A ball follows a path described by the function h(x) = -3x^2 +12x +14 where h(x) is the height of the ball after x seconds. After how many seconds will the ball be at its
highest point?

A. 1 second
B. 2 seconds
C. 3 seconds
D. 4 seconds

Answer:

₹ 81.75

Step-by-step explanation:

Total =  ₹ 120.

He bought one book for ₹ 25.75 and a pen for ₹ 12.50.

    ₹ 25.75 + ₹ 12.50 =  ₹ 38.25

How much money does he have now?​

     ₹ 120 -  ₹ 38.25 =  ₹ 81.75

₹81.75

you just minus 120 from 25.75 and 12.50

Answer:

Step-by-step explanation:

75-52.5=x

Answer:

10x - 3y = -109 .

Step-by-step explanation:

x^2 + y^2 = 109

Implicit Differentiation:

2x + 2y.y' = 0

y' = -2x/2y = -x/y

So at the point (-10,3) the gradient of the tangent is -(-10)/3 = 10/3.

Equation of the tangent:

y - y1 = m(x - x1)

y - 3 = 10/3(x + 10)

y - 3 = 10/3 x + 100/3

3y - 9 = 10x + 100

-109 + 3y = 10x

10x - 3y = -109  

Answer:

[tex](9x-1)^2[/tex]

Step by step explanation:

[tex]81x^2-18x+1\\\\=81x^2 - 9x-9x+1\\\\=9x(9x-1)-(9x-1)\\\\=(9x-1)(9x-1)\\\\=(9x-1)^2[/tex]

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 1

Let's think of the boat and wind as vectors:

Boat Vector --> [tex]\langle28cos36^\circ,28sin36^\circ\rangle[/tex]

Wind Vector --> [tex]\langle10cos22^\circ,10sin22^\circ\rangle[/tex]

Now, let's add the vectors:

[tex]\langle28cos36^\circ+10cos22^\circ,28sin36^\circ+10sin22^\circ\rangle[/tex]

Find the magnitude (the true velocity):

[tex]\sqrt{(28cos36^\circ+10cos22^\circ)^2+(28sin36^\circ+10sin22^\circ)}\approx37.78\approx38[/tex]

Find the direction (angle):

[tex]\theta=tan^{-1}(\frac{28sin36^\circ+10sin22^\circ}{28cos36^\circ+10cos22^\circ})\approx32.32^\circ\approx32^\circ[/tex]

Thus, D is the best answer

Problem 2

Recall that the angle between two vectors is [tex]\theta=cos^{-1}(\frac{u\cdot v}{||u||*||v||})[/tex] where [tex]u\cdot v[/tex] is the dot product of the vectors and [tex]||u||*||v||[/tex] is the product of each vector's magnitude:

[tex]\theta=cos^{-1}(\frac{u\cdot v}{||u||*||v||})\\\\\theta=cos^{-1}(\frac{\langle-82,47\rangle\cdot\langle92,80\rangle}{\sqrt{(-82)^2+(47)^2}*\sqrt{(92)^2+(80)^2}})\\\\\theta=cos^{-1}(\frac{(-82)(92)+(47)(80)}{\sqrt{6724+2209}*\sqrt{8464+6400}})\\\\\theta=cos^{-1}(\frac{(-7544)+3760}{\sqrt{8933}*\sqrt{14864}})\\\\\theta=cos^{-1}(\frac{-3784}{\sqrt{132780112}})\\\\\theta\approx109.17^\circ\approx109^\circ[/tex]

Therefore, C is the best answer

Hello!

Circumference of a Circle

C=πd

or

C=2πr

Plug in the values

C=14π

C≈44 cm

C=2π×3.5

C≈22 cm

Hope everything is clear.

Let me know if you have any questions!
#KeepLearning

;-)

Answer:  x=17

Step-by-step explanation:(6x+x+4x)+(10+17-34)=180

we know this because triangles always equal 180o so the 3 angles equa; 180 so 187/11 + 17

Answer:

  a) 96 = 3.57√h

  b) h ≈ 723.11 m

Step-by-step explanation:

The equation you want to solve is the model with the given values filled in.

  D(h) = 3.57√h . . . . model

  96 = 3.57√h . . . . . equation for seeing 96 km to the horizon

__

We solve this equation by dividing by the coefficient of the root, then squaring both sides.

  96/3.57 = √h

  h ≈ 26.891² ≈ 723.11 . . . . meters above sea level

Dustin would need to have an elevation of 723.11 meters above sea level to see 96 km to the horizon.

Step-by-step explanation:

first we want to convert it to minutes so we multiply 60m/s by 60. 3600m/min or 3.6km/min

we just multiply 50 by 3.6.

therefore the car travels 180km south west

Answer:

i think the answer is d which is 50°

Answer:

[tex]w < 4 \frac{2}{3}[/tex]

Step-by-step explanation:

The inequality symbol for 'less than' is '<'.

Let's relate the cost and the number of each item bought together with an equation.

Total cost

= (number of rolls)(cost of each roll) +(number of bags)(cost per bag of bows)

= w(1.50) +2(1.50)

= 1.5w +3

Now, we can form the inequality.

Since the total cost is less than $10,

1.5w +3< 10

Subtract 3 from both sides:

1.5w < 10 -3

1.5w < 7

Divide both sides by 1.5:

[tex]w < 7 \div \frac{3}{2} [/tex]

[tex]w < 7 \times \frac{2}{3} [/tex]

[tex]w < \frac{14}{3} [/tex]

[tex]\textcolor{red}{w < 4 \frac{2}{3} }[/tex]

Answer:

-9/14

Step-by-step explanation:

Hey there!

Guide:

• Difference means subtract/subtraction
• Product means multiply/multiplication
• Sum means add/addition
• Quotient means divide/division

• Now that we know what “product” means… we can make the question/equation easier to solve.

3/4 × -6/7

= 3(-6) / 4(7)

= -18 / 28

= -18 ÷ 2 / 28 ÷ 2

= -9 / 14

Therefore, your answer is: -9/14

Good luck on assignment & enjoy your day!

~Amphitrite1040:)

Answer:

1.) C2.) C3.) D4.) D5.) C

Step-by-step explanation:

A: 1.9 Kilograms of oranges for rs 665.

B: 12 liters of petrol will make a car travel 120 km.

Answer:

Let y = the amount in Adrianna's savings account

Let x = the number of weeks passed

y = 5x + 20

Step-by-step explanation:

2× (4x-5y) = 21

3x+10y= -53

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

8x-10y = 42

3x+10y=-53

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

11x= -11

{x=-1}

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

3x+10y=-53

x=-1

3(-1)+10y=-53

-3+10y=-53

10y= -50

{y=5}

Answer:

14cm

Step-by-step explanation:

Algebra is the way!

But seriously, let's model this algebraically.

We know the that the area of a triangle is represented by:

[tex]A = \frac{1}{2} base*height[/tex]

we'll use b and h to represent base and height

[tex]A = \frac{1}{2} b*h[/tex]

Okay, so you are given the area of the triangle already. It is [tex]35 cm^2[/tex]. SO you have:

[tex]35 = \frac{1}{2} bh[/tex]

Now here's the tricky bit of the question. You have 2 unknown variables, oh no! How will you solve? The key to this is simple, in a case where you have 2 or more unknown variables, ALWAYS think to yourself, "can I represent this variable in terms of another"?

In this case you can. They tell you what the height of the triangle is, in terms of the length of the base. They tell you, height is 4 cm more than 2 times the base.

So replace h in this equation with 2b + 4.

[tex]35 = \frac{1}{2} (b)(2b+4)\\\\35 = \frac{1}{2}(2b^2 + 4b)\\ 70 = 2b^2 + 4b\\0 = 2b^2 + 4b - 70\\0 = (2)(b^2 + 2b - 35)\\0 = (b +7)(b - 5)\\\\b = 5\\b\ \neq -7[/tex]

Okay, so you know the base of the triangle is 5 cm. The question asks for height, so what you're going to do is refer back to what h is.

We found that [tex]h = 2b + 4[/tex]

sub in 5 into that:

[tex]h = 2(5) + 4 = 14[/tex]

Therefore the height of the triangle is 14cm.

Answer:

Step-by-step explanation:

Givens

Area = 1/2 B* H

B = B

H = 2*B + 4

Area = 35

Solution

35 = 1/2 * B * (2B + 4)                          Multiply both sides by 2

35*2 = 2*(1/2) * B * (2B + 4)                 Combine

70 = B(2B + 4)                                      Remove the Brackets.

70 = 2B^2 + 4B                                    Subtract 70 from both sides.

2B^2 + 4B - 70 = 0                               Divide everything by 2

B^2 + 2B - 35 = 0                                 Factor

(B + 7)(B - 5)

B+7 = 0 will give a minus number. Geometry does not allow minus numbers.

B - 5  = 0

B = 5

But what you want is the height.

h = 2b + 4

h = 2*5 + 4

Answer: h = 14

Answer:

H

Step-by-step explanation:

If you convert these fractions into decimals:
-2 1/2 = -2.5
-2.47 = -2.47
-2/5 = -0.4
5 = 5
21/4 = 5.25
You can see that the order from lest to greatest is answer H

Hope this helps :)

[tex] \Large \bold{SOLUTION\ 1:} [/tex]

[tex] \small \begin{array}{l} \text{First, we need to check if the given differential} \\ \text{equation is exact.} \\ \\ (1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \quad M(x, y) dx + N(x, y) dy = 0 \end{array} [/tex]

[tex] \small \begin{array}{l l}\tt\: M(x,y) = \dfrac{1}{(1 - xy)^2}, & N(x,y) = y^2 + \dfrac{x^2}{(1-xy)^2}\\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{-2(-x)}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^2} + \dfrac{-2(-y)x^2}{(1 - xy)^3} \\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{2x}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x(1 - xy)+2x^2y}{(1 - xy)^3} \\ \\\tt \: & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^3} \end{array} [/tex]

[tex] \small \begin{array}{l} \tt\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x} \implies \text{Differential equation is exact.} \\ \\\tt \dfrac{\partial F}{\partial x} = M(x, y) = \dfrac{1}{(1 - xy)^2} \\ \tt\displaystyle F(x, y) = \int \dfrac{1}{(1 - xy)^2} \partial x = -\dfrac{1}{y} \int \dfrac{1}{(1 - xy)^2}(-y)\partial x \\ \\ \tt\:F(x, y) = \dfrac{1}{y(1 - xy)} + h(y) \\ \\ \tt\dfrac{\partial F}{\partial y} = N(x, y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\\tt \dfrac{\partial}{\partial y}\left[\dfrac{1}{y(1 - xy)} + h(y)\right] = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - xy + y(-x)}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - 2xy}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} + \dfrac{1 - 2xy}{y^2(1 - xy)^2} \\ \\ \tt\:h'(y) = y^2 + \dfrac{x^2y^2 - 2xy + 1}{y^2(1-xy)^2} = y^2 + \dfrac{1}{y^2} \\ \\ h(y) = \dfrac{y^3}{3} - \dfrac{1}{y} + C \\ \\ \tt\text{Substituting to }F(x,y),\text{we get} \\ \\ \dfrac{1}{y(1 - xy)} + \dfrac{y^3}{3} - \dfrac{1}{y} = C \\ \\ \quad \quad \text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array} [/tex]

[tex] \Large \bold{SOLUTION\ 2:} [/tex]

[tex] \small \begin{array}{l} \tt\text{Since we already know that the equation is exact,} \\ \text{we can then continue solving for the solution by} \\ \text{inspection method or by algebraic manipulation.} \\ \\ \tt(1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + y^2 dy + \dfrac{x^2}{(1-xy)^2} dy = 0 \\ \\ \tt\dfrac{dx + x^2dy}{(1-xy)^2} + y^2 dy = 0 \\ \\ \tt\text{Divide both numerator and denominator of the} \\ \tt\text{fraction by }x^2. \end{array} [/tex]

[tex] \small \begin{array}{c}\tt \dfrac{\dfrac{1}{x^2}dx + dy}{\dfrac{(1-xy)^2}{x^2}} + y^2 dy = 0 \\ \tt\\ \tt\dfrac{\dfrac{1}{x^2}dx + dy}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt-\dfrac{\left(-\dfrac{1}{x^2}dx - dy\right)}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt\displaystyle {\large{\int}} -\frac{d\left(\dfrac{1}{x}-y\right)}{\left(\dfrac{1}{x}-y\right)^2} + \int y^2 dy = \int 0 \\ \\ \tt\implies\tt \dfrac{1}{\dfrac{1}{x}-y} + \dfrac{y^3}{3} = C \\ \\\text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array} [/tex]

#CarryOnLearning

#BrainlyMathKnower

[tex]\\ \sf\Rrightarrow y=-4(x+2)^2-1[/tex]

Compare to the vertex form of parabola

So vertex should be at (h,k)=(-2,-1)

Graph attached

The length of the rectangle is 48 yards.

Let's denote the width of the rectangle as "w" yards.

According to the information given, the length of the rectangle is 32 yards less than 10 times its width. So, the length can be represented as: 10w - 32 yards.

The formula for the area of a rectangle is length times width: Area = length × width.

Given that the area of the rectangle is 384 square yards, we can set up an equation:

Area = length × width

384 = (10w - 32) × w

Now, let's solve for the width (w):

384 = 10w² - 32w

0 = 10w² - 32w - 384

Dividing the equation by 2 to simplify:

0 = 5w² - 16w - 192

Now we have a quadratic equation. We can either factor it or use the quadratic formula to solve for "w." Let's use the quadratic formula:

w = (-b ± √(b² - 4ac)) / 2a

For our equation, a = 5, b = -16, and c = -192. Plugging these values into the formula:

w = (16 ± √((-16)² - 4 × 5 × (-192))) / (2 × 5)

w = (16 ± √(256 + 3840)) / 10

w = (16 ± √4096) / 10

w = (16 ± 64) / 10

This gives us two possible solutions for the width:

w = (16 + 64) / 10 = 8 yards

w = (16 - 64) / 10 = -4.8 yards

Since width can't be negative, we'll ignore the second solution.

So, the width of the rectangle is 8 yards. Now, let's find the length using the earlier expression:

Length = 10w - 32

Length = 10 × 8 - 32

Length = 80 - 32

Length = 48 yards

Therefore, the length of the rectangle is 48 yards.

To know more about length:

https://brainly.com/question/2497593

#SPJ3

The tree grew about 3.83 inches between the first and third week.

An equation is an expression that shows the relationship between two or more numbers and variables.

On the first week, the tree was 27 1/2 inches (27.5) tall. On the second week, it was 29 1/4 inches (29.25) tall. On the third week, it was 31 1/3 inches (31.33) tall.

The length grown = 31.33 - 17.5 = 3.83 inches

The tree grew about 3.83 inches between the first and third week.

Find out more on equation at: https://brainly.com/question/2972832

Step-by-step explanation:

a)[tex] \sf \: Factor 6x2−3x \\ \sf {6x}^{2} −3x \\ \sf \: =3x(2x−1) \\ \sf \: Answer:3x(2x−1) \\ [/tex]

b)Let's factor x²+3x−28

⇒x²+3x−28

⇒x² - 4x + 7x - 28

⇒(x² - 4x) + (7x - 28)

⇒x( x - 4) + 7(x - 4)

⇒(x -4) (x + 7)

Answer:

(x−4)(x+7)

Answer:

Its B square pyramid

Step-by-step explanation:

Because

Answer:

an+1 = 4an

Step-by-step explanation:

an+1 = 4an        commonratio:   r=4

2

2x4=8

8x4=32

32x4=128

#4

Refer to the attachment

#2

Mid point:-

Step-by-step explanation:

please mark me as brainlest