11. Which property is illustrated by 5+n=n +5?

Carly works at a local bakery. She needs 8 cups of flour to make a cake and 4 cups of flour to make 16 cupcakes. It she started

AleksAgata [21]

Answer:

All things said, Carly needs 32 cups for compose her bakery items.

Step-by-step explanation:

First, let' see how much flower in all would need to be used for the following delicacies, knowing she started with 72 cups of flower.

4 x 8 = 32 cups for the cake

4:16 = for only 16 cupcakes, so double it 8:32... 8 cups of flour.

Thus, 8+32= 40 & 72-40= 32

All things said, Carly needs 32 cups for compose her bakery items.

I hope this helped!!

~ Penny

3 0

4 years ago

Read 2 more answers

How do you express sin x + cos x in terms of sine only?

frez [133]

Answer:

$\sin x + \sqrt{1-\sin^2x}$

Step-by-step explanation:

Given: sin x + cos x

To change the given trigonometry expression in term of sine only.

Trigonometry identity:-

$\sin^2x+\cos^2x=1$
$\cos x=\sqrt{1-\sin^2x}$

Expression: $\sin x+\cos x$

We get rid of cos x from expression and write as sine form.

Expression: $\sin x + \sqrt{1-\sin^2x}$ $\because \cos x=\sqrt{1-\sin^2x}$

Hence, The final expression is only sine function.

4 0

4 years ago

Find the required measurements of the following trapezoids. b1 = 12 cm b2 = 16 cm h = 9 cm. Compute the area. ? cm2

Nastasia [14]

Area= (b1+b2)/h
= (12+16)/9
= 28/9 = 3.111

5 0

3 years ago

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What is the least common multiple 6 and 12

Zolol [24]

Answer:

12

Step-by-step explanation:

7 0

3 years ago

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Use the substitution x = et to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.

Anika [276]

Answer:

$\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \ }$

Step-by-step explanation:

Hello,

let's follow the advise and proceed with the substitution

first estimate y'(x) and y''(x) in function of y'(t), y''(t) and t

$x(t)=e^t\\\dfrac{dx}{dt}=e^t\\y'(t)=\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=e^ty'(x)y'(x)=e^{-t}y'(t)\\y''(x)=\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(e^{-t}\dfrac{dy}{dt})=-e^{-t}\dfrac{dt}{dx}\dfrac{dy}{dt}+e^{-t}\dfrac{d}{dx}(\dfrac{dy}{dt})\\=-e^{-t}e^{-t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}=-e^{-2t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}e^{-t}\\=e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt})$

Now we can substitute in the equation

$x^2y''(x)+9xy'(x)-20y(x)=0\\ e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\ \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\ \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0\\$