F(x)=-3x+5x^2+8 has____

A natural number is an irrational number sometimes always never

Nady [450]

Answer:

always

Step-by-step explanation:

. Always. Real numbers are irrational numbers. ... Natural numbers are rational numbers.

8 0

4 years ago

Read 2 more answers

the surface area of a cuboid is 95cm² and its lateral surface area is 63cm². find the area of its base

wel

Answer:

The Area of the base is 16 cm² .

Step-by-step explanation:

Given as :

The surface area of the cuboid = x = 95 cm²

The lateral surface area of the cuboid = y = 63 cm²

Let The Area of the base = z cm²

Now, Let The length of cuboid = l cm

The breadth of cuboid = b cm

The height of cuboid = h cm

According to question

∵ The surface area of the cuboid = 2 ×(length × breadth + breadth × height + height × length)

Or, x = 2 ×(l × b + b × h + h × l)

Or, 95 = 2 ×(l × b + b × h + h × l)

Or, (l × b + b × h + h × l) = $\dfrac{95}{2}$ ....1

Similarly

∵lateral surface area of the cuboid = 2 ×(breadth × height + length × height)

Or, y = 2 ×(b × h + l × h)

Or, 2 ×(b × h + l × h) = 63

Or, (b × h + l × h) = $\dfrac{63}{2}$ ......2

Putting value of eq 2 into eq 1

so, (l × b + $\dfrac{63}{2}$ ) = $\dfrac{95}{2}$

Or, l × b = $\dfrac{95}{2}$ - $\dfrac{63}{2}$

Or, l × b = $\dfrac{95 - 63}{2}$

i.e l × b = $\dfrac{32}{2}$

so, l × b = 16

Now, Again

∵The Area of the base = ( length × breadth ) cm²

So, z = l × b

i.e z = 16 cm²

So, The Area of the base = z = 16 cm²

Hence,The Area of the base is 16 cm² . Answer

5 0

3 years ago

Which graph represents the solution to this inequality? 16x-80x<37+27

marin [14]

Answer:

A

Step-by-step explanation:

Hope this helps

8 0

3 years ago

A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient

ankoles [38]

Answer:

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

Solution:

We need to show that the gradient of the curve at A is 1

Here given that ,

$y=(x-a) \sqrt{(x-b)}$ --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

$0=(b+1-a) \sqrt{(b+1-b)}$

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

$y^{\prime}=u^{\prime} y+y^{\prime} u$

so, we get

${u}^{\prime}=1$

$v^{\prime}=\frac{1}{2} \sqrt{(x-b)}$

$y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}$

By putting value of point A and putting value of eq 2 we get

$y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}$

$y^{\prime}=\frac{d y}{d x}=1$

Hence proved that the gradient of the curve at A is 1.

7 0

3 years ago

In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien

diamong [38]

Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = $\frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}$

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2 = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) = $\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)}$ = $\frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)}$ = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

4 0

3 years ago

F(x)=-3x+5x^2+8 has_____ roots