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Marina86 [1]
2 years ago
6

Please help ASAP. Quadrilateral CDEF is inscribed in circle A. If the m∠C = 7x + 11° and m∠E = 7x + 1°, what is the measure of ∠

E?
A) 95
B) 99
C) 81
D) 85
Mathematics
2 answers:
RSB [31]2 years ago
5 0

Answer:

The answer is D

Step-by-step explanation:

∠C + ∠E = 180°(consecutive angles)

7x +11° + 7x + 1° = 180°

14x + 12° = 180°

14x = 168°

x = 12°

∠E = 7x + 1°

= 7(12°) + 1

= 85°

Irina-Kira [14]2 years ago
5 0

Since angle C and angle E are opposite to each other and C and E are vertices of a cyclic quadrilateral,

m∠C + m∠E = 180°

7x + 11° + 7x + 1° = 180°

14x + 12° = 180°

14x = 168°

7x = 84°

∠E = 7x + 1°

∠E = 85°

Therefore, the answer is option D.

I hope this helps! If you have any doubts, feel free to ask.

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Since there are 6 students out of which one needs to be selected, the principal chose two die on which there are six numbers each numbered from 1 , 2, 3, 4, 5, 6.

Since there are two dice, the total possible outcome is 36.

Hence, the probability of getting one number each is 1/36

Hence, the principal used a fair method because each result is an equally likely possible outcome.

Option B is correct.

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A 400 gallon tank initially contains 100 gal of brine containing 50 pounds of salt. Brine containing 1 pound of salt per gallon
posledela

Answer:

The amount of salt in the tank when it is full of brine is 393.75 pounds.

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 <em>Q(t)</em> = Rate at which <em>Q(t)</em> enters the tank – Rate at which <em>Q(t)</em> 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 y<em>(t)</em> be the amount of salt (in pounds) in the tank at time <em>t</em> (in seconds). Then we can represent the situation with the below picture.

Then the differential equation we’re after is

\frac{dy}{dt} = (Rate \:in)- (Rate \:out)\\\\\frac{dy}{dt} = 5 \:\frac{gal}{s} \cdot 1 \:\frac{pound}{gal}-3 \:\frac{gal}{s}\cdot \frac{y(t)}{V(t)}  \:\frac{pound}{gal}\\\\\frac{dy}{dt} =5\:\frac{pound}{s}-3 \frac{y(t)}{V(t)}  \:\frac{pound}{s}

V(t) is the volume of brine in the tank at time <em>t. </em>To find it we know that at time 0 there were 100 gallons, 5 gallons are added and 3 are drained, and the net increase is 2 gallons per second. So,

V(t)=100 + 2t

We can then write the initial value problem:

\frac{dy}{dt} =5-\frac{3y}{100+2t} , \quad y(0)=50

We have a linear differential equation. A first-order linear differential equation is one that can be put into the form

\frac{dy}{dx}+P(x)y =Q(x)

where <em>P</em> and <em>Q</em> are continuous functions on a given interval.

In our case, we have that

\frac{dy}{dt}+\frac{3y}{100+2t} =5 , \quad y(0)=50

The solution process for a first order linear differential equation is as follows.

Step 1: Find the integrating factor, \mu \left( x \right), using \mu \left( x \right) = \,{{\bf{e}}^{\int{{P\left( x \right)\,dx}}}

\mu \left( t \right) = \,{{e}}^{\int{{\frac{3}{100+2t}\,dt}}}\\\int \frac{3}{100+2t}dt=\frac{3}{2}\ln \left|100+2t\right|\\\\\mu \left( t \right) =e^{\frac{3}{2}\ln \left|100+2t\right|}\\\\\mu \left( t \right) =(100+2t)^{\frac{3}{2}

Step 2: Multiply everything in the differential equation by \mu \left( x \right) and verify that the left side becomes the product rule \left( {\mu \left( t \right)y\left( t \right)} \right)' and write it as such.

\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+\frac{3y}{100+2t}\cdot \left(100+2t\right)^{\frac{3}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+3y\cdot \left(100+2t\right)^{\frac{1}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})=5\left(100+2t\right)^{\frac{3}{2}}

Step 3: Integrate both sides.

\int \frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})dt=\int 5\left(100+2t\right)^{\frac{3}{2}}dt\\\\y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }+ C

Step 4: Find the value of the constant and solve for the solution y(t).

50 \left(100+2(0)\right)^{\frac{3}{2}}=(100+2(0))^{\frac{5}{2} }+ C\\\\100000+C=50000\\\\C=-50000

y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }-50000\\\\y(t)=100+2t-\frac{50000}{\left(100+2t\right)^{\frac{3}{2}}}

Now, the tank is full of brine when:

V(t) = 400\\100+2t=400\\t=150

The amount of salt in the tank when it is full of brine is

y(150)=100+2(150)-\frac{50000}{\left(100+2(150)\right)^{\frac{3}{2}}}\\\\y(150)=393.75

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Hey mark me brainlest but ur answer is 27/45


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Vikentia [17]

Answer:

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32 = 2 × 16

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Common from all variable part comes b².

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