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olga_2 [115]
3 years ago
7

Need Help which one is it???

Mathematics
2 answers:
tamaranim1 [39]3 years ago
3 0

Answer:

the blue one but not sure

RUDIKE [14]3 years ago
3 0

Answer:

the third one i think

Step-by-step explanation:

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Log (rootx+1 + root x-1)
amid [387]

Answer:

Step-by-step explanation:

dy/dx = d/dx log root under x-1 / root under x+1

dy/dx = root under x+1 / root under x-1 * 1/(x+1)^2/3 * 1/root under x-1

dy/dx = (x+1)^2/ x-1

7 0
3 years ago
Gemma recently rode her bicycle to visit her friend who lives 6 miles away. On her way there, her average speed was 16 miles per
DochEvi [55]

first speed --- x mph

return speed -- x+16 mph

6/x + 6/(x+16) = 1

times each term by x(x+16)

6(x+16) + 6x = x(x+16)

x^2 + 4x - 96 = 0

(x-8)(x+12) = 0

x = 8 or x is a negative

her first speed was 8 mph

her return speed was 24 mph

check:

6/8 + 6/24 = 1 , that's good!

8 0
3 years ago
Help me with this: How many residents were surveyed?
PtichkaEL [24]
22 you just count all of the dots
7 0
2 years ago
How many solutions do these equations have
Arlecino [84]

Answer:

\large\boxed{\bold{one\ solution}\ (1,\ -4)\to x=1,\ y=-4}

Step-by-step explanation:

\left\{\begin{array}{ccc}y=2x-6\\y=-x-3\end{array}\right\\\\\text{These are linear functions. We only need two points to draw a graph.}\\\\\text{Choice two values of x, substitute to the equation,}\\\text{and calculate the values of y.}\\\\y=2x-6\\for\ x=0\to y=2(0)-6=0-6=-6\to(0,\ -6)\\for\ x=3\to y=2(3)-6=6-6=0\to(3,\ 0)\\\\y=-x-3\\for\ x=0\to y=-0-3=0-3=-3\to(0,\ -3)\\for\ x=-3\to y=-(-3)-3=3-3=0\to(-3,\ 0)\\\\\text{Look at the picture}

\text{The intersection of the line is the solution of the system of equations:}\\\\(1,\ -4)\to x=1,\ y=-4

7 0
3 years ago
In triangle abc, m of acb = 90, cd is perpendicular to ab , m of acd is 60. and bd is 5 cm. find ad
weeeeeb [17]

Let us draw a picture to make the things more clear.

Attached is the image.

We have been given that

\angle acd = 60 ^{\circ}

Therefore, we have

\angle dcb =90- 60= 30 ^{\circ}

Now, in triangle bcd, we have

\tan30 = \frac{5}{cd}\\
\\
\frac{1}{\sqrt 3}=\frac{5}{cd}\\
\\
cd=5\sqrt 3

Now, in triangle acb, we have

tan 60 = \frac{ad}{5\sqrt3} \\
\\
\sqrt 3=  \frac{ad}{5\sqrt3}\\
\\
ad= 5\sqrt3 \times \sqrt 3\\
\\
ad= 5\times 3\\
\\
ad=15

Thus, ad is 15 cm.


4 0
3 years ago
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