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3 years ago

Which of the following is equivalent to the expression below? 9^10*9^x

Mathematics

2 answers:

muminat3 years ago

6 0

Answer:

Step-by-step explanation:

$a^{m}*a^{n}=a^{m+n}\\\\9^{10}*9^{x}=9^{10+x}$

In exponent multiplication, if the bases are same then add the exponents.

liraira [26]3 years ago

6 0

$\small \sf \leadsto 9 {}^{10 +x}$

Step-by-step explanation:

Using exponent's product rule

$\small \sf \: x {}^{n} \times x {}^{m} = x {}^{n + m}$

$\small \sf \leadsto \: 9 {}^{10} \times 9 {}^{x}$

$\small \sf \leadsto 9 {}^{10 + x}$

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irinina [24]

Answer:

number 1 is 2 number 2 is 15 number 4 is 30

Step-by-step explanation:

6 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

The school cafeteria seats 250 students. 203 students are already seated. Write and sole an inequality that represents the addit

Nitella [24]

Answer:

An equation that represents the given situation is $203+x=250$ and the additional number of students that can be seated is 47

Step-by-step explanation:

Number of students already seated in school cafeteria = 203

Let x be the additional number of students that can be seated

So, Total number of seats available in school cafeteria = 203+x

We are given that the school cafeteria seats 250 students

So, 203+x=250

$\Rightarrow x=250-203$

$\Rightarrow x=47$

Hence an equation that represents the given situation is $203+x=250$ and the additional number of students that can be seated is 47

6 0

3 years ago

The value of x + y is negative, and x is a positive integer. What must be true about y? Explain.

sergey [27]

Answer:

Y must be negative

Step-by-step explanation:

Use this equation as an example: 5 + -7 = negative

If you add both of them you end up with -2

-2 is a negative integer

5 0

3 years ago

9. A line passes through (2, –1) and (8, 4). a. Write an equation for the line in point-slope form. b. Rewrite the equation in s

babunello [35]

Answer:

Step-by-step explanation:

(4+1)/(8-2)= 5/6

y + 1 = 5/6(x - 2)

y + 1 = 5/6x - 5/3

y + 3/3 = 5/6x - 5/3

y = 5/6x - 8/3

6(y = 5/6x - 8/3)

6y = 5x - 16

-5x + 6y = -16

8 0

3 years ago