∴ x = -15

* The solution of Ix + 3I = 12 is {-15 , 9}

6- ∵ Ix - 3I = -12

- I I never give negative answer

* The solution of Ix - 3I = -12 is ∅

4 0

3 years ago

Use distance formula d=√(X2 – X1 )2 + (X2 – X1 )2, calculate the distance r from the origin to the point r (-2, √5).

lesya692 [45]

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \\d = \sqrt{(\sqrt{5} - (-2))^{2} + (\sqrt{5} - (-2))^{2}} \\d = \sqrt{(\sqrt{5} + 2)^{2} + (\sqrt{5} + 2)^{2}} \\d = \sqrt{(4 + 4\sqrt{5} + 5) + (4v + 4\sqrt{5} + 5)} \\d = \sqrt{0 + 0} \\d = \sqrt{0} \\d = 0$

7 0

3 years ago

what is this have no clue of how to do it ? ? n+31.53 =62.4 and this one to 9.2+n+8.4=20.8 please help !

mihalych1998 [28]

Ok so to solve to first one u do this:

62.4 - 31.53, which gives u 30.87. and then u add 30.87 and 31.33, and u get 62.4

for the second one u do the same thing.

4 0

3 years ago

Read 2 more answers