Zoe is a party planner. She orders cupcakes and sheet cakes from Creative Cakes whenever she needs beautiful cakes to serve. One sheet cake serves 24 people and 2 dozen cupcakes serve 24 people. For the Benson's bridal shower she ordered 8 dozen cupcakes and 3 sheet cakes and paid $160.81. For the Nygaard's 50th wedding anniversary she ordered 6 dozen cupcakes and 15 sheet cakes and paid $342.33. She is planning a graduation party for her niece's high school class. She wants the party to be nice but she offered to pay for the cakes herself so she wants to choose the most economical plan. Zoe estimates 250 people will attend the party.

Answer:

0

Step-by-step explanation:

Arc sec(1)=0, tan(0)=0

9514 1404 393

Answer:

  (a) The blue line ... solution ... (-6, -2)..

Step-by-step explanation:

The second equation describes a line with negative slope and a y-intercept of -4. This is clearly the red line on the graph.

The blue line represents the equation 2x -3y = -6.

The point of intersection of the two lines is (-6, -2), so that is the solution to the system of equations. This, by itself, is sufficient for you to choose the correct answer.

Answer:

9/4 * 3/4 =  27/16  = 1  [tex]\frac{9}{16}[/tex]

Step-by-step explanation:

Answer:

21

Step-by-step explanation:

Answer:

(3,-2)

Step-by-step explanation:

-5/3x + 3 = 1/3x - 3

-5/3x = 1/3x - 6

-2x = -6

x = 3

y = -5/3(3) + 3

y = -5 + 3

y = -2

Answer:

it is 23

Step-by-step explanation:

first divide

then add

then subtract

Answer:

23

Step-by-step explanation:

Follow PEMDAS

16 ÷ 2 = 8

8 + 20 = 28

28 - 5 = 23

==========================================================

Explanation:

area of rectangle = L*W = 129

L*W = 129

x*(2x-9) = 129

2x^2-9x = 129

2x^2-9x-129 = 0

Apply the quadratic formula. We'll use a = 2, b = -9, c = -129.

[tex]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-9)\pm\sqrt{(-9)^2-4(2)(-129)}}{2(2)}\\\\x = \frac{9\pm\sqrt{1113}}{4}\\\\x \approx \frac{9\pm33.36165464}{4}\\\\x \approx \frac{9+33.36165464}{4}\ \text{ or } \ x \approx \frac{9-33.36165464}{4}\\\\x \approx \frac{42.36165464}{4}\ \text{ or } \ x \approx \frac{-24.36165462}{4}\\\\x \approx 10.59041366\ \text{ or } \ x \approx -6.09041364\\\\[/tex]

We ignore the negative solution because a negative length makes no sense.

The length is approximately L = 10.5904 cm.

The width is 2L-9 = 2*10.5904-9 = 12.1808 cm approximately.

As a quick check,

L*W = 10.5904*12.1808 = 128.99954432

which isn't too far off from 129. We have rounding error which is why we don't perfectly land on the target area value. If you wanted to get closer to the value 129, then use more decimal digits in the approximations of L and W.

----------------------------

If you draw a diagonal in the rectangle, then you form two identical or congruent right triangles.

Focusing on one of those triangles, we have

Apply the pythagorean theorem

a^2+b^2 = c^2

c = sqrt( a^2 + b^2 )

c = sqrt( (10.5904)^2 + (12.1808)^2 )

c = 16.1408940520653

c = 16.14

The diagonal is roughly 16.14 cm long.

First, I'll make f(x) = sin(px) + cos(px) because this expression shows up quite a lot, and such a substitution makes life a bit easier for us.

Let's apply the first derivative of this f(x) function.

[tex]f(x) = \sin(px)+\cos(px)\\\\f'(x) = \frac{d}{dx}[f(x)]\\\\f'(x) = \frac{d}{dx}[\sin(px)+\cos(px)]\\\\f'(x) = \frac{d}{dx}[\sin(px)]+\frac{d}{dx}[\cos(px)]\\\\f'(x) = p\cos(px)-p\sin(px)\\\\ f'(x) = p(\cos(px)-\sin(px))\\\\[/tex]

Now apply the derivative to that to get the second derivative

[tex]f''(x) = \frac{d}{dx}[f'(x)]\\\\f''(x) = \frac{d}{dx}[p(\cos(px)-\sin(px))]\\\\ f''(x) = p*\left(\frac{d}{dx}[\cos(px)]-\frac{d}{dx}[\sin(px)]\right)\\\\ f''(x) = p*\left(-p\sin(px)-p\cos(px)\right)\\\\ f''(x) = -p^2*\left(\sin(px)+\cos(px)\right)\\\\ f''(x) = -p^2*f(x)\\\\[/tex]

We can see that f '' (x) is just a scalar multiple of f(x). That multiple of course being -p^2.

Keep in mind that we haven't actually found dy/dx yet, or its second derivative counterpart either.

-----------------------------------

Let's compute dy/dx. We'll use f(x) as defined earlier.

[tex]y = \ln\left(\sin(px)+\cos(px)\right)\\\\y = \ln\left(f(x)\right)\\\\\frac{dy}{dx} = \frac{d}{dx}\left[y\right]\\\\\frac{dy}{dx} = \frac{d}{dx}\left[\ln\left(f(x)\right)\right]\\\\\frac{dy}{dx} = \frac{1}{f(x)}*\frac{d}{dx}\left[f(x)\right]\\\\\frac{dy}{dx} = \frac{f'(x)}{f(x)}\\\\[/tex]

Use the chain rule here.

There's no need to plug in the expressions f(x) or f ' (x) as you'll see in the last section below.

Now use the quotient rule to find the second derivative of y

[tex]\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]\\\\\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{f'(x)}{f(x)}\right]\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-f'(x)*f'(x)}{(f(x))^2}\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2}\\\\[/tex]

If you need a refresher on the quotient rule, then

[tex]\frac{d}{dx}\left[\frac{P}{Q}\right] = \frac{P'*Q - P*Q'}{Q^2}\\\\[/tex]

where P and Q are functions of x.

-----------------------------------

This then means

[tex]\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} + \left(\frac{f'(x)}{f(x)}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} +\frac{(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2+(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\[/tex]

Note the cancellation of -(f ' (x))^2 with (f ' (x))^2

------------------------------------

Let's then replace f '' (x) with -p^2*f(x)

This allows us to form  ( f(x) )^2 in the numerator to cancel out with the denominator.

[tex]\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*f(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*(f(x))^2}{(f(x))^2} + p^2\\\\-p^2 + p^2\\\\0\\\\[/tex]

So this concludes the proof that [tex]\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2 = 0\\\\[/tex] when [tex]y = \ln\left(\sin(px)+\cos(px)\right)\\\\[/tex]

Side note: This is an example of showing that the given y function is a solution to the given second order linear differential equation.

Answer:

y=5x-1 I think because the snd option doesn't make sense but you should try y =5x-1

Area of parallelogram = b × h

Base = x + 7

Height = x + 3

ATQ

Area = (x + 7) ( x + 3)

x² + 3x + 7x + 21

x² + 10x + 21

Answered by Gauthmath must click thanks and mark brainliest

Answer:

x = 74.74 m

Step-by-step explanation:

you need simple trigonometry to solve it

cosine of an angle is the ratio of the adjacent side to the hypotenuse

cos(42.2) is about 0.74

so X / Hypotenuse = 0.74

we know hypotenuse is 101 m

X / 101 = 0.74

x = 74.74 m

hope this helped!

Answer:

5π/4 radians or 225°

Step-by-step explanation:

Answer:

A = 108 feet²

Step-by-step explanation:

Let the width is b.

Length = 3+b

Perimeter of the rectangle, P = 42 ft

Perimeter = 2(l+b)

42 = 2 (3+b+b)

21 = (3+2b)

21-3 = 2b

18 = 2b

b = 9 feet

Length, l = 3+9 = 12 feet

Area of the rectangle,

A = lb

So,

A = 12 × 9

A = 108 feet²

So, the area of the rectangle is 108 feet².

Answer:

49.09°

Step-by-step explanation:

c = circumference = 2×π×r

= 2× 22/7 ×7 = 44 ft

θ = 6/c × 360°

= 6/44 × 360° = 49.09°

Answer: 85 i think

Step-by-step explanation:

Answer:

D. 3

Step-by-step explanation:

A triangle can be defined as a two-dimensional shape that comprises three (3) sides, three (3) vertices and three (3) angles.

Simply stated, any polygon with three (3) lengths of sides is a triangle.

In Geometry, a triangle is considered to be the most important shape.

Generally, there are three (3) main types of triangle based on the length of their sides and these include;

I. Equilateral triangle: it has all of its three (3) sides and interior angles equal.

II. Isosceles triangle: it has two (2) of its sides equal in length and two (2) equal angles.

III. Scalene triangle: it has all of its three (3) sides and interior angles different in length and size respectively.

In Geometry, an acute angle can be defined as any angle that has its size less than ninety (90) degrees.

Hence, we can deduce that the greatest number of acute angles that a triangle can contain is three (3) because the sum of all the interior angles of a triangle is 180 degrees.

Answer:

6

Step-by-step explanation:

Given :

Sample size, n = 36

Sample variance, s² = 1296

The estimated standard error can be obtained using the relation :

Standard Error, S. E = standard deviation / √n

Standard deviation, s = √1296 = 36

S.E = 36/√36

S.E = 36/6

S.E = 6

Hence, estimated standard error = 6

Answer:

assuming youre asking for line of symmetry, it's a line that cuts a shape exactly in half.  

for example, a square has 4 lines of symmetry

Answer:

You dind't include the answer choices but it should look something like

[tex]50000(.8)^t[/tex]

Answer:

see explanation

Step-by-step explanation:

Using the Sine rule in all 3 questions

(1)

[tex]\frac{a}{sinA}[/tex] = [tex]\frac{b}{sinB}[/tex] , substitute values , firstly calculating ∠ B

[ ∠ B = 180° - (78 + 49)° = 180° - 127° = 53° ]

[tex]\frac{a}{sin78}[/tex] = [tex]\frac{18}{sin53}[/tex] ( cross- multiply )

a sin53° = 18 sin78° ( divide both sides by sin53° )

a = [tex]\frac{18sin78}{sin53}[/tex] ≈ 22.0 ( to the nearest tenth )

(3)

[tex]\frac{c}{sinC}[/tex] = [tex]\frac{a}{sinA}[/tex] , substitute values

[tex]\frac{35}{sinC}[/tex] = [tex]\frac{45}{sin134}[/tex] ( cross- multiply )

45 sinC = 35 sin134° ( divide both sides by 35 )

sinC = [tex]\frac{35sin134}{45}[/tex] , then

∠ C = [tex]sin^{-1}[/tex] ( [tex]\frac{35sin134}{45}[/tex] ) ≈ 34.0° ( to the nearest tenth )

(5)

Calculate the measure of ∠ B

∠ B = 180° - (38 + 92)° = 180° - 130° = 50°

[tex]\frac{a}{sinA}[/tex] = [tex]\frac{b}{sinB}[/tex] , substitute values

[tex]\frac{BC}{sin38}[/tex] = [tex]\frac{10}{sin50}[/tex] ( cross- multiply )

BC sin50° = 10 sin38° ( divide both sides by sin50° )

BC = [tex]\frac{10sin38}{sin50}[/tex] ≈ 8.0 ( to the nearest tenth )

Answer:

see explanation

Step-by-step explanation:

Using the Sine rule in all 3 questions

[tex]\frac{a}{sinA}[/tex] = [tex]\frac{b}{sinB}[/tex] = [tex]\frac{c}{sinC}[/tex]

(2)

[tex]\frac{b}{sinB}[/tex] = [tex]\frac{c}{sinC}[/tex] , substitute values

[tex]\frac{45}{sin133}[/tex] = [tex]\frac{c}{sin26}[/tex] ( cross- multiply )

c × sin133°  = 45 × sin26° ( divide both sides by sin133° )

c = [tex]\frac{45sin26}{sin133}[/tex] ≈ 27.0 ( to the nearest tenth )

(4)

[tex]\frac{b}{sinB}[/tex] = [tex]\frac{c}{sinC}[/tex] , substitute values

[tex]\frac{19}{sinB}[/tex] = [tex]\frac{30}{sin97}[/tex] ( cross- multiply )

30 sinB = 19 sin97° ( divide both sides by 30 )

sinB = [tex]\frac{19sin97}{30}[/tex] , then

∠ B = [tex]sin^{-1}[/tex] ( [tex]\frac{19sin37}{30}[/tex] ) ≈ 38.9° ( to the nearest tenth )

(6)

[tex]\frac{b}{sinB}[/tex] = [tex]\frac{c}{sinC}[/tex], substitute values

[tex]\frac{18}{sin102}[/tex] = [tex]\frac{xAB}{sin45}[/tex] ( cross- multiply )

AB sin102° = 18 sin45° ( divide both sides by sin102° )

AB = [tex]\frac{18sin45}{sin102}[/tex] ≈ 13.0 ( to the nearest tenth )

Answer:

The degree of this polynomial is 2.

Step-by-step explanation:

The degree of a polynomial is the highest of the degrees of the polynomial's monomials (individual terms) with non-zero coefficients. The degree of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer.

The given polynomial is:

[tex]9m^2+11m^2+2m^2[/tex]

The only variable is m.

The power of m in all terms is 2.

So, the degree of this polynomial is 2.

Answer:

b

Step-by-step explanation:

the graph gets translated 5 units above its parent graph of y = x

Answer:

Step-by-step explanation:

B; So this is a transformation problem from the parent function of f(x)=|x| so the function is is moved 3 units down giving it the -3 at the end and is moved to the right 7 units so it would be x-7

Im sorry I don't know the answer to the question

Answer:

The 5-hour decay factor for the number of mg of caffeine in Ben's body is of 0.1469.

Step-by-step explanation:

After consuming the energy drink, the amount of caffeine in Ben's body decreases exponentially.

This means that the amount of caffeine after t hours is given by:

[tex]A(t) = A(0)e^{-kt}[/tex]

In which A(0) is the initial amount and k is the decay rate, as a decimal.

The 10-hour decay factor for the number of mg of caffeine in Ben's body is 0.2722.

1 - 0.2722 = 0.7278, thus, [tex]A(10) = 0.7278A(0)[/tex]. We use this to find k.

[tex]A(t) = A(0)e^{-kt}[/tex]

[tex]0.7278A(0) = A(0)e^{-10k}[/tex]

[tex]e^{-10k} = 0.7278[/tex]

[tex]\ln{e^{-10k}} = \ln{0.7278}[/tex]

[tex]-10k = \ln{0.7278}[/tex]

[tex]k = -\frac{\ln{0.7278}}{10}[/tex]

[tex]k = 0.03177289938 [/tex]

Then

[tex]A(t) = A(0)e^{-0.03177289938t}[/tex]

What is the 5-hour growth/decay factor for the number of mg of caffeine in Ben's body?

We have to find find A(5), as a function of A(0). So

[tex]A(5) = A(0)e^{-0.03177289938*5}[/tex]

[tex]A(5) = 0.8531[/tex]

The decay factor is:

1 - 0.8531 = 0.1469

The 5-hour decay factor for the number of mg of caffeine in Ben's body is of 0.1469.

Answer:

m = 6

c = 0

General Formulas and Concepts:

Algebra I

Slope-Intercept Form: y = mx + c

Step-by-step explanation:

Step 1: Define

y = 6x

↓ Compare to Slope-Intercept Form

Slope m = 6

y-intercept c = 0

Answer:

4

Step-by-step explanation:

24 = 2^3 x 3

100 = 2^2 x 5^2

HCF = 2^2 = 4

We need not consider a whole triangle but just point B.

Before reflection we know that [tex]B(5,9)[/tex].

Reflecting B over [tex]y=1[/tex] is relatively easy. First because its a reflection over the horizontal line the only coordinates that will change are y coordinates, while x coordinate will not change so half of the reflection is already done for us,

[tex]B(5,a)[/tex]

Now to what has changed, well currently the distance between 9 and 1 on the y axis is 8 up. But because we are reflecting the a must now be 8 down from 1 which means [tex]1 - 8 = -7[/tex] so our point is now [tex]\boxed{B(5,-7)}[/tex].

Hope this helps :)

9514 1404 393

Answer:

  (5, -7)

Step-by-step explanation:

Reflection across the line y = c is accomplished by the transformation ...

  (x, y) ⇒ (x, 2c -y)

For c=1 and point B, we have ...

  B(5, 9) ⇒ B'(5, 2·1 -9) = B'(5, -7)

The image of point B is (5, -7).