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2 years ago

4

Assume the two lines ab and xy intersect as in the diagram below. which of the following statements are true. Check all that App

ly.

2 answers:

Feliz [49]2 years ago

5 0

Answer:

Correct answer options are A and D.

Step-by-step explanation:

Vertical angles are those that are opposite and equal formed when two lines cross.

∠ARY and ∠XRB are vertical angles

Perpendicular lines are those that intersect to form a right-angle at 90°

Line AB and Line XY are perpendicular lines

geniusboy [140]2 years ago

3 0

Answer:

Option A and Option D.

Step-by-step explanation:

In the given question AB and XY are two lines which intersect each other at point P.

A). ∠ARY and ∠XRB are vertical angles. True.

B). ∠ARY and ∠XRB are supplementary. False.

∠ARY + ∠XRB = 180° but they are not the adjacent angles formed at a point.

They are the vertical angles.

C).∠ARY and ∠XRB are complementary. False.

Since these angle are vertical angles.

D). AB and XY are perpendicular. True.

It's given in the question ∠BRY = 90°

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What is the radical form of the expression?<br><br> (5x3y4)37<br> ↡

hram777 [196]

Option 4: $\sqrt[7]{125x^9y^{12}}$ is the right answer

Step-by-step explanation:

Given expression is:

$(\sqrt{5x^3y^4})^{\frac{3}{7}$

In order to convert an exponent into radical form, the power should be in the form of 1/x where x is any number

so,

in case of 3/7, it will be broken down

$(\sqrt{5x^3y^4})^{3*\frac{1}{7}}$

The 1/7 will be converted into base of radical while 3 will be he exponent

$\sqrt[7]{(5x^3y^4)^3}$

Multiplying exponents

$=\sqrt[7]{5^3x^{3*3}y^{4*3}}\\=\sqrt[7]{125x^9y^{12}}$

Hence,

Option 4: $\sqrt[7]{125x^9y^{12}}$ is the right answer

Keywords: Radicals, Exponents

Learn more about exponents at:

#LearnwithBrainly

4 0

2 years ago

mrs sogiba is catering for her sons birthday party. she has invited some of her sons friends to the party. the number of friends

Alex

The only ones that would work are if the factor is composite and if 1 is added to it it can be divided into 48. So 15 would work. The factor 15 is composite, and if 1 is added it can be divided into 48.Therefore the son is 15.

7 0

2 years ago

Read 2 more answers

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

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

6 0

2 years ago

By how many factors of 10 did you multiply 0.008? Why?

Tamiku [17]

Answer: 3

Step-by-step explanation:

Move the decimal back to its point of origin

7 0

2 years ago

A student claims that 3x + 2y = 5xy. Use a visual representation to show whether this is true or false

Ket [755]

Answer:

It is true, except when x = 2/5, since it is an asymptote.

Step-by-step explanation:

3x=5xy-2y

3x=y(5x-2)

3x/5x-2=y

5x-2=0 -------> 5x=2 --------> x=2/5

7 0

2 years ago