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3 years ago

Answer this question plzzzz ASAPPPPP!!!!!!!!

Mathematics

2 answers:

NISA [10]3 years ago

6 0

Its the bottom one...

notka56 [123]3 years ago

5 0

The last answer...............

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Solve for x: x³=23 Choose all that apply: ∛23 3 1 23/3

rosijanka [135]

Your answer is A. I have attached a photo of an explanation

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2 years ago

Read 2 more answers

If Richard can paint their living room in 4 hours, and Vanessa can paint the same living room in 5 hours, then how long will it

Tju [1.3M]

Answer:

2 hrs 13 mins

Step-by-step explanation:

If Richard can paint the living room in 4 hrs, then in 1 hour he can paint 1/4 of the living .

If Vanessa can paint the living room in 5 hours, then in one hour, she can paint 1/5 of the living room.

In one hour Richard and Vanessa can paint 1/4 + 1/5 = 1/y

y is the number of hours it will take both Richard and Vanessa to paint the living room. 1/y is how much of the house they can paint in one hour.

1/4 + 1/5 = 1/y

(5 + 4)/20. ( find the LCM)

9/20 = 1/y

9y= 20

y = 20/9

y = 2.22hrs

y = 2 hrs 13 mins

4 0

3 years ago

The class has collected $75.50 for a pizza party. A large pizza costs $9.50. Which one-step-inequality could be used to find the

Veseljchak [2.6K]

Answer:

7 large pizzas

Step-by-step explanation:

multiply 9.50 by every number on the calculator and get 9.50×7=66.5

or just divide 75.50 by 9.50

or idk

4 0

3 years ago

On tax-free weekend, Ben buys school supplies totaling $47.50. He has a sale coupon for 15% off his entire purchase. What will B

Thepotemich [5.8K]

47.50 x .15= 7.125
47.50-7.125=40.375

3 0

2 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

3 years ago