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taurus [48]
3 years ago
7

Write the number 345000 into other forms

Mathematics
2 answers:
Alona [7]3 years ago
5 0
300,000 + 40,000 + 5,000

Three hundred and fourty-five thousand.

Hope this helped ;)
Tomtit [17]3 years ago
4 0
345,000

300,000+40,000+5,000

<span>Three hundred forty-five thousand

Hope I helped!!</span>
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What is the solution to y-12=20
vredina [299]

Hey there!

y-12=20

Add 12 to both sides. (do the inverse of subtraction, which is addition)

y=32

Check answer.

32-12=20 <=== checks out :)

I hope this helps!

~kaikers

4 0
3 years ago
Read 2 more answers
Find the exact value of cos 157.5°.
Nonamiya [84]
Cos 157.5º=-cos (180º-157.5º)=-cos 22.5=-cos(45/2)

cos Ф/2=⁺₋√((1+cosФ)/2).

In this case 157.5º is in the second quadrant, therefore we use the following equation:
cos Ф/2=-√((1+cosФ)/2).          (we will have a negative number)

cos 157.5º=-cos (45/2)=-√((1+cos 45º)/2)
=-√((1+√2/2)/2)
=-√((2+√2)/4)
=-√(2+√2) / 2   (≈-0.92387...)

Answer: cos 157.5º= -√(2+√2) / 2

3 0
3 years ago
Work out 6z - 4z .what is the answer?
Reika [66]

Answer:

4 is the answer

Step-by-step explanation:

you can - or + the equation only then when the letter is same z and z both are same so you can - them and get the answer.

4 0
2 years ago
What is the domain of the function y=vx<br> 2 0 03X&lt;&lt;<br> 1
Tema [17]

Answer:

<u>Option 3:  0 ≤ x ≤ ∞ </u>

Step-by-step explanation:

The domain of a function is the set of all possible x values for the function.

The given function is y = √x

The domain for the square root function should be : x ≥ 0

So, the domain = [0,∞)

Or, it can be written as inequality

So, 0 ≤ x ≤ ∞

So, the answer is option 3.

6 0
3 years ago
f(x) = 2<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align="absmiddle" class="latex
loris [4]

Answer:

No answer is possible

Step-by-step explanation:

First, we can identify what the parabola looks like.

A parabola of form ax²+bx+c opens upward if a > 0 and downward if a < 0. The a is what the x² is multiplied by, and in this case, it is positive 2. Therefore, this parabola opens upward.

Next, the vertex of a parabola is equal to -b/(2a). Here, b (what x is multiplied by) is 1 and a =2, so -b/(2a) = -1/4 = -0.25.

This means that the parabola opens upward, and is going down until it reaches the vertex of x=-0.25 and up after that point. Graphing the function confirms this.

Given these, we can then solve for when the endpoints of the interval are reached and go from there.

The first endpoint in -2 ≤ f(x) ≤ 16 is f(x) = 2. Therefore, we can solve for f(x)=-2 by saying

2x²+x-4 = -2

add 2 to both sides to put everything on one side into a quadratic formula

2x²+x-2 = 0

To factor this, we first can identify, in ax²+bx+c, that a=2, b=1, and c=-2. We must find two values that add up to b=1 and multiply to c*a = -2  * 2 = -4. As (2,-2), (4,-1), and (-1,4) are the only integer values that multiply to -4, this will not work. We must apply the quadratic formula, so

x= (-b ± √(b²-4ac))/(2a)

x = (-1 ± √(1-(-4*2*2)))/(2*2)

= (-1 ± √(1+16))/4

= (-1 ± √17) / 4

when f(x) = -2

Next, we can solve for when f(x) = 16

2x²+x-4 = 16

subtract 16 from both sides to make this a quadratic equation

2x²+x-20 = 0

To factor, we must find two values that multiply to -40 and add up to 1. Nothing seems to work here in terms of whole numbers, so we can apply the quadratic formula, so

x = (-1 ± √(1-(-20*2*4)))/(2*2)

= (-1 ± √(1+160))/4

= (-1 ± √161)/4

Our two values of f(x) = -2 are (-1 ± √17) / 4 and our two values of f(x) = 16 are (-1 ± √161)/4 . Our vertex is at x=-0.25, so all values less than that are going down and all values greater than that are going up. We can notice that

(-1 - √17)/4 ≈ -1.3 and (-1-√161)/4 ≈ -3.4 are less than that value, while (-1+√17)/4 ≈ 0.8 and (-1+√161)/4 ≈ 2.9 are greater than that value. This means that when −2 ≤ f(x) ≤ 16 , we have two ranges -- from -3.4 to -1.3 and from 0.8 to 2.9 . Between -1.3 and 0.8, the function goes down then up, with all values less than f(x)=-2. Below -3.4 and above 2.9, all values are greater than f(x) = 16. One thing we can notice is that both ranges have a difference of approximately 2.1 between its high and low x values. The question asks for a value of a where a ≤ x ≤ a+3. As the difference between the high and low values are only 2.1, it would be impossible to have a range of greater than that.

7 0
2 years ago
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