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3 years ago

Please I need help ASAP, please tell me how you know

Mathematics

1 answer:

mixas84 [53]3 years ago

7 0

Answer: D it the For the way they have as examples

what do you need help on. what the problem

You might be interested in

Plz help me find these functions!

Iteru [2.4K]

Answer:

$\sin \alpha = \frac{5}{13}$ , $\cos \alpha = \frac{12}{13}$ , $\tan \alpha = \frac{5}{12}$ , $\cot \alpha = \frac{12}{5}$ , $\sec \alpha = \frac{13}{12}$ , $\csc \alpha = \frac{13}{5}$

Step-by-step explanation:

The angle $\alpha$ is the angle opposite to the side of length 5 and adjacent to the side of length 12. From Trigonometry we remember the following relationships:

$\sin \alpha = \frac{5}{\sqrt{5^{2}+12^{2}}}$

$\sin \alpha = \frac{5}{13}$

$\cos \alpha = \frac{12}{\sqrt{5^{2}+12^{2}}}$

$\cos \alpha = \frac{12}{13}$

$\tan \alpha = \frac{5}{12}$

$\cot \alpha = \frac{12}{5}$

$\sec \alpha = \frac{\sqrt{5^{2}+12^{2}}}{12}$

$\sec \alpha = \frac{13}{12}$

$\csc \alpha = \frac{\sqrt{5^{2}+12^{2}}}{5}$

$\csc \alpha = \frac{13}{5}$

4 0

3 years ago

2x+ 8y = -3 3x+ 6y = -4 Choose all answers that apply

Anvisha [2.4K]

multiply the top equation by 3, multiply the bottom equation by 4, then subtract the bottom equation from the top equation

multiply the top equation by 3, multiply the bottom equation by -2, then add the equations

4 0

3 years ago

A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in the solution. Water containing1 lb

devlian [24]

Answer:

(a) The amount of salt in the tank at any time prior to the instant when the solution begins to overflow is $\left(1-\frac{4000000}{\left(200+t\right)^3}\right)\left(200+t\right)$ .

(b) The concentration (in lbs per gallon) when it is at the point of overflowing is $\frac{121}{125}\:\frac{lb}{gal}$ .

(c) The theoretical limiting concentration if the tank has infinite capacity is $1\:\frac{lb}{gal}$ .

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).

The main equation that we’ll be using to model this situation is:

Rate of change of Q(t) = Rate at which Q(t) enters the tank – Rate at which Q(t) exits the tank

where,

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x (concentration of substance in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x (concentration of substance in liquid exiting)

Let $C$ be the concentration of salt water solution in the tank (in $\frac{lb}{gal}$ ) and $t$ the time (in minutes).

Since the solution being pumped in has concentration $1 \:\frac{lb}{gal}$ and it is being pumped in at a rate of $3 \:\frac{gal}{min}$ , this tells us that the rate of the salt entering the tank is

$1 \:\frac{lb}{gal} \cdot 3 \:\frac{gal}{min}=3\:\frac{lb}{min}$

But this describes the amount of salt entering the system. We need the concentration. To get this, we need to divide the amount of salt entering the tank by the volume of water already in the tank.

$V(t)$ is the volume of brine in the tank at time t. To find it we know that at t = 0 there were 200 gallons, 3 gallons are added and 2 are drained, and the net increase is 1 gallons per second. So,

$V(t)=200+t$

Therefore,

The rate at which $C(t)$ enters the tank is

$\frac{3}{200+t}$

The rate of the amount of salt leaving the tank is

$C\:\frac{lb}{gal} \cdot 2 \:\frac{gal}{min}+C\:\frac{lb}{gal} \cdot 1\:\frac{gal}{min}=3C\:\frac{lb}{min}$

and the rate at which $C(t)$ exits the tank is

$\frac{3C}{200+t}$

Plugging this information in the main equation, our differential equation model is:

$\frac{dC}{dt} =\frac{3}{200+t}-\frac{3C}{200+t}$

Since we are told that the tank starts out with 200 gal of solution, containing 100 lb of salt, the initial concentration is

$\frac{100 \:lb}{200 \:gal} =0.5\frac{\:lb}{\:gal}$

Next, we solve the initial value problem

$\frac{dC}{dt} =\frac{3-3C}{200+t}, \quad C(0)=\frac{1}{2}$

$\frac{dC}{dt} =\frac{3-3C}{200+t}\\\\\frac{dC}{3-3C} =\frac{dt}{200+t} \\\\\int \frac{dC}{3-3C} =\int\frac{dt}{200+t} \\\\-\frac{1}{3}\ln \left|3-3C\right|=\ln \left|200+t\right|+D\\\\$

We solve for C(t)

$C(t)=1+D(200+t)^{-3}$

D is the constant of integration, to find it we use the initial condition $C(0)=\frac{1}{2}$

$C(0)=1+D(200+0)^{-3}\\\frac{1}{2} =1+D(200+0)^{-3}\\D=-4000000$

So the concentration of the solution in the tank at any time t (before the tank overflows) is

$C(t)=1-4000000(200+t)^{-3}$

(a) The amount of salt in the tank at any time prior to the instant when the solution begins to overflow is just the concentration of the solution times its volume

$(1-4000000(200+t)^{-3})(200+t)\\\left(1-\frac{4000000}{\left(200+t\right)^3}\right)\left(200+t\right)$

(b) Since the tank can hold 500 gallons, it will begin to overflow when the volume is exactly 500 gal. We noticed before that the volume of the solution at time t is $V(t)=200+t$ . Solving the equation

$200+t=500\\t=300$

tells us that the tank will begin to overflow at 300 minutes. Thus the concentration at that time is

$C(300)=1-4000000(200+300)^{-3}\\\\C(300)= \frac{121}{125}\:\frac{lb}{gal}$

(c) If the tank had infinite capacity the concentration would then converge to,

$\lim_{t \to \infty} C(t)= \lim_{t \to \infty} 1-4000000\left(200+t\right)^{-3}\\\\\lim _{t\to \infty \:}\left(1\right)-\lim _{t\to \infty \:}\left(4000000\left(200+t\right)^{-3}\right)\\\\1-0\\\\1$

The theoretical limiting concentration if the tank has infinite capacity is $1\:\frac{lb}{gal}$

4 0

3 years ago

Which ratio is equivlant to 3:4

Crank

Answer:

There is multiple. You didnt give options so heres all of them.

Step-by-step explanation:

6 : 8 9 : 12 12 : 16 15 : 20 18 : 24 21 : 28 24 : 32 27 : 36 30 : 40 33 : 44 36 : 48 39 : 52 42 : 56 45 : 60 48 : 64 51 : 68 54 : 72 57 : 76 60 : 80 63 : 84 66 : 88 69 : 92 72 : 96 75 : 1007 8 : 104 81 : 108 84 : 112 87 : 116 90 : 120 93 : 124 96 : 128 99 : 132 102 : 136 105 : 140 108 : 144 111 : 148 114 : 152 117 : 156 120 : 160 123 : 164 126 : 168 129 : 172 132 : 1761 35 : 180 138 : 184 141 : 188 144 : 192 147 : 196 150 : 200 153 : 204 156 : 208 159 : 2121 62 : 216 165 : 220 168 : 224 171 : 228 174 : 232 177 : 236 180 : 240 183 : 244 186 : 248 189 : 252 192 : 256 195 : 260 198 : 264 201 : 268 204 : 272 207 : 276 210 : 280 213 : 284 216 : 288 219 : 292 222 : 296 225 : 300 228 : 304 231 : 308 234 : 312 237 : 316 240 : 320 243 : 324246 : 328249 : 332252 : 336255 : 340258 : 344261 : 348264 : 352267 : 356270 : 360273 : 364276 : 368279 : 372282 : 376285 : 380288 : 384291 : 388

etc etc

4 0

3 years ago

Will mark brainliest!!! Solve for x

antiseptic1488 [7]

Answer:

3.5

Step-by-step explanation:

49÷14 is how you get x

4 0

3 years ago