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crimeas [40]
3 years ago
11

Please help, is there a number between 5.0007 and 5.03???

Mathematics
1 answer:
zheka24 [161]3 years ago
5 0

Yes, 5.02 for example.

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Please help.. I will give points
Daniel [21]

A. 8^9/2= around 11585.2375; (sqrt8)^9= around 11585.2375. So choice A shows a pair of equivalent expressions.

B. (3sqrt125)^9=1,953,125; 125^9/3=1,953,125. So choice B also shows a pair of equivalent expressions.

C. 12^2/7= around 2.03394; (sqrt12)^7= around 5,985.96759. So choice C does not show a pair of equivalent expressions.

D. 4^1/5=around 1.31951; (sqrt4)^5=32. So choice D also doesn't show a pair of equivalent expressions.

Your answers are A and B.

I hope this helps ;)

4 0
4 years ago
Read 2 more answers
Please help explain how to solve problem 16?
Anarel [89]
Here is the problem.....√(15x + 10) = 2x+3

to remove the square root, we do the opposite which is to square everything.

(√(15x + 10))² = (2x + 3)²      (the square negates the square root)
15x + 10 = (2x +3)(2x + 3)    (use the distributive property to continue)
15x + 10 = 4x² + 6x + 6x + 9  (combine like terms)
15x + 10 = 4x² + 12x + 9      (subtract 15x and 10 from each side)
-15x - 10          -15x - 10
0 = 4x² - 3x - 1                      (factor completely)
(x - 1) (4x + 1)                    (set each to equal 0)

x - 1 = 0                4x + 1 = 0
x = 1                      4x = -1
                               x = -1/4

place both into the equation to check for reasonableness...we see the negative number is not reasonable, but the x value of 1 is a solution.

answer is 1
4 0
3 years ago
The temperature went from -12°F to 18°F. How much did the temperature change?
Triss [41]

you have negative -12

turn this into a positive (12)

12 plus 18

= 30

8 0
3 years ago
Topic: The Quadratic Formula
Finger [1]

Answer:

Step-by-step explanation:

The quadratic formula for a equation of form

ax²+bx + c = 0 is

x= \frac{-b +- \sqrt{b^2-4ac} }{2a}

For the first equation,

x²+3x-4=0,

we can match that up with the form

ax²+bx + c = 0

to get that

ax² =  x²

divide both sides by x²

a=1

3x = bx

divide both sides by x

3 = b

-4 = c

. We can match this up because no constant multiplied by x could equal x² and no constant multiplied by another constant could equal x, so corresponding terms must match up.

Plugging our values into the equation, we get

x= \frac{-3 +- \sqrt{3^2-4(1)(-4)} }{2(1)} \\= \frac{-3+-\sqrt{25} }{2} \\ = \frac{-3+-5}{2} \\= -8/2 or 2/2\\=  -4 or 1

as our possible solutions

Plugging our values back into the equation, x²+3x-4=0, we see that both f(-4) and f(1) are equal to 0. Therefore, this has 2 real solutions.

Next, we have

x²+3x+4=0

Matching coefficients up, we can see that a = 1, b=3, and c=4. The quadratic equation is thus

x= \frac{-3 +- \sqrt{3^2-4(1)(4)} }{2(1)}\\= \frac{-3 +- \sqrt{9-16} }{2}\\= \frac{-3 +- \sqrt{-7} }{2}\\

Because √-7 is not a real number, this has no real solutions. However,

(-3 + √-7)/2 and (-3 - √-7)/2 are both possible complex solutions, so this has two complex solutions

Finally, for

4x² + 1= 4x,

we can start by subtracting 4x from both sides to maintain the desired form, resulting in

4x²-4x+1=0

Then, a=4, b=-4, and c=1, making our equation

x=\frac{-(-4) +- \sqrt{(-4)^2-4(4)(1)} }{2(4)} \\= \frac{4+-\sqrt{16-16} }{8} \\= \frac{4+-0}{8} \\= 1/2

Plugging 1/2 into 4x²+1=4x, this works as the only solution. This equation has one real solution

7 0
3 years ago
Please help!!!!!!!!!!!!!!
vfiekz [6]

Answer:

Step-by-step explanation:

A p - 4 is the answer to your question

8 0
4 years ago
Read 2 more answers
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