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2 years ago

Pls help me rn bec sue tmr pls pls

Mathematics

1 answer:

blagie [28]2 years ago

3 0

Answer:

Number of 1-cm cube 24. Volume of solid 24 cm^2

Step-by-step explanation:

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the market bought grapes for $0.87 a pound and sold them for $1.15 a pound what is the percent of increase rounded to the neares

kherson [118]

((1.15 - 0.87)/0.87)*100% = 32.2%

3 0

2 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

Ricardo bought 3 pounds of red grapes, 4 pounds of green grapes, and 1 pound of black grapes. What is the ratio of green grapes

ivann1987 [24]

Answer:

4:3

Step-by-step explanation:

6 0

3 years ago

What is the solution to this equation?

Radda [10]

Answer:

9. This is because you have to add 13 to -4, which gives you positive 9. Hope this helped!

5 0

2 years ago

I need help right now. report cards are coming in soon and i need to get my grades up

saul85 [17]

Answer:

x = 10

Step-by-step explanation:

3x - 5 = 2x + 5

3x - 5 + 5 = 2x + 5 + 5

3x = 2x + 10

3x - 2x = 2x - 2x + 10

1x = 10

x = 10

4 0

2 years ago

Read 2 more answers