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3 years ago

Find the absolute value of negative 15. please help :)

Mathematics

2 answers:

sammy [17]3 years ago

8 0

Answer:

the answer is 15.

Step-by-step explanation:

SVEN [57.7K]3 years ago

8 0

Answer:

Step-by-step explanation: because the absolute value of any number is always positive. like the absolute value of 5 is 5

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Help pls all shaded parts need answers

pickupchik [31]

Answer:

x,x,12,12 ( for the paragraph part)

Step-by-step explanation:

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3 years ago

What is the interquartile range of the data set? 5,6,7,8,9,10,11

sasho [114]

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3 0

3 years ago

Evaluate the function.

Natali5045456 [20]

As you may already be familiar, these functions f(x) and g(x) are piecewise. They consist of multiple functions with different domains.

1. For #1, the given input is f(0). Since 0≤1, you should use the first equation to solve. f(0)=3(0)-1 ➞ f(0)=-1
2. Continue to evaluate the given input for the domains given. 1≤1, therefore f(1)=3(1)-1➞f(1)=2
3. 5>1, therefore f(5)=1-2(5)➞f(5)=-9
4. -4≤1; f(-4)=3(-4)-1➞f(-4)=-13
5. -3<0<1; g(0)=2
6. -3≤-3; g(-3)=3(-3)-1➞g(-3)=-10
7. 1≥1; g(1)=-3(1)➞g(1)=-3
8. 3≥1; g(3)=-3(3)➞g(3)=-9
9. -5≤-3; g(-5)=3(-5)-1➞g(-5)=-16

Hope this helps! Good luck!

5 0

3 years ago

Read 2 more answers

Identify the functions that are continuous on the set of real numbers and arrange them in ascending order of their limits as x t

Studentka2010 [4]

Answer:

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

Step-by-step explanation:

1. $f(x)=\frac{x^2+x-20}{x^2+4}$

The denominator of f is defined for all real values of x

Therefore, the function is continuous on the set of real numbers

$\lim_{x\rightarrow 5}\frac{x^2+x-20}{x^2+4}=\frac{25+5-20}{25+4}=\frac{10}{29}=0.345$

3. $h(x)=\frac{3x-5}{x^2-5x+7}$

$x^2-5x+7=0$

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function h is defined for all real values.

$\lim_{x\rightarrow 5}\frac{3x-5}{x^2-5x+7}=\frac{15-5}{25-25+7}=\frac{10}{7}=1.43$

2. $g(x)=\frac{x-17}{x^2+75}$

The denominator of g is defined for all real values of x.

Therefore, the function g is continuous on the set of real numbers

$\lim_{x\rightarrow 5}\frac{x-17}{x^2+75}=\frac{5-17}{25+75}=\frac{-12}{100}=-0.12$

4. $i(x)=\frac{x^2-9}{x-9}$

x-9=0

x=9

The function i is not defined for x=9

Therefore, the function i is not continuous on the set of real numbers.

5. $j(x)=\frac{4x^2-7x-65}{x^2+10}$

The denominator of j is defined for all real values of x.

Therefore, the function j is continuous on the set of real numbers.

$\lim_{x\rightarrow 5}\frac{4x^2-7x-65}{x^2+10}=\frac{100-35-65}{25+10}=0$

6. $k(x)=\frac{x+1}{x^2+x+29}$

$x^2+x+29=0$

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function k is defined for all real values.

$\lim_{x\rightarrow 5}\frac{x+1}{x^2+x+29}=\frac{5+1}{25+5+29}=\frac{6}{59}=0.102$

7. $l(x)=\frac{5x-1}{x^2-9x+8}$

$x^2-9x+8=0$

$x^2-8x-x+8=0$

$x(x-8)-1(x-8)=0$

$(x-8)(x-1)=0$

$x=8,1$

The function is not defined for x=8 and x=1

Hence, function l is not defined for all real values.

8. $m(x)=\frac{x^2+5x-24}{x^2+11}$

The denominator of m is defined for all real values of x.

Therefore, the function m is continuous on the set of real numbers.

$\lim_{x\rightarrow 5}\frac{x^2+5x-24}{x^2+11}=\frac{25+25-24}{25+11}=\frac{26}{36}=\frac{13}{18}=0.722$

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

6 0

3 years ago

Round 5049 correct to 1 significant figure

Step2247 [10]

5000

Addition (+) and subtraction (-) round by the least number of decimals.
Multiplication (* or ×) and division (/ or ÷) round by the least number of significant figures.
Logarithm (log, ln) uses the input's number of significant figures as the result's number of decimals.
Antilogarithm (n^x.y) uses the power's number of decimals (mantissa) as the result's number of significant figures.
Exponentiation (n^x) only rounds by the significant figures in the base.
To count trailing zeros, add a decimal point at the end (e.g. 1000.) or use scientific notation (e.g. 1.000 × 10^3 or 1.000e3).
Zeros have all their digits counted as significant (e.g. 0 = 1, 0.00 = 3).
Rounds when required, after parentheses, and on the final step.

- BRAINLIEST answerer ❤️

7 0

2 years ago

Read 2 more answers