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3 years ago

Please answer this question

Mathematics

2 answers:

telo118 [61]3 years ago

5 0

The answer is the last one

aleksklad [387]3 years ago

3 0

The last one is correct

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Drag each capacity to match it to an equivalent capacity.

weqwewe [10]

Answer:

2 cups ⇔ 16 fluid ounces
96 ounces ⇔ 6 pints
12 pints ⇔ 24 cups
4 quarts ⇔ 1 gallon

<em>Also see attached</em>

3 0

2 years ago

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Seriously need help with this question or I will fail my senior year

nlexa [21]

Answer:

you cheater

Step-by-step explanation:

Tring to cheat because you don't feel like doing it. Hopefully you fail so you can learn your lesson

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2 years ago

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What are the solutions of the quadratic equation? 4x^2+34x+60=0

insens350 [35]

Answer:

Option 3) -6,-5/2

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$4x^{2} +34x+60=0$

$a=4\\b=34\\c=60$

substitute in the formula

$x=\frac{-34(+/-)\sqrt{34^{2}-4(4)(60)}} {2(4)}$

$x=\frac{-34(+/-)\sqrt{196}} {8}$

$x=\frac{-34(+/-)14} {8}$

$x_1=\frac{-34(+)14} {8}=-\frac{20}{8}=-\frac{5}{2}$

$x_2=\frac{-34(-)14} {8}=-6$

6 0

3 years ago

If u is a unit vector, find u · v and u · w. (assume v and w are also unit vectors.) equilateral triangle

serg [7]

solution:

we know that ,

u.v = ΙuΙ ΙvΙcosθ

here,

θ =60° (since the given triangle is equilateral triangle)

u.v = ΙuΙ ΙvΙcos60°

= 1 x 1 x 1/2

u.v = 1/2

now, u.w = ΙuΙ ΙwΙcosθ

= ΙuΙ x cos(60x2)

u.w = -1/2

5 0

3 years ago

A cold drink is poured out at 52°F. After 2 minutes of sitting in a 72°F room, its temperature has risen to 55°F. Find an equati

I am Lyosha [343]

Answer:

The model for the temperature of the drink can be written as

$T=72-20e^{-0.08t}$

Step-by-step explanation:

For a cold drink in a hotter room, we can say that the rate of change of temperature of the drink is proportional to the difference of temperature between the drink and the room.

We can model that in this way

$\frac{dT}{dt}=k*(T_r-T)$

If we rearrange and integrate

$\int\frac{dT}{(T-Tr)} =-k*\int dt\\\\ln(T-T_r)=-kt+C1\\\\T-T_r=Ce^{-kt}\\\\T=T_r+Ce^{-kt}$

We know that at time 0, the temperature of the drink was 52°F. Then we have:

$T=T_r+Ce^{-kt}\\\\52=72+Ce^0=72+C\\\\C=-20$

We also know that at t=2, T=55°F

$T=T_r+Ce^{-kt}\\\\55=72-20e^{-k*2}\\\\e^{-k*2}=(72-55)/20=0.85\\\\-2k=ln(0.85)=-0.1625\\\\k=0.08$

The model for the temperature of the drink can be written as

$T=72-20e^{-0.08t}$

7 0

4 years ago