Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Ming throws a stone off a bridge into a river below. The stone's height (in meters above the water), x seconds after Ming threw it, is modeled by: h(x)=-5(x-1)^(2)+45 How many seconds after being thrown will the stone reach its maximum height? seconds

Ming throws a stone off a bridge into a river below.\newlineThe stone's height (in meters above the water), xx seconds after Ming threw it, is modeled by:\newlineh(x)=−5(x−1)2+45h(x)=-5(x-1)^{2}+45\newlineHow many seconds after being thrown will the stone reach its maximum height?\newlineseconds\text{seconds}

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 1right chevron icon
Solve quadratic equations: word problemsright chevron icon

Full solution

Q. Ming throws a stone off a bridge into a river below.\newlineThe stone's height (in meters above the water), xx seconds after Ming threw it, is modeled by:\newlineh(x)=−5(x−1)2+45h(x)=-5(x-1)^{2}+45\newlineHow many seconds after being thrown will the stone reach its maximum height?\newlineseconds\text{seconds}
  1. Given Equation: We are given the equation for the height of the stone above the water as a function of time: h(x)=−5(x−1)2+45 h(x) = -5(x - 1)^2 + 45 . To find the time at which the stone reaches its maximum height, we need to identify the vertex of the parabola represented by this equation. Since the coefficient of the squared term is negative (−5 -5 ), the parabola opens downwards, and the vertex will give us the maximum height.
  2. Identifying the Vertex: The vertex form of a parabola is given by h(x)=a(x−h)2+kh(x) = a(x - h)^2 + k, where (h,k)(h, k) is the vertex of the parabola. In our equation, h(x)=−5(x−1)2+45h(x) = -5(x - 1)^2 + 45, we can see that the vertex is at (h,k)=(1,45)(h, k) = (1, 45). This means that the maximum height is reached at x=1x = 1 second.
  3. Finding the Maximum Height: Since the question asks for the time after being thrown that the stone reaches its maximum height, and we have found that this occurs at x=1x = 1 second, we can conclude that the stone reaches its maximum height 11 second after being thrown.

More problems from Solve quadratic equations: word problems

Question
Solve for v v .\newlinev2+8v+12=0 v^2+8v+12=0\newlineWrite each solution as an integer, proper fraction, or improper fraction in simplest form. If there are multiple solutions, separate them with commas.\newlinev= v=_____
Get tutor helpright-arrow

Posted 3 months ago

Question
Solve for u u .\newlineu2=9 u^2 = 9 \newlineWrite your answers as integers or as proper or improper fractions in simplest form.\newlineu= u = _____ or u= u = _____
Get tutor helpright-arrow

Posted 9 months ago

Question
Complete the square. Fill in the number that makes the polynomial a perfect-square quadratic.\newlinek2−20k+_____k^2 - 20k + \_\_\_\_\_
Get tutor helpright-arrow

Posted 8 months ago

Question
Solve by completing the square.\newlineu2−24u−23=0u^2 - 24u - 23 = 0\newlineWrite your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.\newline`u` = ________ or `u` =______
Get tutor helpright-arrow

Posted 5 months ago

Question
Find the discriminant. \newline2q2−9q+8=02q^2 - 9q + 8 = 0\newline______
Get tutor helpright-arrow

Posted 5 months ago

Question
Solve the system of equations.\newliney=22x−38 y = 22x - 38 \newliney=x2+11x−14 y = x^2 + 11x - 14 \newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(_,_) (\_,\_) \newline(_,_) (\_,\_)
Get tutor helpright-arrow

Posted 5 months ago

Question
Solve for aa.\newline(a+5)(a+1)=0(a + 5)(a + 1) = 0\newlineWrite your answers as integers or as proper or improper fractions in simplest form.\newlinea=a = _____ or a=a = _____
Get tutor helpright-arrow

Posted 8 months ago

Question
Let p=x2−7 p=x^{2}-7 . Which equation is equivalent to (x2−7)2−4x2+28=5 (x^{2}-7)^{2}-4x^{2}+28=5 in terms of p p ? Choose 11 answer: \newline(A) p2−4p+23=0 p^{2}-4p+23=0 \newline(B) p2+4p−5=0 p^{2}+4p-5=0 \newline(C) p2−4p−5=0 p^{2}-4p-5=0 \newline(D) p2+4p+23=0 p^{2}+4p+23=0
Get tutor helpright-arrow

Posted 8 months ago

Question
Let m=2x+3m=2x+3. Which equation is equivalent to (2x+3)2−14x−21=−6(2x+3)^{2}-14x-21=-6 in terms of mm? Choose 11 answer: \newline(A) m2+7m+6=0m^{2}+7m+6=0 \newline(B) m2−7m−15=0m^{2}-7m-15=0 \newline(C) m2−7m+6=0m^{2}-7m+6=0 \newline(D) m2+7m−15=0m^{2}+7m-15=0
Get tutor helpright-arrow

Posted 8 months ago

Question
Let m=x2+3m=x^{2}+3. Which equation is equivalent to (x2+3)2+7x2+21=−10(x^{2}+3)^{2}+7x^{2}+21=-10 in terms of mm? Choose 11 answer:\newline(A) m2−7m+31=0m^{2}-7m+31=0\newline(B) m2+7m+31=0m^{2}+7m+31=0\newline(C) m2−7m+10=0m^{2}-7m+10=0\newline(D) m2+7m+10=0m^{2}+7m+10=0
Get tutor helpright-arrow

Posted 9 months ago

View all (43)

Related topics

Algebra - Order of OperationsAlgebra - Distributive Property`X` and `Y` AxesGeometry - Scalene TriangleCommon MultipleGeometry - Quadrant