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blondinia [14]
1 year ago
7

Let angle G be an acute angle and tan G = 0.43 . Use technology to approximate the measure of angle G to the nearest 10th of a d

egree.
Mathematics
1 answer:
klasskru [66]1 year ago
6 0

We need to use the calculator, using the tan⁻¹ function:

\begin{gathered} \tan(G)=0.43 \\ \tan^{-1}(\tan(G))=\tan^{-1}(0.43) \\ G=\tan^{-1}(0.43) \end{gathered}

Using the calculator:

G=\tan^{-1}(0.43)=23.26770481

To the nearest tenth:

G=23.3º

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Vinil7 [7]

Answer: 2x - 5y = -17

Step-by-step explanation: Let's use the point-slope formula.

The point-slope formula is written y - y₁ = m(x - x₁).

In this formula, (x₁, y₁) is our point and <em>m</em> is our slope.

Now let's plug all our information into the formula.

So we have y - 3 = 2/5(x + 1).

Notice that I put x + 1 inside the parenthses and this

is because we have x minus a negative 1 which means x + 1.

Our next step would be to distribute this 2/5through both

terms inside the parenthses to get 2/5x + 2/5 on the right side.

So we have y - 3 = 2/5x + 2/5.

Now in standard form, we cannot have any fractions.

We need to get rid of them by multiplying

both sides of the equation by 5.

When we do this, we get 5y - 15 = 2x + 2.

Now we just move our number to the right by

adding 15 to both sides of the equation.

That gives us 5y = 2x + 17.

Move our 2x to the left side by subtracting

2x from both sides to get -2x + 5y = 17.

In standard form, the coefficient of the x term must be positive.

So our last step is to divide both sides

by -1 so that our answer is 2x - 5y = -17.

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Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!
IgorC [24]
Hey there, hope I can help!

-4x^2+9y^2+32x+36y-64=0

\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \  9y^2+32x+36y-4x^2=64

\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \  -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4
-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9
-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}

\mathrm{Convert}\:x\:\mathrm{to\:square\:form}
-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)

\mathrm{Convert\:to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)

\mathrm{Convert}\:y\:\mathrm{to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)

\mathrm{Convert\:to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)

\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \  -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1

Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1

For me I used
\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1
As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1

Hope this helps!
4 0
3 years ago
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