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3 years ago

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Need help with this question

1 answer:

Sedbober [7]3 years ago

6 0

Answer:

B

Step-by-step explanation:

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Write each equation in standard form. identify A,B,C.

aleksley [76]

Write in y=ax²+bx+c form
solve for y

given
$\frac{x+5}{3}=-2y+4$
solve for y

easy
minus 4 both sides
$\frac{x+5}{3}-4=-2y$
times -1/2 to both sides
$\frac{x+5}{-6}+2=y$
$y=\frac{x+5}{-6}+2$
$y=\frac{-x}{6}-\frac{6}{5}+2$
2=10/5
$y=\frac{-1}{6}x-\frac{6}{5}+\frac{10}{5}$
$y=\frac{-1}{6}x+\frac{4}{5}$
y=ax²+bx+c
$y=0x^2+\frac{-1}{6}x+\frac{4}{5}$

a=0
b= $\frac{-1}{6}$
c= $\frac{4}{5}$

3 0

3 years ago

Help Please.<br><br> If 7m - 3 = 53. What is the value of 11m + 2?

Pachacha [2.7K]

Answer:

90.

Step-by-step explanation:

7m - 3 = 53

m would be 8 because 56 - 3 = 53.

11m + 2 equals 90 because 11 times 8 (8 is m) is 88, which added by two is 90.

4 0

3 years ago

F(x)=2x-3, solve when x=0,-1,2 <br> answers <br> A-3, -1,1 <br> B-3,-5,1 <br> C3,-1,1 <br> D-3,-1,7

guajiro [1.7K]

Answer:

b

Step-by-step explanation:

8 0

3 years ago

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

Please help me with this

Fittoniya [83]

160•.60=96
160-96=64
the discount is 96 and the amount after the discount is 64 i think

7 0

3 years ago