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2 years ago

First to answer gets brainliest and five stars

Mathematics

2 answers:

Dovator [93]2 years ago

8 0

Answer:

length a = 5

length b = 4

Step-by-step explanation:

puteri [66]2 years ago

3 0

Answer:

A=5 B=4
volume = 88

Step-by-step explanation:

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4-4x-9x how many terms

11Alexandr11 [23.1K]

Answer:

3 terms

Step-by-step explanation:

4 4x and 9x

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2 years ago

I need help on this the full answer

Elan Coil [88]

Answer:

t > 3

Step-by-step explanation:

1 hour = 194

582 / 194 = 3

She needs to ride for 3 hours to burn 582 calories

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2 years ago

What is the smallest value in the stem and leaf plot below?

Alexeev081 [22]

Listing the data from the Stem and Leaves table as raw data we have:

02, 03, 03, 07, 10, 10, 17, 34, 34, 35, 35

Hence the smallest value is 02 = 2

Answer: A

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2 years ago

Answer this for me pleasee i need help

Fiesta28 [93]

Answer:

x = 5 and x = -3

Step-by-step explanation:

hope it helps

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3 years ago

Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?

9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

$f(x)=\frac{(x-1)(x^2-4)}{x^2-a}$

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that $x=\sqrt{a}$ will annulate the denominator, i.e

$(\sqrt{a})^2-a = a-a = 0$ . But, if a = 1 we have:

$f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}$

so, the value $x=\sqrt{a} = \sqrt{1} = 1$ won't annulate the denominator.

Now, for a = 4 we have:

$f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1$

so, the value $x=\sqrt{a} = \sqrt{4} = 2$ won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0

2 years ago