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2 years ago

A race car driver won a 300 mile race with a speed of 183.2 mph find the drivers time

Mathematics

1 answer:

Ostrovityanka [42]2 years ago

3 0

Answer:

The time taken by the driver is: t = 1.6 seconds

Step-by-step explanation:

Given

Distance covered by a race car = d = 300 mile

Speed of a car = s = 183.2 mph

To determine

The time of the driver = t = ?

Using the formula to determine the time of the driver

distance = speed x time

d = st

t = d/s

substitute d = 300 and s = 183.2

t = 300 / 183.2

t = 1.6 seconds

Therefore, the time taken by the driver is: t = 1.6 seconds

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What is the x-intercepts of y= -2(x-3)[2]+2?

mrs_skeptik [129]

Answer: (x,y)=(14/4,0)

Step-by-step explanation:

y=-2(x-3)(2)+2

Clear the brackets

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

y=-4x+12+2

y=-4x+14

To get the x-intercept y=0

0=-4x+14

4x=14

x=14/4

(x,y)=(14/4,0)

8 0

3 years ago

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estimate the square root of 14 to the nearest integer

Feliz [49]

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3 years ago

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What is the length of the curve with parametric equations x = t - cos(t), y = 1 - sin(t) from t = 0 to t = π? (5 points)

zzz [600]

Answer:

B) 4√2

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Parametric Differentiation

Integration

Integrals
Definite Integrals
Integration Constant C

Arc Length Formula [Parametric]: $\displaystyle AL = \int\limits^b_a {\sqrt{[x'(t)]^2 + [y(t)]^2}} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \left \{ {{x = t - cos(t)} \atop {y = 1 - sin(t)}} \right.$

Interval [0, π]

Step 2: Find Arc Length

[Parametrics] Differentiate [Basic Power Rule, Trig Differentiation]: $\displaystyle \left \{ {{x' = 1 + sin(t)} \atop {y' = -cos(t)}} \right.$
Substitute in variables [Arc Length Formula - Parametric]: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{[1 + sin(t)]^2 + [-cos(t)]^2}} \, dx$
[Integrand] Simplify: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx$
[Integral] Evaluate: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}$

Topic: AP Calculus BC (Calculus I + II)

Unit: Parametric Integration

Book: College Calculus 10e

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3 years ago

Is y=x^3 a solution of the differential equation yy'=x^5+y

shepuryov [24]

No; we have $y=x^3\implies y'=3x^2$ . Substituting these into the DE gives

$3x^5=x^5+x^3$

which reduces to $x^3=0$ , true only for $x=0$ .

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3 years ago

Use the gcf and the distributive property to express the sum as a product 20 35

Maslowich

Common factor of 20= 1,2,4,5,10.......
common factor of 35=1,5,7.............
so here we find that 1 and 5 r the common factors
but the GCF of 20 and 35 is 5
so,
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3 years ago