# Mary Wants To Hang A Mirror In Her Room. The Mirror And Frame Must Have An Area Of 7 Square Feet. The Mirror Is 2 Feet Wide And 3 Feet Long. Which Quadratic Equation Can Be Used To Determine The Thickness Of The Frame, X? Square With An Inner Frame With Height Of 2ft On The Left Frame And Width Of 3ft On The Top. Arrow On The Bottom Frame With An X And An Arrow On The Right Frame With An X. X2 + 14x − 2 = 0 2×2 + 10x − 7 = 0 3×2 + 12x − 7 = 0 4×2 + 10x − 1 = 0

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## Mary wants to hang a mirror in her room. The mirror and frame must have an area of 7 square feet. The mirror is 2 feet wide and 3 feet long. Which quadratic equation can be used to determine the thickness of the frame, x? Square with an inner frame with height of 2ft on the left frame and width of 3ft on the top. Arrow on the bottom frame with an x and an arrow on the right frame with an x. x2 + 14x − 2 = 0 2×2 + 10x − 7 = 0 3×2 + 12x − 7 = 0 4×2 + 10x − 1 = 0

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If h(x) = 4x 8 and j(x) = x, solve h[j(3)] and select the correct… – Answer: Option (b) is correct. We have to solve for h[j(3)] and choose the correct option from given options. Function composition is when one function is inside of another function and we evaluate the value of a function by plug one into into another function.MrSmoot MrSmoot. Answer: h[j(3)] = -20. Step-by-step explanation Step-by-step explanation: h(x) = 4x − 8 and j(x) = −x. We have to solve h[j(3)]. The solid shown below is made of 10 small cubes of side 2 cm each.algebra questions and answers. Transcribed Image Text from this Question. Question 10(Multiple Choice Worth 1 points) (07 09) If h(x) 5x – 3 and j(x) –2x, solve hj(2)] and select the correct answer below O 17 0 23 9 17 23.

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Solved: Question 10(Multiple Choice Worth 1 Points)… | Chegg.com – The mirror is 2 feet wide and 3 feet long. Which quadratic equation can be used to determine the thickness of the frame, x? what bout If h(x) = 4x − 8 and j(x) = −x, solve h[j(3)] and select the correct answer below.

**Age Word Problems – MathHelp.com – Algebra Help** – .

**TRANSLATING WORDS INTO ALGEBRAIC EXPRESSIONS!** – .

**Find the Interval That a Linear First Order Differential Equation Has a Unique Solution** – – [Narrator] Welcome to

a lesson on determining the intervals for which

a linear first order differential equation

would have unique solutions as well as the interval for

which the differential equation would have a unique solution

containing an initial value.

If we have a first order

differential equation in this form here with

this initial condition if P of x and f of x are

continuous on an open interval from a to b, then there

exists a unique solution for every x in the interval. So it follows that if the

interval contains x sub zero from the initial condition

then there exists a unique solution on the

interval that satisfies the initial value problem. So a couple things to notice here… This is function P of x. This is function f of x and

x sub zero is the x value from the initial condition. Also notice this does not

tell us what the solution is or how to find it, it just tells us if there's a unique solution. Also y sub zero from the initial

condition does not affect the interval and the interval

containing x sub zero is sometimes called the

interval of validity. So the main idea is we

want to find the interval on which both P of x and f

of x are continuous and this will be the interval for

which unique solutions exist. So let's take a look at some examples. We want to find the intervals for which the DE has unique solutions. Then state the interval

containing the initial condition. The first step is to

recognize that we do have a linear first order

differential equation. However, it's currently

not in the correct form or the form given here below. We want our first term here to be dy dx. So we're going to divide

everything by x to begin with. So we'll have the differential

equation dy dx plus, I'm going to write this as

three divided by x times y so it fits this form equals four x. Now we should recognize

that P of x is equal to three divided by x and

f of x is equal to four x. Now we'll find where P of x is continuous, find where f of x is continuous, and then find the intersection

of those two intervals. Well, for P of x, we

know x can't equal zero because we'd have division by zero. So P of x is continuous

from negative infinity to zero or from zero to infinity. F of x is continuous for

all real values of x. Therefore, we can say

that f of x is continuous from negative infinity

to positive infinity. So now our main goal

is to find the interval for which both of these

functions are continuous or the intersection of

these two intervals. Well, the intersection

of these two intervals would just be the interval

for which P of x is continuous or this interval here. Which means this is the interval for which the original differential equation would have unique solutions

for all values of x in this interval. So that's the first part of the question and now if we look at the

given initial condition, we have y of one equals two. So we want to find the interval

containing x equals one which would be the second interval here and because it's in this interval, we know that this initial

value problem is going to have a unique solution. Remember these intervals

represent x values only. Let's take a look at another example. Same question, different

differential equation. So the first step is to put

the differential equation in the correct form. So here we're going to divide everything by the quantity x minus three. So P of x is going to be

equal to natural log x divided by the quantity x minus three and f of x is going to be equal to two x divided by the quantity x minus three. Now we'll start by determining

where P of x is continuous. Now there are two things to consider here. We know x can't equal three

because we'd have division by zero but the domain for

natural log x is when x is greater than zero or the

interval from zero to infinity. So P of x will be

continuous on this interval as long as x doesn't equal three. So P of x is continuous

on the open interval from zero to three or

from three to infinity. So it's where natural log x is continuous except we must exclude three. Now, looking at f of x, the

numerator is a linear function which is always continuous but notice we would have division by

zero when x equals three. So we know x can't equal three if we want f of x to be continuous. So f of x is continuous

from negative infinity to three or from three to infinity which means the differential equation will have unique solutions

on the intersection of these two intervals or when both P of x and f of x are continuous. So comparing these intervals,

notice that the intersection of these two intervals

would just be the interval for P of x. We would have to exclude the interval from negative infinity to zero. So we'd have the open

interval from zero to three or the open interval

from three to infinity. Now going back to our initial condition, we have y of one equals two. So we're going to find the interval that contains x equals one which would be the open interval from zero to three and because it's in this interval, we know that this initial value problem has a unique solution. Okay, let's take a look at one more. Same question, different

differential equation. Notice how this time,

the differential equation is in the correct form

so we should recognize that P of x is equal

to tangent x and f of x is equal to sin x. For P of x, remember

tangent theta is equal to y over x on the unit circle. So if we look at the coordinate

plane of the unit circle, P of x will be undefined or discontinuous whenever the angle has a terminal

side where x equals zero. So x equals zero is actually the y axis. So when x equals pi over

two or three pi over two, P of x will be discontinuous

but it will be continuous in between these two angles. So we could start by saying

that P of x is continuous on the open interval from pi

over two to three pi over two but for the same reason why

P of x would be discontinuous at these two angles, it's

going to be discontinuous at any co-terminal angle

to these two as well. Notice how these angles

are just pi over two plus or minus multiples of pi. So we'd also have to include the interval from three pi over two to

five pi over two and so on. There's going to be an

infinite number of intervals. Moving to the left, we could also start at negative pi over two

and go to pi over two. Again, and so on. So these are the

intervals for which P of x would be continuous. For f of x equals sin x,

there are no restrictions. X can be any real number. So f of x is continuous

from negative infinity to positive infinity which

means the differential equation is going to have unique

solutions on the intersection of these two intervals which

would just be the interval where P of x is continuous. So I'm going to go ahead and state this as the intervals where P of x is continuous and then looking at our initial condition, we have y of pi equals zero. So we're going to find the interval that contains x equals pi which

would be this interval here and because it's in this interval, we know this initial value

problem has a unique solution. And I'll go ahead and write

the interval down here. Pi over two, three pi over two

and this is an open interval. Okay, I hope you found this helpful. .