All categories

3 years ago

Which of the following statements is not true?

1 answer:

7 0

In order for us to know which statements are true, we need to go over the choices.

A. True
0.6 = 3/5
0.6 = 3/5
60 / 100 = 3 / 5

B. True
0.125 = 1/8
0.125 = 125/1000 = 1/8

C. FALSE.
2.015 = 2 15/1000 = 2 3/200

The statement that is false is Choice <span>C. 2.015 = 2 1/100</span>

You might be interested in

ONLY ANSWER IF YOU HAVE THE ANSWER P L E A S E

Nadusha1986 [10]

Answer:

Linear: greater than 2011

Quadratic: lesser than 2011

Cubic: lesser than 2011

Step-by-step explanation:

At x = 2

Linear regression equation:

y = -6777.54(2) + 110001.04

y = 96445.96

More than 2011

Quadratic: y = -251.10(2)² - 2760.02(2) + 98618.06

y = 92093.62

lesser than 2011

Cubic: y = -6.23(2)² - 101.67(2)² - 3747.49(2) + 100142.23

y = 92190.73

lesser than 2011

3 0

3 years ago

which statement about the points (2, 5) and (-2, 5) is true plot the points on a coordinate plane to help you answer the questio

iVinArrow [24]

Answer:

A, (2,5) and (-2,5) are reflections of each other on the x axis

5 0

2 years ago

Jeremy makes a drawing of a walkway shaped like three rectangles. The total area of the walkway is 24 square meters. What is the

nadya68 [22]

Answer:

Step-by-step explanation:

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

Problem 3

Murljashka [212]

Answer:

$-\frac{2}{5}$

Step-by-step explanation:

Given expression:

2x + 5y = -10

The equation of a straight line is;

y = mx + c

y and x are the coordinates

m is the slope

c is the intercept

Now;

let us write the given expression in slope intercept format;

2x + 5y = -10

5y = -2x - 10

y = $-\frac{2x}{5}$ - 2

So, the slope of the line is $-\frac{2}{5}$

8 0

3 years ago