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3 years ago

Question 8. I need help with

Mathematics

1 answer:

ziro4ka [17]3 years ago

4 0

Answer:B is the right answer. Everything is divisible by 2x so divide 2x by all three components and leave remaining components letters in parentheses)

Step-by-step explanation:

You might be interested in

If ABC is reflected

Anna [14]

Answer:

option b is the answer of your question

4 0

3 years ago

1/3x(-6+27y-51z)simplify

Naya [18.7K]

Final result :

9y - 17z - 2
Step by step solution :

Step 1 :

1
Simplify —
3
Equation at the end of step 1 :

1
— • (27y - 51z - 6)
3
Step 2 :

Pulling out like terms :

3.1 Pull out like factors :

27y - 51z - 6 = 3 • (9y - 17z - 2)

Final result :

9y - 17z - 2

8 0

3 years ago

Read 2 more answers

10) A rhombus has a<br> diagonal 8.6 mm long<br> and an area of 81.7 mm. what is the<br> perimeter?

BaLLatris [955]

Answer:

P=41.72

Step-by-step explanation:

S=ACxDB/2

81.7=8.6xDB/2

81.7=4.3xDB|:4.3

19(mm)=DB

DO=19/2=9.5

OC=8.6/2=4.3

(O is the center of the rhombus, where two diagonals meet)

a²+b²=c² (DO²+OC²=DC²)

9.5²+4.3²=c²

90.25+18,49=c²

√108,74=√c²

c≈10.43

P=4c

P=4x10.43

P=41.72

Hope it helps:)

6 0

2 years ago

Which radical expression is in simplified form A. 11y/sqrt3 B. sqrt6/5y C. sqrt17/sqrt4 D. sqrt25/18

Mkey [24]

Answer: $\frac{\sqrt{6}}{5y}$

Step-by-step explanation:

The given radical expressions are:

$A. \frac{11y}{\sqrt{3}}\\\text{It is not completely simplified ,}\\\text{since denominator is radical, we need to rationalize it first to simplify. }\\\\B.\frac{\sqrt{6}}{5y}\\\text{It is completely simplified ,}\\\text{since its denominator is not radical and 6 is not a perfect square. }\\\\C. \frac{\sqrt{17}}{\sqrt{4}}\\\text{It is not completely simplified ,}\\\text{since denominator is radical, we need to rationalize it first to simplify. }\\\\$ $\\\\D. \frac{\sqrt{25}}{18}=\frac{5}{18}\\\text{Hence, expression D was not completely simplified .}$

8 0

2 years ago

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

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

pounds of salt.

7 0

1 year ago