Answer:
Since I cannot draw here, (right?) I will tell you were to draw the point.
Exactly at - 1,5, which is exactly the second small dash when counting from the left.
Step-by-step explanation:
Answer:
9 songs
Explanation:
Given:
The slideshow is 31 whole number 1/2 minutes long
convert to an improper fraction
31×2+1/2 = 63/2
The songs are each 3 whole number 1/2 minutes long
convert to an improper fraction
3×2+1/2 = 7/2
To find the number of songs that Marcus is picking to play during a slideshow, divide the slideshow by the long of one song
(63/2) / (7/2)
= 9 songs.
![b\in(0,\ 1)\ \cup\ (1,\ \infty)\\x > 0\\y\in\mathbb{R}\\--------------------------\\\\y=\log_bx\\\\For\ (1,\ 0)\to x=1,\ y=0.\ Substitute:\\\\\log_b1=0\to b^0=1\to b\in(0,\ 1)\ \cup\ (1,\ \infty)\\\\For\ (116,\ 2)\to x=116,\ y=2.\ Substitute:\\\\\log_b116=2\to b^2=116\to b=\sqrt{116}\\\to b=\sqrt{4\cdot29}\to b=\sqrt4\cdot\sqrt{29}\to b=2\sqrt{29}\\\\For\ (4,\ -1)\to x=4,\ y=-1.\ Substitute:\\\\\log_b4=-1\to b^{-1}=4\to b=\dfrac{1}{4}](https://tex.z-dn.net/?f=b%5Cin%280%2C%5C%201%29%5C%20%5Ccup%5C%20%281%2C%5C%20%5Cinfty%29%5C%5Cx%20%3E%200%5C%5Cy%5Cin%5Cmathbb%7BR%7D%5C%5C--------------------------%5C%5C%5C%5Cy%3D%5Clog_bx%5C%5C%5C%5CFor%5C%20%281%2C%5C%200%29%5Cto%20x%3D1%2C%5C%20y%3D0.%5C%20Substitute%3A%5C%5C%5C%5C%5Clog_b1%3D0%5Cto%20b%5E0%3D1%5Cto%20b%5Cin%280%2C%5C%201%29%5C%20%5Ccup%5C%20%281%2C%5C%20%5Cinfty%29%5C%5C%5C%5CFor%5C%20%28116%2C%5C%202%29%5Cto%20x%3D116%2C%5C%20y%3D2.%5C%20Substitute%3A%5C%5C%5C%5C%5Clog_b116%3D2%5Cto%20b%5E2%3D116%5Cto%20b%3D%5Csqrt%7B116%7D%5C%5C%5Cto%20b%3D%5Csqrt%7B4%5Ccdot29%7D%5Cto%20b%3D%5Csqrt4%5Ccdot%5Csqrt%7B29%7D%5Cto%20b%3D2%5Csqrt%7B29%7D%5C%5C%5C%5CFor%5C%20%284%2C%5C%20-1%29%5Cto%20x%3D4%2C%5C%20y%3D-1.%5C%20Substitute%3A%5C%5C%5C%5C%5Clog_b4%3D-1%5Cto%20b%5E%7B-1%7D%3D4%5Cto%20b%3D%5Cdfrac%7B1%7D%7B4%7D)
Different values of b.
<h3>Answer: There is no logarithmic function whose graph goes through given points.</h3><h3 />
Maybe the second point is ![\left(\dfrac{1}{16},\ 2\right)](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7B1%7D%7B16%7D%2C%5C%202%5Cright%29)
Substitute:
![\log_b\dfrac{1}{16}=2\to b^2=\dfrac{1}{16}\to b=\sqrt{\dfrac{1}{16}}\to b=\dfrac{1}{4}](https://tex.z-dn.net/?f=%5Clog_b%5Cdfrac%7B1%7D%7B16%7D%3D2%5Cto%20b%5E2%3D%5Cdfrac%7B1%7D%7B16%7D%5Cto%20b%3D%5Csqrt%7B%5Cdfrac%7B1%7D%7B16%7D%7D%5Cto%20b%3D%5Cdfrac%7B1%7D%7B4%7D)
<h3>Then we have the answer:</h3>
![\boxed{y=\log_{\frac{1}{4}}x}](https://tex.z-dn.net/?f=%5Cboxed%7By%3D%5Clog_%7B%5Cfrac%7B1%7D%7B4%7D%7Dx%7D)
Your answer is D. 16x² - 56xy + 49y².
A perfect square trinomial is the result of a squared binomial, like (a + b)². Using this example, the perfect square trinomial would be a² + 2ab + b², as that is what you get when you expand the brackets.
Therefore, to determine which of these is a perfect square trinomial, we have to see if it can be factorised into the form (a + b)².
I did this by first square rooting the 16x² and 49y² to get 4x and 7y as our two terms in the brackets. We automatically know the answer isn't A or B as you cannot have a negative square number.
Now that we know the brackets are (4x + 7y)², we can expand to find out what the middle term is, so:
(4x + 7y)(4x + 7y)
= 16x² + (7y × 4x) + (7y × 4x) + 49y²
= 16x² + 28xy + 28xy + 49y²
= 16x² + 56xy + 49y².
So we know that the middle number is 56xy. Now we assumed that it was (4x + 7y)², but the same 16x² and 49y² can also be formed by (4x - 7y)², and expanding this bracket turns the +56xy into -56xy, forming option D, 16x² - 56xy + 49y².
I hope this helps!