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anygoal [31]
1 year ago
15

Fill in the blank to make the statement true: -18 ___ -19

Mathematics
1 answer:
LenaWriter [7]1 year ago
6 0

Answer:

3. >

Step-by-step explanation:

-18 is closer to zero than -19 is. So therefore, -18 is greater than -19.

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Which inequality is shown in this graph (0,2) (-1,-2)
vivado [14]

Answer:

C

Step-by-step explanation:

The linear inequality has the same properties of its equation as slope intercept does. This means y\leq mx+b where m is the slope and b is the y-intercept. The y-intercept here is 2 and it has a positive slope of 4. This means only C or D are options. Then test a point to determine in the shaded region to determine the sign.

(0,0) --->0\leq 4(0)+2 --->0\leq 2  is true so (0,0) is in the shaded portion. C is the solution.

5 0
3 years ago
Read 2 more answers
Identify the leading term and leading coefficient of the polynomial expression. −4x6+9x2−9x5−3 Leading Term: Answer Leading Coef
LenKa [72]

Answer:

yay

Step-by-step explanation:

na

8 0
3 years ago
Can you solve -4x+5=-2x-1
Lesechka [4]

Answer:

5

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
PLEASE HELP!!!!!!!!!
Irina-Kira [14]

Answer:

114

Step-by-step explanation:

The external angle measure is equal to the half of difference between the larger intercepted arc (5x-12) and the smaller intercepted arc (3x)

32 = (5x-12 -3x)/2

combine like terms in parenthesis and multiply both sides by 2

32·2 = 2x-12

multiply 32·2 = 64, and add 12 to both sides

64 + 12 = 2x

add 64+12 = 76, and divide both sides by 2 to isolate x

76/2 = x

38 = x

The measure of arc AB is 3x.

Substitute x for 38

Measure of arc AB = 3x = 3·38 = 114

7 0
3 years ago
Find the Taylor series for f(x)=sin(x) centered at c=π/2.sin(x)=∑ n=0 [infinity]On what interval is the expansion valid? Give yo
agasfer [191]

Answer:

sinx =1-\frac{1}{2} (x-\frac{\pi}{2} )^2+\frac{1}{24} (x-\frac{\pi}{2} )^4-\frac{1}{720} (x-\frac{\pi}{2} )^6+\frac{1}{40320} (x-\frac{\pi}{2} )^8+...

Step-by-step explanation:

given that f(x) = sin x

we have to find the Taylor series for that

f(x) = sin x   : f( = 1\\f'(x) = cos x :(f'\frac{\pi}{2})=0\\f"(x) = -sinx :f" (\frac{\pi}{2}) =-1\\f^4 (x) = -cosx : f^4 (\frac{\pi}{2}) =0

and so on.

i.e. 2nd, 4th, 6th terms would be 0

and also 1st, 5th, 9th terms would be positive for f value and 3rd, 9th,... would be negative

Using the above we can write Taylor series as

f(x) = f(a)+\frac{f'(a)}{1!} (x-\frac{\pi}{2}) +...+f^n(a) /n! (x- \frac{\pi}{2})^n+...

sinx =1-\frac{1}{2} (x-\frac{\pi}{2} )^2+\frac{1}{24} (x-\frac{\pi}{2} )^4-\frac{1}{720} (x-\frac{\pi}{2} )^6+\frac{1}{40320} (x-\frac{\pi}{2} )^8+...

This is valid for all real values of x.

x ∈(-\infty, infty)

3 0
3 years ago
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