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3 years ago

An object is launched from a platform.

Mathematics

2 answers:

uysha [10]3 years ago

7 0

Answer:

6seconds

Step-by-step explanation:

Given the height of an object modeled by the expression

h(x)=-5x^2+20x+60

The height of the object on the ground is zero i.e h(x) = 0

On substituting

h(x)=-5x^2+20x+60

0 =-5x^2+20x+60

Multiply through by -1

5x^2-20x- 60 = 0

Divide through by 5

x^2 - 4x - 12 = 0

x^2 - 6x + 2x - 12 = 0

x(x-6)+2(x-6) = 0

(x-6)(x+2) = 0

x - 6 = 0 and x+2 = 0

x = 6 and -2

Since time cannot be negative, hence it takes 6seconds after launch for the object land to on the ground

gregori [183]3 years ago

6 0

Answer:

Step-by-step explanation:

It was correct on Khan.

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Answer:

5.28 if you round to the nearest hundredth. Drop a best answer and a like if it helped

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3 years ago

Quadrilateral PQRS is a square whos side length is 10. Let X and Y be points outside the square so that XQ = YS = 6 and XP = YR

erik [133]

Answer:

392

Step-by-step explanation:

Triangles XQP and YRS are right triangles because triples 6, 8, 10 are Pythagorean triples.

Extend lines XQ, YR, YS and XP and mark their intersection as A and B.

Quadrilateral XAYB is a square because all right triangles PXQ, QAR, RYS and SBP are congruent (by ASA postulate) and therefore

all angles of the quadrilateral XAYB are right angles

all sides of XAYB are congruent and equal to 6 + 8 = 14 units.

Segment XY is the diagonal of the square XAYB, by Pythagorean theorem,

$XY^2=XA^2+AY^2\\ \\XY^2=14^2+14^2\\ \\XY^2=196+196\\ \\XY^2=392$

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Lelu [443]

15min:2h
15min:120min
15/120
3/24
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3 years ago

Jake needs to put gas in the delivery van regular gasoline is 3.20 per gallon he only had a $20 bill on him how many gallons can

Fed [463]

Your max amount for this is $20. Let's say x represents the number of gallons.

20 ≥ 3.2x
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3 years ago

Read 2 more answers

Which of the following expressions are equivalent? Justify your reasoning.. . a.4√x3. . 1. x−1. . b.10√x5•x4•x2. . c.x. 1. 3. •x

sladkih [1.3K]

A ) $\sqrt[4]{ x^{3} } = x^{3/4}$
B ) $\frac{1}{ x^{-1} } = x$
C ) $\sqrt[10]{ x^{5} * x^{4}* x^{2} } = \sqrt[10]{ x^{11} }= x^{11/10}$
D ) $x^{1/3} * x^{1/3} * x^{1/3} = x$
Therefore: B = D

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3 years ago