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White raven [17]
3 years ago
10

What inequality is on the number line?​

Mathematics
2 answers:
a_sh-v [17]3 years ago
4 0

Answer:

x is >_ -5

Step-by-step explanation:

greater than or equal to

mafiozo [28]3 years ago
3 0

Answer:

In algebra, the inequality will refer to a number, or range of numbers, which are either greater than, greater than or equal to, less than, or less than or equal to a fixed value. This can be shown on a number line using lines and circles. The line indicates the range of possible values. The circle is the end point of the line.

Step-by-step explanation:

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K(t)=10t−19 k=-7 k=-7=<br> Please help
igomit [66]

Answer:

t = 6/5

Step-by-step explanation:

Step 1: Define

k(t) = 10t - 19

k(t) = -7

Step 2: Substitute and Evaluate

-7 = 10t - 19

12 = 10t

t = 6/5

6 0
3 years ago
Next Activity
astra-53 [7]
700%10=70

70•25=1,750 Calories

Brainliest would be cool, hope this helped
7 0
3 years ago
In the derivation of Newton’s method, to determine the formula for xi+1, the function f(x) is approximated using a first-order T
dimaraw [331]

Answer:

Part A.

Let f(x) = 0;

suppose x= a+h

such that f(x) =f(a+h) = 0

By second order Taylor approximation, we get

f(a) + hf'±(a) + \frac{h^{2} }{2!}f''(a) = 0

h = \frac{-f'(a) }{f''(a)} ± \frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}

So, we get the succeeding equation for Newton's method as

x_{i+1} = x_{i} + \frac{1}{f''x_{i}}  [-f'(x_{i}) ± \sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ]

Part B.

It is evident that Newton's method fails in two cases, as:

1.  if f''(x) = 0

2. if f'(x)² is less than 2f(x)f''(x)    

Part C.

In case  x_{i+1} is close to x_{i}, the choice that shouldbe made instead of ± in part A is:

f'(x) = \sqrt{f'(x)^{2} - 2f(x)f''(x)}  ⇔ x_{i+1} = x_{i}

Part D.

As given x_{i+1} = x_{i} = h

or                 h = x_{i+1} - x_{i}

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

Also,             (x_{i+1}-x_{i})² = -(x_{i+1}-x_{i})(f(x_{i})/f'(x_{i}))

So,                f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0

It becomes   h = -f(a)/f'(a) + (h/2)[f''(a)f(a)/(f(a))²]

Also,             x_{i+1} = x_{i} -f(x_{i})/f'(x_{i}) + [(x_{i+1} - x_{i})f''(x_{i})f(x_{i})]/[2(f'(x_{i}))²]

6 0
3 years ago
3x-2y=-39 and x+3y=31 solution using elimination
Kobotan [32]
This is hard to write... I hope this makes sense to you...
3x-2y=-39
x+3y= 31
We want to use elimination therefor, we need to either get our x or our y to add together to get zero. 
To do this, we will multiply -3 to (x=3y=31) 
NEW EQUATION
3x-2y=-39 PLUS
-3x-9y= -93 Equals
Answer: -11y= -54 
y= 4.9
Plug in to solve for x (put new y in)
3x-2(4.9)=-39
x= -9.73 
5 0
3 years ago
Read 2 more answers
X^2+4x-1/(x+2)^2=0 solve for x
dalvyx [7]
<span>x^2+4x-1/(x+2)^2=0
x^2 + 4x - 1 = 0
</span>
\frac{x^2+10x-7}{x^2+10x+25} =0\\x^2+10x-7=0\\x= \frac{-b\pm\sqrt{b^2-4ac}}{2a};
where a = 1, b = 4 and c = -1

x=\frac{-4\pm\sqrt{4^2-(4\times1\times(-1)}}{2\times1}\\=\frac{-4\pm\sqrt{16+4}}{2}=\frac{-4\pm\sqrt{20}}{2}\\=\frac{-4\pm\sqrt{4\times5}}{2}=\frac{-4\pm2\sqrt{5}}{2}=-2\pm\sqrt{5}\\=-2+\sqrt{5}&#10; \ or \ -2-\sqrt{5}\\x=0.24 \ or \ x = -4.24
5 0
3 years ago
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