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3 years ago

Determinate the domain of the following function

Mathematics

1 answer:

velikii [3]3 years ago

4 0

The domain of this function is any x-value that doesn’t cause the denominator to equal zero and cause the function to be indeterminate.

In this case, x cannot be zero, and x cannot be the square root of 5.

Brainliest always accepted! :)

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What is the total of the following lengths 1.6m + 2.87M+4.04m

grandymaker [24]

The answer would be 8.51m if you add all together

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3 years ago

Read 2 more answers

Question #2

Elanso [62]

Do you have anything

8 0

3 years ago

Could someone please help mee Step by step

Keith_Richards [23]

Let Q be the point (a, b). Since the midpoint of P (2, 3) and Q (a, b) is (4, 6), we have

(2 + a)/2 = 4

(3 + b)/2 = 6

Solve for a and b :

(2 + a)/2 = 4

2 + a = 8

a = 6

(3 + b)/2 = 6

3 + b = 12

b = 9

So the coordinates of Q are (6, 9).

8 0

3 years ago

One packet of biscuits requires 2 15/16 and 1 7/8 cups of sugar. Estimated total quantity of both ingredients used in 10 such pa

frez [133]

Answer:

Hey Mate! Here's your answer.

Step-by-step explanation:

One packet of biscuits requires 2½ cups of flour = \frac{5}{2}25

One packet of biscuits requires 1 \times \frac{2}{3}1×32

One packet of biscuits requires 1 \times \frac{2}{3}1×32cups of sugar. = \frac{5}{3}35

Total ingradiants for one packet of biscuits =

Total ingradiants for one packet of biscuits =( \frac{5}{2} + \frac{5}{3} ) = \frac{15 + 10}{6} = \frac{25}{6}(25+35)=615+10=625

Then, total quantity of both ingredients used in 10 such packets of biscuits =

Then, total quantity of both ingredients used in 10 such packets of biscuits =10 \times ( \frac{25}{6} ) = 5 \times ( \frac{25}{3} ) = \frac{125}{3} = 41 \times \frac{2}{3}10×(625)=5×(325)=3125=41×32

Hope it helped you!

7 0

2 years ago

I need help fast!! -14x^-2y^3/8x^-5y^5

bija089 [108]

-14x^-2y^3/8x^-5y^5

= -14/8 x^(-2+5) y^(3-5)

= -7/4 x^3 y ^-2

= -7x^3 / 4y^2

Answer

-7x^3

------------

4y^2

6 0

3 years ago