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3 years ago

how to determine which quantity is independent and which quantity is dependent when considering a situation

1 answer:

7 0

An easy way to think of independent and dependent variables is, when you're conducting an experiment, the independent variable is what you change, and the dependent variable is what changes because of that. You can also think of the independent variable as the cause and the dependent variable as the effect.

I hope this helps!

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Rick is building a scale model of the Alamo for his Texas History class. In this model, he is using a scale in which 3 inches re

motikmotik

Answer:

18 inches

Step-by-step explanation:

Rick is building a scale model of the Alamo.

We are given a scale that 3 in. = 15 ft.

The length of the Alamo is 90 feet.

What is the length of the scale model in inches?

So we know that for every 15 feet, 3 inches are actually used in the project and we need a total of 90 feet. To solve, we need to find how many times 15 goes into 90, in which the number it multiplies to will apply to the inches.

15 * ? = 90

90 / 15

It takes 15 6 times to get to 90, meaning that we need to multiply 3 inches 6 times to find the actual length of the scale model in inches.

3 * 6

18 inches.

8 0

3 years ago

Help me pleaseee pleaseee

Darya [45]

Answer:

x = 3

m<PAW = 1

Step-by-step explanation:

☆They are supplementary angles.

☆Supplementary angles equal 180 degrees.

$- 11x + 17 3 - 1x + 43 = 180 \\ - 12x + 216 = 180 \\ \frac{ \: \: \: \: \: \: \: \: \: \: \: \: \: - 216 = - 216}{ \frac{ - 12x}{ - 12} = \frac{ - 36}{ - 12} } \\ x = 3$

☆Plug in for m<AEZ

$- 1(3) + 4 \\ - 3 + 4 \\ 1$

☆m<PAW is corresponding to m<AEZ

☆Corresponding angles are congruent.

☆m<PAW equals 1.

7 0

3 years ago

What is the solution for x in the equation 5/3x + 4 = 2/3x

marta [7]

The answer is -4! Hope this helped you!

7 0

4 years ago

5x(x+6)=−50 i need to find the xs by completing the square

FrozenT [24]

5x(x+6)=-50 X(x+6)=-10 X square +6x=-10 (X+3) square=-10-9 (X+3) square=-19 So this equation has no solution

7 0

3 years ago

Find the exact value of the expression.

Tresset [83]

$\sin(a-b)=\sin a \cos b-\cos a \sin b$ . If we let $a=\cos^{-1} \left(\frac{5}{6} \right)$ and $b=\tan^{-1} \left(\frac{1}{2} \right)$ , then the given expression is equal to:

$\sin \left(\cos^{-1} \left(\frac{5}{6}} \right) \right) \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)-\cos\left(\cos^{-1} \left(\frac{5}{6} \right) \right) \sin \left( \tan^{-1} \left(\frac{1}{2} \right) \right)$

Using the Pythagorean identities $\sin^{2} x+\cos^{2} x=1$ and $\tan^{2} x+1=\sec^{2} x$ ,

$1) \sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\cos^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\frac{25}{36}=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{11}{36}\sin \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{\sqrt{11}}{6}$

$2) \tan^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{1}{4}+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{5}{4}=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\sec \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{\sqrt{5}}{2}\\\implies \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$

$\cos^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\frac{4}{5}+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{5}\\\left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$

This means we can write the original expression as:

$\left(\frac{\sqrt{11}}{6} \right) \left(\frac{2\sqrt{5}}{5} \right)-\left(\frac{5}{6} \right) \left(\frac{\sqrt{5}}{5} \right)\\=\frac{2\sqrt{11}\sqrt{5}}{30}-\frac{5\sqrt{5}}{30}\\=\boxed{\frac{\sqrt{5}(2\sqrt{11}-5)}{30}}$

6 0

2 years ago

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