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3 years ago

ASAP I AM TIMED

Mathematics

1 answer:

strojnjashka [21]3 years ago

5 0

X=3 falls in the third piecewise function,
f(x)=x+1
f(3)=3+1=4

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Draw and classify the polygon with the given vertices. Find the perimeter and the area of the polygon. M(-2,5), N(3,-2), P(-2,-2

Nadusha1986 [10]

Answer:

Perimeter = 20.6cm to 1dp = 20.60cm to 2dp

Area = Base = 8.6/2 = 4.3 x 4.069 = 17.5cm^2

Height is found 4.07 cos( 54.47deg) 5/8.6 = 4.069cm

Step-by-step explanation:

Right angle side triangle, sides

= MN =8.6cm

= NP=5cm

= PM=7cm

= 25sq + 49sq = 74sq

MN^2 =√74 = 8.60232526704 = 8.6cm

P = 8.6 + 5 + 7 = 20.6cm

5 0

3 years ago

Read 2 more answers

in rectangle ABCD, BD and AC are diagonals and intersect at E. If BD = 84 units and AE = 6y, what is the value of y

goldfiish [28.3K]

Answer:

y = 7

Step-by-step explanation:

The diagonals of a rectangle are congruent and bisect each other, then

AC = BD , that is

AE + EC = BD [ AE = EC = 6y ]

6y + 6y = 84

12y = 84 ( divide both sides by 12 )

y = 7

3 0

2 years ago

At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce

Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$ . By the Mean Value Theorem, there is a number c such that $0 < c$ with $v'(c)=60 \:{\frac{mi}{h^2}}$ . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly $60 \:{\frac{mi}{h^2}}$ .

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $v(0 \:h) = 30 \:{\frac{mi}{h} }$ and $v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} }$ (note that 20 minutes is $20/60=1/3$ of an hour), so the average rate of change of v on the interval $[0 \:h, \frac{1}{3} \:h]$ is

$\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in $(0 \:h, \frac{1}{3} \:h)$ at which $v'(c)=60 \:{\frac{mi}{h^2}}$ .