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3 years ago

(⅔)-⁴ (two over three to the power minus 4)I need answer asap pleaseeeee

Mathematics

1 answer:

Fudgin [204]3 years ago

8 0

Answer:

81/16

Step-by-step explanation:

(⅔)-⁴

81/16

= 5.0625

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In the diagram below, ST is parallel to QR. angle P is 40 degrees, and QST is 2R+8. Find the measure of STR in degrees.

lozanna [386]

Answer:

148º

Step-by-step explanation:

Use the information given:

The measure of a straight line is 180º. So angle PST=180-(2R+8)
Since ST and QR are parallel, angle PST=angle PQR. Now, angle PQR=180-(2R+8)
The sum of a triangle's interior angles is 180º. So 40+180-2R-8+R=180º.
Combine like terms, 212-R=180
Subtract, -R=-32
Convert to positive, R=32º

With R=32º, we can find angle QST and PQR. Angle QST=2(32)+8=64+8=72º. So angle PQR=180º-72º=108º. The sum of a quadrilateral's interior angles is 360º:

32º+72º+108º+STR=360º
Combine like terms, 212º+STR=360º
Subtract, STR=148º

3 0

3 years ago

Can u answer this for me

I am Lyosha [343]

200-70=130/10=13. so 13 weeks

6 0

3 years ago

Which is not a representation of the equation y = 0.5x - 2?

Studentka2010 [4]

Answer:

D is incorrect so ur answer is D

8 0

3 years ago

A tank contains 60 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

MissTica

Answer:

(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L

Step-by-step explanation:

(a) Initial amount of salt in tank

The tank initially contains 60 kg of salt.

(b) Amount of salt after 4.5 h

$\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into tank}\\\text{and }r_{0} =\text{rate of salt going out of tank}$

(i) Set up an expression for the rate of change of salt concentration.

$\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with no salt, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{3 L}}{\text{1 min}} \times \dfrac {A\text{ kg}}{\text{1000 L}} =\dfrac{3A}{1000}\text{ kg/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -0.003A \text{ kg/min}$

(ii) Integrate the expression

$\dfrac{\text{d}A}{\text{d}t} = -0.003A\\\\\dfrac{\text{d}A}{A} = -0.003\text{d}t\\\\\int \dfrac{\text{d}A}{A} = -\int 0.003\text{d}t\\\\\ln A = -0.003t + C$

(iii) Find the constant of integration

$\ln A = -0.003t + C\\\text{At t = 0, A = 60 kg/1000 L = 0.060 kg/L} \\\ln (0.060) = -0.003\times0 + C\\C = \ln(0.060)$

(iv) Solve for A as a function of time.

$\text{The integrated rate expression is}\\\ln A = -0.003t + \ln(0.060)\\\text{Solve for } A\\A = 0.060e^{-0.003t}$

(v) Calculate the amount of salt after 4.5 h

a. Convert hours to minutes

$\text{Time} = \text{4.5 h} \times \dfrac{\text{60 min}}{\text{1h}} = \text{270 min}$

b.Calculate the concentration

$A = 0.060e^{-0.003t} = 0.060e^{-0.003\times270} = 0.060e^{-0.81} = 0.060 \times 0.445 = \text{0.0267 kg/L}$

c. Calculate the volume

The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.

The volume added in 4.5 h is

$\text{Volume added} = \text{270 min} \times \dfrac{\text{3 L}}{\text{1 min}} = \text{810 L}$

Total volume in tank = 1000 L + 810 L = 1810 L

d. Calculate the mass of salt in the tank

$\text{Mass of salt in tank } = \text{1810 L} \times \dfrac{\text{0.0267 kg}}{\text{1 L}} = \textbf{21.6 kg}$

(c) Concentration at infinite time

$\text{As t $\longrightarrow \, -\infty,\, e^{-\infty} \longrightarrow \, 0$, so A $\longrightarrow \, 0$.}$

This makes sense, because the salt is continuously being flushed out by the fresh water coming in.

The graph below shows how the concentration of salt varies with time.

3 0

3 years ago

What is the slope of the line that passes through (1, -5) and (3, -8)?

aleksklad [387]

Answer:

The slope is : -1.5

3 0

3 years ago

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(⅔)-⁴ (two over three to the power minus 4)​I need answer asap pleaseeeee

(⅔)-⁴ (two over three to the power minus 4)I need answer asap pleaseeeee