Answer:
x=30, <A=60, <C=30
Step-by-step explanation:
A triangle always has its total angles added up to 180 degrees, so we can set up the following equation:
90+2x+x=180
90+3x=180
3x=90
x=30
Since x=30, this means <C is 30 degrees. For <A, it's twice as much as <C, so 2*30=60.
So x=30, <A=60, <C=30
Hope this helped!
Given:
The figure of a right angle triangle.
To find:
The value which is equal to r÷q.
Solution:
In the given right triangle, the length of the hypotenuse is q units.
For angle x degrees, perpendicular is p units and base is r units.
For angle y degrees, perpendicular is r units and base is p units.
In a right angle triangle,
![\sin \theta=\dfrac{Perpendicular}{Hypotenuse}](https://tex.z-dn.net/?f=%5Csin%20%5Ctheta%3D%5Cdfrac%7BPerpendicular%7D%7BHypotenuse%7D)
![\cos \theta=\dfrac{Base}{Hypotenuse}](https://tex.z-dn.net/?f=%5Ccos%20%5Ctheta%3D%5Cdfrac%7BBase%7D%7BHypotenuse%7D)
So,
![\sin y=\dfrac{r}{q}](https://tex.z-dn.net/?f=%5Csin%20y%3D%5Cdfrac%7Br%7D%7Bq%7D)
![\cos x=\dfrac{r}{q}](https://tex.z-dn.net/?f=%5Ccos%20x%3D%5Cdfrac%7Br%7D%7Bq%7D)
Therefore, the value of r÷q is equal to siny or cosx. So, the correct option is C.
The resulting function after all transformations have been carried out is; -7log6(-x) - 2.
<h3>Which function results from all the Transformations?</h3>
The transformations which ensued on the function given are as follows;
- Stretched vertically by a factor of 7; Hence, we have; 7f(x) = 7log6(x).
- Reflected over the x-axis; -7f(x) = -7log6(x).
- Reflected over the y-axis; -7f(x) = -7log6(-x).
- Shifted up by 2 units; -7f(x) -2 = -7log6(-x) - 2.
On this note, it follows that the resulting function upon all the Transformations is; = -7log6(-x) - 2.
Read more on transformations;
brainly.com/question/1616371
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![\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \qquad \begin{cases} A=\textit{current amount}\\ P=\textit{original amount deposited}\to &\$7000\\ r=rate\to 5\%\to \frac{5}{100}\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, means 4 times} \end{array}\to &4\\ t=years\to &10 \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BCompound%20Interest%20Earned%20Amount%7D%0A%5C%5C%5C%5C%0AA%3DP%5Cleft%281%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D%0A%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Bcurrent%20amount%7D%5C%5C%0AP%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cto%20%26%5C%247000%5C%5C%0Ar%3Drate%5Cto%205%5C%25%5Cto%20%5Cfrac%7B5%7D%7B100%7D%5Cto%20%260.05%5C%5C%0An%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%0A%5Ctextit%7Bquarterly%2C%20means%204%20times%7D%0A%5Cend%7Barray%7D%5Cto%20%264%5C%5C%0A%0At%3Dyears%5Cto%20%2610%0A%5Cend%7Bcases%7D)
how about monthly? well, there are 12 months in a year, so it will the compound cycle is 12, thus