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melamori03 [73]
2 years ago
7

Help needed ASAP will give you BRAINLIEST and 5 stars rate

Mathematics
1 answer:
iVinArrow [24]2 years ago
5 0
I believe it is context clues. The reader typically looks for context clues to understand the reading.
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What are the coordinates of the point that is of the way from A to B?
Leokris [45]

Answer:

The coordinates are 3,1

Step-by-step explanation:

i did the assignment and it js correct

8 0
1 year ago
Which of these is the algebraic expression for "3 less than the product of 4 and some number?"
svet-max [94.6K]

Answer:

4s - 3

hope this helps!

Step-by-step explanation:

7 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
Can someone please help me with this? Mhanifa? I will mark brainliest
den301095 [7]

Answer:

Option C 20

GOOD LUCK FOR THE FUTURE! :)

8 0
2 years ago
Find the value of x that will make A||B<br> A-<br> 2x + 50<br> 5x - 80<br> B-<br> x = [?]
LenKa [72]

Answer:

15

Step-by-step explanation:

If A||B then the sum of given angles must be equal to 180°

2x + 5 + 5x - 80 = 180 add like terms

7x - 75 = 180 subtract 75 from both sides

7x = 105 divide both sides by 7

x = 15

8 0
3 years ago
Read 2 more answers
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