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liubo4ka [24]
2 years ago
8

I need help write expression in exponential from

Mathematics
1 answer:
podryga [215]2 years ago
5 0
3. 5x5x5. i am not sure if i am correct but i am assuming it is that.
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15) What is the measure, in degrees, of an angle that is equivalent to 3/360 of a circle
blsea [12.9K]

There are exactly 360° in a circle, so the angle measure would be 3/360 * 360 = 3°.

5 0
3 years ago
A circle has a diameter of 6 centimeters what is the approximate area of the circle​
KiRa [710]

Answer:

28.26 cm^{2}

Step-by-step explanation:

The radius is half of the diameter. 6/2 = 3, so the radius of this circle is 3 cm.

The formula for the area of a circle is: πr^{2}

Plugging 3 into the radius, we have:

3^{2} × π =

9 × π

We can round π to 3.14

9 × 3.14 = 28.26

7 0
2 years ago
Solve the following quadratics. State the FACTORS AND SOLUTIONS. 1. 2x^2 - 7x + 3 2. 3x^2 + 7x +2
tekilochka [14]

Answer:

1. x = 3, 1/2 (solutions); (x - 3)(2x - 1) (factors)

2. x = -1/3, -2 (solutions); (3x + 1)(x + 2) (factors)

Step-by-step explanation:

<u>1. 2x^2 - 7x + 3</u>

To solve problem 1, you will need to identify your a, b, and c values in this quadratic function.

Since this problem is in standard form, it will be easy to identify these values. The standard form of a quadratic function is ax^2 + bx + c.

The a value is 2, the b value is -7, and the c value is 3 if we use our standard form and see which numbers are plugged into it.

Since we know that

  • a = 2
  • b = -7
  • c = 3

we can use the quadratic formula: x = \frac{-b~\pm~\sqrt{b^2~-~4ac} }{2a}

Substitute the a, b, and c values into the quadratic formula: x=\frac{-(-7)\pm\sqrt{(-7)^2-4(2)(3)} }{2(2)}

Now simplify using the laws of pemdas: x=\frac{7\pm\sqrt{(49)-(24)} }{4}

Simplify even further: x=\frac{7\pm\sqrt{(25)} }{4} \rightarrow x=\frac{7\pm (5) }{4}

Now split this equation into two equations to solve for x: x=\frac{12 }{4} ~~and~~ x=\frac{2 }{4}

12/4 can be simplified to 3, and 2/4 can be simplified to 1/2.

This means your solutions to problem 1 is 3, 1/2.

\boxed {x=3,\frac{1}{2} }

There is also another way to solve for the quadratic functions, and this was by factoring.

If you factor 2x^2 - 7x + 3 using the bottoms-up method, you will get (x - 3)(2x - 1).

After factoring, solving for the solutions is simple because all you have to do is set each factor to 0.

  • x - 3 = 0
  • 2x - 1 = 0

After solving for x by adding 3 to both sides, or by adding 1 to both sides then dividing by 2, you will end up with the same solutions: x = 3 and x = 1/2.

<u>2. 3x^2 + 7x + 2</u>

To save time I'll be using the bottoms-up factoring method, but remember to refer back to problem 1 (quadratic formula) if you prefer that method.

Factor this quadratic function using the bottoms-up method. After factoring you will have (3x + 1)(x + 2). These are your factors.

Now to solve for x and find the solutions of the quadratic function, you will set both factors equal to 0.

  • 3x + 1 = 0
  • x + 2 = 0

Solve.

<u>First factor:</u> 3x + 1 = 0

Subtract 1 from both sides.

3x = -1

Divide both sides by 3.

x = -1/3

<u>Second factor:</u> x + 2 = 0

Subtract 2 from both sides.

x = -2

Your solutions are x = -1/3 and x = -2.

\boxed {x = -\frac{1}{3} , -2}

7 0
3 years ago
A twelve inch candle and an 18 inch candle are lit at 6 pm. The 12-in. candle burns 0.5 inches every hour. The 18 inch candle bu
Greeley [361]
 <span>18 - 2t = 12 - 0.5t 
6 = 1.5t 
t = 6/1.5 
t = 4 hours 
time required to have the same height = 4hours 
At what time ? 
answer : at 10pm 
------------------- 
12inch candle lasts longer ..</span>
3 0
3 years ago
Multiply.4 3/10× 2 1/3Enter your answer in the box as a mixed number in simplest form.
faust18 [17]

Answer:

vfvv j  mv fvkfvnfjvnsjvnfjsvndfjkv

Step-by-step explanation:

cc dscds j

4 0
2 years ago
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