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2 years ago

PLEASE HELP ME WITH THESE!! i will give brainliest to best answer!!!

Mathematics

1 answer:

dimaraw [331]2 years ago

8 0

Answer:

$1) {11}^{12}$

$2) {13}^{10}$

$3) {14}^{19}$

$4) {2}^{8}$

$5) {7}^{14}$

$6) {3}^{12}$

$7) {17}^{16}$

$8) {5}^{9}$

$9) {16}^{6}$

$10) {18}^{9}$

$11) {9}^{14}$

$12) {8}^{10}$

$13) {12}^{13}$

$14) {17}^{17}$

$15) {5}^{6}$

$16) {3}^{4}$

$17) {19}^{12}$

$18) {20}^{9}$

$19) {4}^{7}$

$20) {6}^{9}$

$21) {10}^{15}$

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At a restaurant the cost for a breakfast taco and a small glass of milk is $2.10. The cost for 2 tacos and 3 small glasses of mi

tangare [24]

Answer:

The price for a small glass of milk is $0.95, and the price for a breakfast taco is $1.15.

Step-by-step explanation:

Let the price of a taco be x and the price for a small glass of milk be y.

$x + y = 2.1$

$2x + 3y = 5.15$

Multiply the first equation by 2:

$2x + 2y = 4.2$

Subtract the new third equation from the second equation:

$(2x + 3y) - (2x + 2y) = 5.15 - 4.2$

$y = 0.95$

Hence, the price for a small glass of milk is $0.95.

Substitute y into the first equation.

$x + 0.95 = 2.1$

$x = 1.15$

Therefore, the price for a breakfast taco is $1.15.

Hope this helped!

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2 years ago

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AB and BC share point B to form AC. If AC=7 and BC=5, then what is the length of AB?

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$|AC|=7;\ |BC|=5\\\\|AC|=|AB|+|BC|\\\\|AB|+5=7\ \ \ \ \ |subtract\ 5\ from\ both\ sides\\\\\boxed{|AB|=2}$

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3 years ago

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A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$