HELP. Need help on this

Answer:

y"= 2 wich is positive

Step-by-step explanation:

Step-by-step explanation:

Our equation is: y=(x-3)²

x should be replaced by x-4

y=(x-3)²

y=[(x-4)-3]²

y=(x-4-3)²

y=(x-7)²

The graph is still a parabola but with a different vertex

The vertex here is :

y= (x-7)²

y= x²-14x-49

y'= 2x-14

solve y'=0

2x-14=0

2x=14

x=7

You can easily find it without derivating by dividing -14 by -2

since: x²-14x-49

a=1 b= -14  c=-49

-b/2a = 14/2 = 7

the image of 7 is:

y=(7-7)² = 0

so the coordinates of the new vertex are (7,0) and it's a maximum

since y">0

y'= 2x-14

y"= 2 wich is positive

Answer:

Step-by-step explanation:

It's 5/7

You get that by having a calculator that does that. If you don't then the way to do it is multiply the numerator and denominator by 1.25

4 * 1.25 = 5

5.6 * 1.25 = 7

Answer:

gang nem

Step-by-step explanation:

Answer:

64

Step-by-step explanation:

The Ratio of Blue to White drops is 2:8

16*4=64,... 16:64 Have a nice day!

Answer:

a. 24.12 ft³/hr b. 0.0768 ft/hr

Step-by-step explanation:

a. Find the rate at which the volume of water in the pool is increasing at time t=6 hours.

The net rate of change of volume of the cylinder dV/dt = volume flow rate in - volume flow rate out

Since volume flow rate in = P(t) and volume flow rate out = R(t),

dV/dt = P(t) - R(t)

[tex]\frac{dV}{dt} = P(t) - 18e^{0.04t}[/tex]

We need to find the rate of change of volume when t = 6.

From the table when t = 6, P(6) = 47 ft³/hr

Also, substituting t = 6 into R(t), we have R(6)

[tex]\frac{dV}{dt} = P(t) - 18e^{0.04t}[/tex]

[tex]\frac{dV}{dt} = 47 - 18e^{0.04X6}\\\frac{dV}{dt} = 47 - 18e^{0.24}\\\frac{dV}{dt} = 47 - 18 X 1.27125\\\frac{dV}{dt} = 47 - 22.882\\\frac{dV}{dt} = 24.118 ft^{3}/hr[/tex]

dV/dt ≅ 24.12 ft³/hr

So, the rate at which the water level in the pool is increasing at t = 6 hours is 24.12 ft³/hr

b. How fast is the water level in the pool rising at t=6 hours?

Since the a rate at which the water level is rising is dV/dt and the volume of the cylinder is V = πr²h where r = radius of cylinder = 10 ft and h = height of cylinder = 5 feet

dV/dt = d(πr²h)/dt = πr²dh/dt since the radius is constant and dh/dt is the rate at which the water level is rising.

So, dV/dt = πr²dh/dt

dh/dt = dV/dt ÷ πr²

Since dV/dt = 24.12 ft³/hr and r = 10 ft,

Substituting the values of the variables into the equation, we have that

dh/dt = dV/dt ÷ πr²

dh/dt = 24.12 ft³/hr ÷ π(10 ft)²

dh/dt = 24.12 ft³/hr ÷ 100π ft²

dh/dt = 0.2412 ft³/hr ÷ π ft²

dh/dt = 0.2412 ft³/hr

dh/dt = 0.0768 ft/hr

So, the arate at which the water level is rising at t = 6 hours is 0.0768 ft/hr

Answer:

375 guests

Step-by-step explanation:

costs: 1500 + 8g

income: 12g

They must be equal

1500+8g = 12g

Subtract 8g from each side

1500 +8g-8g =12g-8g

1500= 4g

Divide by 4

1500/4 = 4g/4

375 = g

Answer:

i will say methods try yourself:

How to Determine Maximum Value

How to Determine Maximum ValueIf your equation is in the form ax2 + bx + c, you can find the maximum by using the equation:

How to Determine Maximum ValueIf your equation is in the form ax2 + bx + c, you can find the maximum by using the equation:max = c - (b2 / 4a).

How to Determine Maximum ValueIf your equation is in the form ax2 + bx + c, you can find the maximum by using the equation:max = c - (b2 / 4a).The first step is to determine whether your equation gives a maximum or minimum. ...

How to Determine Maximum ValueIf your equation is in the form ax2 + bx + c, you can find the maximum by using the equation:max = c - (b2 / 4a).The first step is to determine whether your equation gives a maximum or minimum. ...-x2 + 4x - 2.

How to Determine Maximum ValueIf your equation is in the form ax2 + bx + c, you can find the maximum by using the equation:max = c - (b2 / 4a).The first step is to determine whether your equation gives a maximum or minimum. ...-x2 + 4x - 2.Since the term with the x2 is negative, you know there will be a maximum point.

.

.

.

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plzz follow and make brainlist

Answer:

a

Step-by-step explanation:

Lateral faces that are parallel is not a property of a regular pyramid. The correct option is A.

Any pyramid whose base is a regular polygon and whose lateral edges are all of the same lengths is said to be regular.

The properties of a regular pyramid are:-

Lateral faces that are congruent isosceles triangles. Lateral edges that are congruent and the volume of the pyramid is equal to one-third the product of the area of its base and its altitude.

Hence, option A is correct.

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Answer:

Sorry I dont really understand wish I could help:(

Step-by-step explanation:

Answer:

[tex]\sqrt{10}[/tex]

[tex]\sqrt{5 } ^{2} + \sqrt{5 } ^{2} = x^{2}[/tex]

[tex]x^{2} =10[/tex]

Step-by-step explanation:

Answer:

A.  243

Step-by-step explanation:

[tex] 81^\frac{5}{4} = (3^4)^\frac{5}{4} = 3^{4 \times \frac{5}{4}} = 3^5 = 243 [/tex]

Answer: A

I just simplified it to 3^5, and that is also 243.

Area of parallelogram = b × h

Base = x + 7

Height = x + 3

ATQ

Area = (x + 7) ( x + 3)

x² + 3x + 7x + 21

x² + 10x + 21

Answered by Gauthmath must click thanks and mark brainliest

Answer:

strong and fexible .

variable

Answer:

no solution

Step-by-step explanation:

a negative number cannot be square rooted

Answer:

"not possible". no such thing as a negative squared number

Answer:

Answer:f(2)=-3(2)^2+2*2+11

Answer:f(2)=-3(2)^2+2*2+11 =-3*4+4+11

Answer:f(2)=-3(2)^2+2*2+11 =-3*4+4+11 =-12+4+11

Answer:f(2)=-3(2)^2+2*2+11 =-3*4+4+11 =-12+4+11 =3

hence, f(2)=3

9514 1404 393

Answer:

  KM = 8

Step-by-step explanation:

The ratio of long side to short side is the same for all of the triangles in this geometry:

  KM/ML = JM/KM

  KM² = ML·JM = 4(20-4) = 64

  KM = √64 = 8

The length of KM is 8 units.

Answer:

Tom, 10.25

Step-by-step explanation:

Distance covered by Tom=4.5*(4 1/2+5)=81/4=42.75km

Distance covered by Jerry=6.5*(5)=32.5km

Tom is ahead from Jerry by 10.25km

9514 1404 393

Answer:

  E.  384 cu ft

Step-by-step explanation:

The sum of length and width is half the perimeter, so the width is ...

  (40/2) -12 = 8 . . . feet

The depth is half that, so is 8/2 = 4 feet.

The volume is ...

  V = LWH = (12 ft)(8 ft)(4 ft) = 384 ft³

9514 1404 393

Answer:

  10 and 24

Step-by-step explanation:

We know that some of the Pythagorean triples that appear in math problems are (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17).

These have (perimeter, area) values of (12, 6), (30, 30), (56, 84), (40, 60).

For some scale factor n, we want (p·n, a·n²) = (60, 120). Of the triangles listed, we see that the (5, 12, 13) triangle scaled by n=2 will satisfy the problem requirements. (30·2, 30·2²) = (60, 120)

The side lengths are 10 and 24.

__

Check

For the side lengths we found, the perimeter is 10+24+26 = 60; the area is 1/2(10)(24) = 120. The hypotenuse is 2×13 = 26 = 24+2.

__

In the attached, one side is x, the other is y. The hypotenuse is (x+2). The square root equation comes from ...

  x² +y² = (x+2)²   ⇒   y² = (x² +4x +4) -x²   ⇒   y² = 4x +4 = 4(x +1)

_____

Additional comment

The graph shows the solution of the various constraints. At least, the combination of constraints will give a quadratic equation in x. They can be combined in a way that gives a cubic equation in x. Either way, we prefer the graphical or "guess and check" approach (above) as being easier to do.

Using the third equation in the attachment to write an expression for y, we have ...

  y = 58 -2x

Substituting that into the second equation gives ...

  (x(58 -2x)/2 = 120

  29x -x² = 120

  x² -29x +120 = 0

  (x -5)(x -24) = 0 . . . . x = 5 or 24.

The root x=5 is a legitimate solution to the pair of equations we chose to solve. The line y=58-2x intersects the hyperbola xy/2 = 120 in two places. However, (x, y) = (5, 48) does not satisfy the hypotenuse requirement that x+2 > y.

There are 50 different possible debating teams that could be selected as obtained using COMBINATION.

Since there are EQUAL number of juniors and seniors ;

Then we have 5 of each.

Here, the order of arrangement DOES NOT matter, Hence, we use COMBINATION

since the team MUST contain 4 SENIORS and 2 JUNIORS

4 Seniors from 5  = 5C4 = 5

2 Juniors from 5 = 5C2 = 10

Hence, (5C4 * 5C2) = 5 * 10 = 50

Hence, there are 50 different possible debating teams that could be selected.

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Answer:

too complex:<

Step-by-step explanation:

120cm+120cm=240cm (2 squirrels + 2 air spaces)

90cm+90cm=180cm (2 rats + 2 air spaces)

240cm-180cm=(2 squirrels + 2 air spaces) - (2 rats + 2 air spaces)

                       =2 squirrels + 2 air spaces - 2 rats - 2 air spaces

                       =2 squirrels - 2 rats

                       =60cm

1 squirrel - 1 rat = 60cm divided by 2

                         = 30cm

120cm + 90cm = squirrel + air space + rat + air space

                        = 210cm

I've no idea!! This qn is too challenging!!

But i hope the above workings might help you in a way or another:>

The table is 105cm tall.

Two or more expressions with an Equal sign is called as Equation.

Let the table is x

Squirrel is y

Rat is z

From 1st diagram

x+y-z=120...(1)

From 2nd diagram

x+z-y=90...(2)

Add 1 and 2

x+y-z+x+z-y=120+90

2x=210

Divide both sides by 2

x=105

Hence, the table is 105cm tall.

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9514 1404 393

Answer:

  a. 5/10

  b. 2.5/10

  c. 1/10

  d. 4/10

Step-by-step explanation:

To find the numerator, multiply each fraction by 10.

  a. (1/2)(10) = 5, so 1/2 = 5/10

  b. (1/4)(10) = 2.5, so 1/4 = (2.5)/10 or (5/2)/10

  c. (3/30)(10) = 1, so 3/30 = 1/10

  d. (12/30)(10) = 4, so 12/30 = 4/10

Answer:

-1/2

Step-by-step explanation:

We can use the slope formula

m = ( y2-y1)/(x2-x1)

   = ( -2 - -4)/( -11 - -7)

   =( -2+4)/( -11+7)

   = 2 / -4

  = -1/2

Answer:

please ask in English

Step-by-step explanation:

then I can help

Answer:

1. Degree = 1, monomial

2. Degree = 2, monomial

3. degree = 2, trinomial

4. Degree = 2, binomial

5. Degree  = 2, binomial

Step-by-step explanation:

Answer:

In mathematics, a real number is a value of a continuous quantity that can represent a distance along a line (or alternatively, a quantity that can be represented as an infinite decimal expansion).

Answer:

distance=√[(x2-x1)²+(y2-y1)²]

√[{6-(-2)}²+ (-8-(-4))²]

√(64+16)

√[100]

10

Points given

Distance:-

[tex]\\ \sf \longmapsto \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

[tex]\\ \sf \longmapsto \sqrt{(-8+4)^2+(6+2)^2}[/tex]

[tex]\\ \sf \longmapsto \sqrt{(-4)^2+(8)^2}[/tex]

[tex]\\ \sf \longmapsto \sqrt{64+16}[/tex]

[tex]\\ \sf \longmapsto \sqrt{80}[/tex]

[tex]\\ \sf \longmapsto 8.4[/tex]

Let the ratio be m:n

We know

[tex]\boxed{\sf (x,y)=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}[/tex]

[tex]\\ \sf\longmapsto (4,5)=\left(\dfrac{7m+2n}{m+n},\dfrac{8m+3n}{m+n}\right)[/tex]

Now

.[tex]\\ \sf\longmapsto \dfrac{7m+2n}{m+n}=4\dots(1)[/tex]

[tex]\\ \sf\longmapsto \dfrac{8m+3n}{m+n}=5\dots(2)[/tex]

Adding both

[tex]\\ \sf\longmapsto \dfrac{7m+2n+8m+3n}{m+n}=4+5[/tex]

[tex]\\ \sf\longmapsto \dfrac{7m+8m+2n+3n}{m+n}=9[/tex]

[tex]\\ \sf\longmapsto \dfrac{15m+5n}{m+n}=9[/tex]

[tex]\\ \sf\longmapsto 15m+5n=9(m+n)[/tex]

[tex]\\ \sf\longmapsto 15m+5n=9m+9n[/tex]

[tex]\\ \sf\longmapsto 15m-9m=9n-5n[/tex]

[tex]\\ \sf\longmapsto 6m=4n[/tex]

[tex]\\ \sf\longmapsto \dfrac{m}{n}=\dfrac{6}{4}[/tex]

[tex]\\ \sf\longmapsto \dfrac{m}{n}=\dfrac{3}{2}[/tex]

[tex]\\ \sf\longmapsto m:n=3:2[/tex]

Option B is correct

Answer: -2

Step-by-step explanation:

g(-4) + f(1) = 7 + (-9) = 7 - 9 = -2

Answer:

378 and 380

Step-by-step explanation:

The two even consecutive numbers that add up to 758 are going to be very close to half of 758. This is because two half of 758 are going to be the most similar addends of 758. This is important because the answers will be consecutive and therefore, must also be very similar. To solve, first, divide 758 by 2. This is 379, which is not an even number. So, to find the needed addends subtract and add 1 to 379. Both of these will be even and consecutive. These two numbers are 378 and 380. Then, to check you, can add them and see that they do sum 758.

Answer:

Step-by-step explanation:

Let the first number = x

Let the second number = x + 2

x + x + 2 = 758                      Collect like terms

2x + 2 = 758                         Subtract 2

2x = 758 - 2                          Combine

2x = 756                                Divide by 2

2x/2 = 756/2

x = 378

The first number is 378

The second number 380

If your teacher is really fussy, you can do it this way.

Let the first number = 2x

Let the second number = 2x + 2

The reason for this is to guarantee that both numbers were even to start with.

2x + 2x+2 = 758                    Combine like terms

4x + 2 = 758                          Subtract 2

4x = 756                                Divide by 4

x = 756/4

x = 189

Therefore 2x = 378

2x + 2 = 380                 Just as before.