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2 years ago

9

given the functions f(x) = 3x2, g(x) = x2 − 4x + 5, and h(x) = –2x2 + 4x + 1, rank them from least to greatest based on their ax

is of symmetry.

1 answer:

Arturiano [62]2 years ago

8 0

The answer is <span>f(x), g(x), h(x)
(well, that's my opinion)
:) I hope this helped u :)</span>

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Does anyone know how to solve this?

KATRIN_1 [288]

Answer:

<30,-20>

Step-by-step explanation:

4u + 2v

4< 6,-4> + 2< 3,-2>

<24,-16> + < 6,-4>

<30,-20>

4 0

1 year ago

(a) Find the remainder when 15! is divided by 17. (b) Find the remainder when 2(26!) is divided by 29, Thatermine whether 17 is

dlinn [17]

Answer:

Step-by-step explanation:

a) We have 15! as the product of 1 to 15 natural numbers. Since 17 is prime there will be no factor common to these

By actual division we find

15! (mod 17) =16

From this we deduce

even 16! mod 17 = 16 = -1

According to Wilson theorem

(17-1)! = -1 mod 17

Thus verified 17 is prime

Hence 15! (mod 17) =-1=16

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

b) 2(26!) is divided by 29

Since 29 is prime

(29-1)! = -1 mod 29

28! = -1 mod 29 = 28

When divided this gives 25 as remainder

4 0

2 years ago

The volume of a right circular cone with radius r and height h is V = pir^2h/3.

Scorpion4ik [409]

The question is incomplete. The complete question is :

The volume of a right circular cone with radius r and height h is V = pir^2h/3. a. Approximate the change in the volume of the cone when the radius changes from r = 5.9 to r = 6.8 and the height changes from h = 4.00 to h = 3.96.

b. Approximate the change in the volume of the cone when the radius changes from r = 6.47 to r = 6.45 and the height changes from h = 10.0 to h = 9.92.

a. The approximate change in volume is dV = _______. (Type an integer or decimal rounded to two decimal places as needed.)

b. The approximate change in volume is dV = ___________ (Type an integer or decimal rounded to two decimal places as needed.)

Solution :

Given :

The volume of the right circular cone with a radius r and height h is

$V=\frac{1}{3} \pi r^2 h$

$dV = d\left(\frac{1}{3} \pi r^2 h\right)$

$dV = \frac{1}{3} \pi h \times d(r^2)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

a). The radius is changed from r = 5.9 to r = 6.8 and the height is changed from h = 4 to h = 3.96

So, r = 5.9 and dr = 6.8 - 5.9 = 0.9

h = 4 and dh = 3.96 - 4 = -0.04

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (5.9)(4)(0.9)+\frac{1}{3} \pi (5.9)^2 (-0.04)$

$dV=44.484951 - 1.458117$

$dV=43.03$

Therefore, the approximate change in volume is dV = 43.03 cubic units.

b). The radius is changed from r = 6.47 to r = 6.45 and the height is changed from h = 10 to h = 9.92

So, r = 6.47 and dr = 6.45 - 6.47 = -0.02

h = 10 and dh = 9.92 - 10 = -0.08

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (6.47)(10)(-0.02)+\frac{1}{3} \pi (6.47)^2 (-0.08)$

$dV=-2.710147-3.506930$

$dV= -6.22$

Hence, the approximate change in volume is dV = -6.22 cubic units

8 0

2 years ago

Which data set could be represented by the box plot shown below?

DochEvi [55]

Answer:

Step-by-step explanation:22

8 0

2 years ago

Read 2 more answers

Zigmanuir [339]

Gradient of the line formed;
m= $\frac{change in y}{change in x}$
m= $\frac{-2-1}{6--5}$
m= $\frac{-3}{11}$
y=mx+c
y= $\frac{-3}{11} x+c$
Replacing for x and y;
1=-3/11*(-5)+c
1=15/11+c
c=1-15/11
c=-4/11
y= $\frac{-3}{11} x- \frac{4}{11}$

4 0

2 years ago