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Ede4ka [16]
3 years ago
12

Bella pushes a box to her right. In which direction does the box need to move for Bella to have done work n on it

Mathematics
2 answers:
Luden [163]3 years ago
8 0

Answer:

left

Step-by-step explanation:

if she pushes it right then she would have to push it left back to her to work on it

KIM [24]3 years ago
3 0

It needs to move to the left

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What is the exponent for 9x9x9x9? help plz
Alenkinab [10]

Answer:

9^4

Step-by-step explanation:

There is 4 nines.

9*9*9*9 = 6561

9^4 = 6561

9^3 = 729

5 0
3 years ago
Kareen ran 4.16 x 10^3 miles long
Savatey [412]
Which means she ran 4160 miles long
if this isn't the answer you wanted, please be more precise 
6 0
3 years ago
Read 2 more answers
Given: △ABC, AB=5sqrt2 <br> m∠A=45°, m∠C=30°<br> Find: BC and AC
Marysya12 [62]

BC is 10 units and AC is 5+5\sqrt{3} units

Step-by-step explanation:

Let us revise the sine rule

In ΔABC:

  • \frac{AB}{sin(C)}=\frac{BC}{sin(A)}=\frac{AC}{sin(B)}
  • AB is opposite to ∠C
  • BC is opposite to ∠A
  • AC is opposite to ∠B

Let us use this rule to solve the problem

In ΔABC:

∵ m∠A = 45°

∵ m∠C = 30°

- The sum of measures of the interior angles of a triangle is 180°

∵ m∠A + m∠B + m∠C = 180

∴ 45 + m∠B + 30 = 180

- Add the like terms

∴ m∠B + 75 = 180

- Subtract 75 from both sides

∴ m∠B = 105°

∵ \frac{AB}{sin(C)}=\frac{BC}{sin(A)}

∵ AB = 5\sqrt{2}

- Substitute AB and the 3 angles in the rule above

∴ \frac{5\sqrt{2}}{sin(30)}=\frac{BC}{sin(45)}

- By using cross multiplication

∴ (BC) × sin(30) = 5\sqrt{2} × sin(45)

∵ sin(30) = 0.5 and sin(45) = \frac{1}{\sqrt{2}}

∴ 0.5 (BC) = 5

- Divide both sides by 0.5

∴ BC = 10 units

∵ \frac{AB}{sin(C)}=\frac{AC}{sin(B)}

- Substitute AB and the 3 angles in the rule above

∴ \frac{5\sqrt{2}}{sin(30)}=\frac{AC}{sin(105)}

- By using cross multiplication

∴ (AC) × sin(30) = 5\sqrt{2} × sin(105)

∵ sin(105) = \frac{\sqrt{6}+\sqrt{2}}{4}

∴ 0.5 (AC) = \frac{5+5\sqrt{3}}{2}

- Divide both sides by 0.5

∴ AC = 5+5\sqrt{3} units

BC is 10 units and AC is 5+5\sqrt{3} units

Learn more:

You can learn more about the sine rule in brainly.com/question/12985572

#LearnwithBrainly

6 0
3 years ago
What is the solution to the following equation?
lions [1.4K]

Answer:

D. x = 7

Step-by-step explanation:

NOTE : <u><em>there should be an equal sign somewhere in the given expression.</em></u>

suppose the equation is the following:

3(x-4)-5 = x-3

………………………………………………………

3(x - 4) - 5 = x - 3

⇔ 3x - 12 - 5 = x - 3

⇔ 3x - 17 = x - 3

⇔ 3x - 17 + 17= x - 3 + 17

⇔ 3x = x + 14

⇔ 2x = 14

⇔ x = 14/2

⇔ x = 7

8 0
2 years ago
One zero of x^3 - 4x = 0 is 0 what are the other zeros of the function
myrzilka [38]
So you have x^3 - 4x = 0. What you can do is pull out an x from both x^3 and - 4x so it looks like this:

x( {x}^{2} - 4) = 0

Then you can find a number that makes the part inside the parentheses turn into zero. For beginners, it may be easier to write it out seperately and solve for x.

{x}^{2} - 4 = 0

We need to solve for x, so the first step is to add 4 to both sides, so we get something like this:

{x}^{2} = 4

Then, we can square root both sides to get rid of the power on the x, so it looks like this:

x = \sqrt{4}

Now, every square root has two answers, a positive and a negative. If we look at the bottom example:

{2}^{2} = 4

{( - 2)}^{2} = 4

We can see that both -2 and 2 to the power of two will equal to 4.

So finally, we get:

x = - 2 \: and \: 2

These are the other 'Zero's for the original function. If you are not sure of what a 'Zero' is, it is where the function crosses over the x-axis on a graph.
5 0
2 years ago
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