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zysi [14]
4 years ago
13

You know the length and width of a scale model. what additional information do you need to find the scale of the model?

Mathematics
1 answer:
DENIUS [597]4 years ago
5 0
What the model is scaled by
You might be interested in
Which statement is true about the equations-3x 4y 12 and tx 1?
nydimaria [60]

We need to solve -3x+4y=12 for x

Let's start by adding -4y to both sides

-3x+4y-4y=12-4y

-3x=-4y+12

x = (-4y+12)/-3

x= 4/3 y -4

Now substitute 4/3 y -4 for x in 1/4 x - 1/3 y =1

1/4 x -1/3 y =1

1/4 (4/3 y -4) -1/3 y =1

Use the distributive property

(1/4)(4/3 y) + (1/4)(-4) -1/3 y =1

1/3 y -1 - 1/3 y =1

Now combine like terms

(1/3y -1/3y) + (-1) =1

= -1

-1 = 1

Now add 1 to both sides

0=2

So there are no Solutions

The answer is C


I hope that's help Will :)

7 0
3 years ago
Match the following rational expressions to their rewritten forms.
Nookie1986 [14]

Answer:

Answer image is attached.

Step-by-step explanation:

Given rational expressions:

1.\ \dfrac{x^2+x+4}{x-2}\\2.\ \dfrac{x^2-x+4}{x-2}\\3.\ \dfrac{x^2-4x+10}{x-2}\\4.\ \dfrac{x^2-5x+16}{x-2}

And the rewritten forms:

(x-2)+\dfrac{6}{x-2}\\(x+3)+\dfrac{10}{x-2}\\(x+1)+\dfrac{6}{x-2}\\(x-3)+\dfrac{10}{x-2}

We have to match the rewritten terms with the given expressions.

Let us consider the rewritten terms and let us solve them one by one by taking LCM.

(x-2)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x-2)^{2}+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+4+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+10}{x-2}

So, correct option is 3.

(x+3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x+3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{(x^2+3x-2x-6)+10}{x-2}\\\Rightarrow \dfrac{x^2+x+4}{x-2}

So, correct option is 1.

(x+1)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x+1)(x-2)+6}{x-2}\\\Rightarrow \dfrac{x^{2} +x-2x-2+6}{x-2}\\\Rightarrow \dfrac{x^{2} -x+4}{x-2}

So, correct option is 2.

(x-3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x-3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{x^2-3x-2x+6+10}{x-2}\\\Rightarrow \dfrac{x^2-5x+16}{x-2}

So, correct option is 4.

The answer is also attached in the answer area.

7 0
3 years ago
If dy dx equals cosine squared of the quantity pi times y over 4 and y = 1 when x = 0, then find the value of x when y = 3.
Afina-wow [57]
Answer: x = -\frac{8}{\pi}

Explanation: 
Note that

\frac{dy}{dx} = \cos^2 \left ( \frac{\pi y}{4} \right )
\\
\\ \frac{dy}{\cos^2 \left ( \frac{\pi y}{4} \right )} = dx
\\
\\ \int{\frac{dy}{\cos^2 \left ( \frac{\pi y}{4} \right )}} = \int dx
\\
\\ \boxed{x = \int{\sec^2 \left ( \frac{\pi y}{4} \right )dy}}

To evaluate the integral in the boxed equation, let u = \frac{\pi y}{4}. Then, 

du = \frac{\pi}{4} dy 
\\ \Rightarrow \boxed{dy = \frac{4}{\pi}du} 

So,

x = \int{\sec^2 \left ( \frac{\pi y}{4} \right )dy}
\\
\\ = \int{\sec^2 u \left ( \frac{4}{\pi}du \right )}
\\
\\ = \frac{4}{\pi}\int{\sec^2 u du}
\\
\\ = \frac{4}{\pi} \tan u + C
\\
\\ \boxed{x = \frac{4}{\pi} \tan \left ( \frac{\pi y}{4} \right ) + C}\text{  (1)}

Since y = 1 when x =0, equation (1) becomes

0 = \frac{4}{\pi} \tan \left ( \frac{\pi (1)}{4} \right ) + C 
\\ 
\\ 0 = \frac{4}{\pi} (1) + C 
\\
\\ \frac{4}{\pi} + C = 0
\\
\\ \boxed{C = -\frac{4}{\pi}}

With the value of C, equation (1) becomes

\boxed{x = \frac{4}{\pi} \tan \left ( \frac{\pi y}{4} \right ) -\frac{4}{\pi} }

Hence, if y = 3, 

x = \frac{4}{\pi} \tan \left ( \frac{\pi y}{4} \right ) -\frac{4}{\pi} 
\\
\\ x = \frac{4}{\pi} \tan \left ( \frac{\pi (3)}{4} \right ) -\frac{4}{\pi}  
\\
\\ x = \frac{4}{\pi} (-1) -\frac{4}{\pi}  
\\
\\ \boxed{x = -\frac{8}{\pi}}
6 0
3 years ago
What is an equivalent equation of 200x+50=8000?
11111nata11111 [884]
You could just switch the order so it is 50 + 200x = 8000. Or you could subtract 50 on both sides to get 200x = 7950.
8 0
3 years ago
Read 2 more answers
Write 7 as a fraction with the 9 in the denominator
mel-nik [20]
So 7=7/1
to put a 9 in the bottom and make the fraction stay the same, multiply it by 1
or 9/9 since 9/9=1 so
7/1 times 9/9=(7 times 9)/(1 times 9)=63/9
answe ris 63/9
6 0
3 years ago
Read 2 more answers
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