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kakasveta [241]
3 years ago
12

Which of the following is an example of a chemical change?

Mathematics
2 answers:
SCORPION-xisa [38]3 years ago
7 0

The answer is B.

To find the answer you can immediately cross out A and C. You're left with a mixture of ingredients that turn into a cake or choclate melting. Chocolate melting is a physical change so by using the process of elimination you can conclude that B is the correct answer.

Mariulka [41]3 years ago
4 0

Answer:

The correct answer is option B.

Step-by-step explanation:

We call chemical change to the process where two or more substances are transformed into other substances.

During this change, the substances will have a modification in the structure of their molecules and their bonds.

In the case of the example we can see how different substances (flour, sugar, eggs and butter) will change the shape of their structure to form another substance: the cake. This cake will receive the name of <em>product</em>.

Given this information, we can say that the correct answer is option B.

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NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 11z​
Roman55 [17]

Answer:

(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}

(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}

Step-by-step explanation:

Given system of equations:

\begin{cases}y=x^2-9\\y=4x-4\end{cases}

To solve by the method of substitution, substitute the first equation into the second equation and rearrange so that the equation equals zero:

\begin{aligned}x^2-9&=4x-4\\x^2-4x-9&=-4\\x^2-4x-5&=0\end{aligned}

Factor the quadratic:

\begin{aligned}x^2-4x-5&=0\\x^2-5x+x-5&=0\\x(x-5)+1(x-5)&=0\\(x+1)(x-5)&=0\end{aligned}

Apply the <u>zero-product property</u> and solve for x:

\implies x+1=0 \implies x=-1

\implies x-5=0 \implies x=5

Substitute the found values of x into the <u>second equation</u> and solve for y:

\begin{aligned}x=-1 \implies y&=4(-1)-4\\y&=-4-4\\y&=-8\end{aligned}

\begin{aligned}x=5 \implies y&=4(5)-4\\y&=20-4\\y&=16\end{aligned}

Therefore, the solutions are:

(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}

(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}

6 0
1 year ago
Read 2 more answers
Which expression is greater than 1/2
polet [3.4K]
Anything greater than 1/2 can be shown as >1/2
8 0
3 years ago
Please help me 33/2 + 3y/5 = 7y/10 + 15
8090 [49]

We are given with an equation in <em>variable y</em> and we need to solve for <em>y</em> . So , now let's start !!!

We are given with ;

{:\implies \quad \sf \dfrac{33}{2}+\dfrac{3y}{5}=\dfrac{7y}{10}+15}

Take LCM on both sides :

{:\implies \quad \sf \dfrac{165+6y}{10}=\dfrac{7y+150}{10}}

<em>Multiplying</em> both sides by <em>10</em> ;

{:\implies \quad \sf \cancel{10}\times \dfrac{165+6y}{\cancel{10}}=\cancel{10}\times \dfrac{7y+150}{\cancel{10}}}

{:\implies \quad \sf 165+6y=7y+150}

Can be <em>further written</em> as ;

{:\implies \quad \sf 7y+150=165+6y}

Transposing <em>6y </em>to<em> LHS</em> and <em>150</em> to<em> RHS </em>

{:\implies \quad \sf 7y-6y=165-150}

{:\implies \quad \bf \therefore \quad \underline{\underline{y=15}}}

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