Graph the system of equations below on a piece of paper. What is the solution?
y = x – 1
y = 3x + 1

 

A. (2, 1)
 

B. (1, –2)
 

C. (–1, –1)
 

D. (–1, –2)

The derivative[tex]\( \frac{dw}{dt} \) is \( \frac{7t^6 e^{4-t}}{2+9t} - \frac{t^7 e^{4-t}}{2+9t} - \frac{9t^7 e^{4-t}}{(2+9t)^2} \).[/tex]

To find [tex]\( \frac{dw}{dt} \)[/tex] using the chain rule for the given function[tex]\( w = \frac{x e^y}{z} \), where \( x = t^7 \), \( y = 4 - t \), and \( z = 2 + 9t \)[/tex], follow these steps:

1. **Express ( w ) in terms of ( t ):**

  Substitute ( x ), ( y ), and ( z ) into ( w ):

[tex]\[ w = \frac{x e^y}{z} = \frac{(t^7) e^{(4 - t)}}{2 + 9t} \][/tex]

2. **Apply the chain rule:**

  The chain rule states that for a function ( w(t) ) defined implicitly by ( w = f(x(t), y(t), z(t)) ), the derivative [tex]\( \frac{dw}{dt} \)[/tex] is given by:

[tex]\[ \frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} + \frac{\partial w}{\partial z} \cdot \frac{dz}{dt} \][/tex]

3. **Compute partial derivatives of ( w ) with respect to ( x ), ( y ), and ( z ):**

[tex]\( \frac{\partial w}{\partial x} = \frac{e^y}{z} \)[/tex]  

  [tex]\( \frac{\partial w}{\partial y} = \frac{x e^y}{z} \)[/tex]  

[tex]\( \frac{\partial w}{\partial z} = -\frac{x e^y}{z^2} \)[/tex]

4. **Compute [tex]\( \frac{dx}{dt} \), \( \frac{dy}{dt} \), and \( \frac{dz}{dt} \):**[/tex]

[tex]\( \frac{dx}{dt} = 7t^6 \)[/tex]  

[tex]\( \frac{dy}{dt} = -1 \)[/tex]

[tex]\( \frac{dz}{dt} = 9 \)[/tex]

5. **Substitute these into the chain rule formula:**

[tex]\[ \frac{dw}{dt} = \frac{e^y}{z} \cdot 7t^6 + \frac{x e^y}{z} \cdot (-1) + \left(-\frac{x e^y}{z^2}\right) \cdot 9 \][/tex]

6. **Substitute[tex]\( x = t^7 \), \( y = 4 - t \), \( z = 2 + 9t \)[/tex] into the expression:**

[tex]\( e^y = e^{4 - t} \)[/tex]

  Substitute these values into the formula for [tex]\( \frac{dw}{dt} \):[/tex]

[tex]\[ \frac{dw}{dt} = \frac{e^{4 - t}}{2 + 9t} \cdot 7t^6 - \frac{t^7 \cdot e^{4 - t}}{2 + 9t} - \frac{9t^7 \cdot e^{4 - t}}{(2 + 9t)^2} \][/tex]

Therefore, [tex]\( \frac{dw}{dt} \)[/tex] is:

[tex]{\frac{dw}{dt} = \frac{7t^6 e^{4 - t}}{2 + 9t} - \frac{t^7 e^{4 - t}}{2 + 9t} - \frac{9t^7 e^{4 - t}}{(2 + 9t)^2} } \][/tex]

Answer:

Hence, AB=12.

Step-by-step explanation:

We are given that the perpendicular bisector of side AB of ∆ABC intersects side BC at point D.

this means that side AE=BE.

Also we could clear;ly observe that

ΔBED≅ΔAED

( since AE=BE, side ED common, ∠BED=∠AED

so by SAS congruency the two triangles are congruent)

Now we are given that:

the perimeter of ∆ABC is 12 cm larger than the perimeter of ∆ACD.

i.e. AB+AC+BC=AC+AD+CD+12

AB+BC=AD+CD+12

as AD=BD

this means that AD+CD=BD+CD=BC

AB+BC=BC+12

AB=12

Hence AB=12

Answer:

The required length of [tex]AB[/tex] is [tex]12\rm\;{cm}[/tex].

Step-by-step explanation:

Given: The perpendicular bisector of side [tex]AB[/tex] of [tex]\bigtriangleup{ABC}[/tex] intersects side [tex]BC[/tex] at point [tex]D[/tex] and the perimeter of  [tex]\bigtriangleup{ACD}[/tex].

From the figure,

[tex]AE=BE[/tex]         .......(1)              (as [tex]DE[/tex] is perpendicular bisector of side [tex]AB[/tex])

Now, In [tex]\bigtriangleup{BED}[/tex] and [tex]\bigtriangleup{AED}[/tex]

     [tex]AE=BE[/tex]                                     ( from equation 1 )

[tex]\angle {BED} =\angle {AED}[/tex]                               ( Both [tex]90^\circ[/tex] )

    [tex]ED=ED[/tex]                                     ( Common side)

[tex]\bigtriangleup{BED}\cong\bigtriangleup{AED}[/tex]                              ( by SAS congruence rule)

      [tex]BD=AD[/tex]    .........(2)                   (by CPCT)

As per question,

The perimeter of ∆ABC is with 12 cm larger than the perimeter of ∆ACD.

[tex]AB+BC+AC=AC+CD+AD+12[/tex]

         [tex]AB+BC=AD+CD+12\\AD+CD=BD+CD\\AB+BC=BC+12\\[/tex]

                   [tex]AB=12\rm\;{cm}[/tex]

Hence, the length of [tex]AB[/tex] is [tex]12\rm\;{cm}[/tex].

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Answer:

5836200.

Step-by-step explanation:

Given  :  5836197 .

To find : Round 5836197 to the nearest hundred.

Solution : We have given 5836197

Step 1 : First, we look for the rounding place which is the hundreds place.

Step 2 : Rounding place is 97.

Step 3 : 97  is greater than  50 then it would be rounded up mean next number to 97 would be increase to 1 and 97 become 00.

Step 4 : 5836200.

Therefore, 5836200.

Answer: There are 95 students who scored between 35 and 61.

Step-by-step explanation:

Since we have given that

The following data : 11,35,61,70,79.

So, the median of this data would be = 61

First two data belongs to "First Quartile " i.e. Q₁

and the second quartile is the median i.e. 61.

The last two quartile belongs to "Third Quartile" i.e. Q₃

And we know that each quartile is the 25th percentile.

And we need "Number of students who scored between 35 and 61."

So, between 35 and 61 is 25% of total number of students.

So, Number of students who scored between 35 and 61 is given by

[tex]\dfrac{25}{100}\times 380\\\\=\dfrac{1}{4}\times 380\\\\=95[/tex]

Hence, There are 95 students who scored between 35 and 61.

The number of students who scored between 35 and 61 is 95

The 5 number summary is the value of the ;

The total Number of students = 380

The lower quartile (Lower 25%) = 35

The median (50%) = 61

This means that 25% of the total students scored between 35 and 61.

Hence, 95 students scored between 35 and 61.

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To find a, b, c for the solution:

Let's start by writing down the expression for the function x(t) and its derivative:

We have:

x(t) = at² + bt + c
and
x'(t) = 2at + b

Using x' and x into the differential equation x′ + 2x = t² + 4t + 7 gives us:

2at + b + 2*(at² + bt + c) = t² + 4t + 7
Expanding this gives:
2at² + 2bt + b + 4at + 2c = t² + 4t + 7

By equating the coefficients of equivalent powers of t on both sides, we get three equations:

For t² :
2a = 1
So, a = 1/2

For t:
2b + 4a = 4
Substitute a = 1/2 into the equation gives b = 1 - 2 = -1

For the constant term:
b + 2c = 7
Substituting b = -1 gives c = 4.

So the solution is a = 1/2, b = -1, c = 4.

So the specific solution of this differential equation is given by x(t) = (1/2)t² - t + 4.

The reduction factor of 0.8 means the lengths in the reduced triangle are 0.8 times those of the original.

Then the original 6 inch length is reduced to 0.8×6 inches = 4.8 inches in the reduced triangle.

Answer:

4.8 inches

Step-by-step explanation:

The scale factors are used to convert a figure into another one with similar characteristics but different lengths, in this example is a triangle, and in order to calculate the measure of the height you just have to multiply the original height by the scale factor:

6 inches * scale factor

6 inches* 0.8= 4.8 inches

So the resultant triangle will have a height of 4.8 inches.

Answer:

B

Step-by-step explanation:

Compute for the total balance:

total balance = $415 + $215 = $630

Then we compute for the total credit limit:

total credit limit = $1,000 + $750 = $1,750

The credit utilization would simply be the percentage ratio of total balance over total credit limit. That is:

credit utilization = ($630 / $1,750) * 100%

credit utilization = 36%

Answer:= 5/6

Step-by-step explanation:hope this helps

Answer: n=296

Step-by-step explanation: 9 x 23 + 3 x 39 - 28 + n = 296  

Hope this helps! <3

Answer:

Two column proofs are organized into statement and reason columns. Each statement must be justified in the reason column. The reason column will typically include "given", vocabulary definitions, and theorems.

Therefore, what can be used as a reason in a two-column proof are:

Postulates

Definitions

sin(a+b) = -1/7 +√455/42 = 0.8721804464845457

cos(a-b) = √13/42 - √35/7 =  -0.7761476987942811

tan(a+b )= (6√13/13 + √35) / (1 - 6√455/13) = -0.525

Given sin(a) = 6/7 and cos(b) = -1/6, with a in quadrant II and b in quadrant III, we need to utilize trigonometric identities to find sin(a+b), cos(a-b), and tan(a+b).

Firstly, since a is in quadrant II, cos(a) is negative. We use the identity sin²(a) + cos²(a)=1 to find cos(a):

Similarly, since b is in quadrant III, sin(b) is also negative. We use the identity sin²(b) + cos²(b)=1 to find sin(b):

Now we can use the angle addition and subtraction formulas:

1. sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

2. cos(a - b) = cos(a)cos(b) + sin(a)sin(b)

3. tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a)tan(b))

146.54 units are the total units will she sell her ninth year.

Here, we have,

periodic exponential decay is:

[tex]A = I(1-r)^{t/p}[/tex]

where,

A = amount

I = initial amount

r = rate

t = time

p = period

we know, the first year she sold 400 units,

thus year 0, 400 units, t = 0, A = 400.

so, we get,

periodic exponential decay is:

[tex]A = I(1-r)^{t/p}[/tex]

=> 400 = I × 1

=> 400 = I

therefore then, when, rate = 20 %

we get,

periodic exponential decay is:

[tex]A = I(1-r)^{t/p}[/tex]

[tex]= > A = 400(1-0.2)^{t/2}[/tex]

now, how many units will it had decreased by the 9th year?  t = 9

so, we get,

[tex]= > A = 400(1-0.2)^{9/2}[/tex]

=> A = 146.54

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After factoring x from both terms, you can factor the difference of two squares.

= x(x² -64)

= x(x -8)(x +8)

_____

It is worth remembering the "speciall form" that is the difference of two squares:

... a² - b² = (a -b)(a +b)

Answer:

x(x -8)(x +8)

Step-by-step explanation:

The coordinate of point C on line CE will be (8, 4).

What is Coordinates?

A pair of numbers which describe the exact position of a point on a cartesian plane by using the horizontal and vertical lines is called the coordinates.

Given that;

D is the midpoint of CE.

E has coordinates (-3,-2), and D has coordinates (2 1/2, 1).

Now, By the definition of midpoint;

Let the coordinate of point C = (x, y)

Then,

((x + (-3))/2 , (y + (-2))/2) = (2 1/2, 1)

By comparison we get;

x + (-3) / 2= 2 1/2

x - 3 = 2 (5/2)

x - 3 = 5

x = 3 + 5

x = 8

And, (y + (-2))/2 = 1

y - 2 = 2

y = 2 + 2

y = 4

Thus, The coordinate of point C = (x, y) = (8, 4)

So, The coordinate of point C on line CE will be (8, 4).

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