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Keith_Richards [23]
3 years ago
13

=49^{2x-2}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
ioda3 years ago
8 0
DONT CLICK ON THAT LINK PLEASE! But the answer is x= -5
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50-0<br> ———(12.0-0) +0 = <br> 12.0-0 <br><br> Just a warm up
vagabundo [1.1K]

\frac{50}{12} x 12

the 12 in the numerator (top) and the 12 in the denominator (bottom) cross out so you are left with 50.

Answer: 50

8 0
4 years ago
Read 2 more answers
Express cos6mcos2m as a sum or difference
raketka [301]

Given :

An expression (cos 6m)(cos 2m) .

To Find :

We need to express it in terms as sum or difference.

Solution :

We know,

cos( A + B ) = cosA cos B - sin A sin B

cos( A - B ) = cosA cos B + sin A sin B

Adding both the equations we get :

2cos A cos B = cos( A + B) + cos( A - B )

or

cos A cos B = cos( A + B) + cos( A - B )/2

Putting value of A = 6m and B = 2m in above equation, we get :

(cos 6m)(cos 2m) = cos( 6m + 2m ) + cos( 6m - 2m )/2

(cos 6m)(cos 2m) = cos(8m) + cos(4m)/2

Hence, this is the required solution.

4 0
3 years ago
How do the graphs of the function f(x)=(3/2)x and g(x)=(2/3)x compare
allochka39001 [22]

Answer:

f is a line through 0 and (2,3)

g is a line through 0 and (3,2)

Step-by-step explanation:

f and g are proportional linear functions. They each have the form y=mx+b where b=0 and m is a fraction. The m or slope tell us the form the function takes on the graph. It is a straight line through 0 with steepness m.

f(x)=\frac{3}{2} x has steepness 3/2. Starting at (0,0) or the origin, move up three grid lines on the y-axis and then over 2 grid lines. Mark your point at (2,3). Then connect and draw arrows on the end.


g(x)=\frac{2}{3}x has steepness 2/3. Starting at (0,0) or the origin, move up two grid lines on the y-axis and over 3 grid lines. Mark your point at (3,2). Then connect and draw he arrows on the end.

6 0
3 years ago
Read 2 more answers
Round 25,678 to the nearest ten thousand
Art [367]
25,678 rounded to the nearest 10,000 would be 30,000
5 0
4 years ago
The vector v has initial point P(1, 0) and terminal point Q that is on the y-axis and above the initial point. Find the coordina
zloy xaker [14]

Answer:

The terminal point is Q(0,4)

Step-by-step explanation:

We are given the following in the question:

Initial point of vector, P(1,0)

The terminal point Q lies on the y-axis, thus Q have coordinates of the form:

Q(0,y)

The magnitude of the vector is \sqrt{65}

Magnitude is given by:

v = \sqrt{(x_2-x_2)^2+(y_2-y_1)^2}\\\sqrt{65}=\sqrt{(0-1)^2+(y-0)^2}\\\sqrt{65}=\sqrt{1+y^2}\\65 = 1+y^2\\y^2 = 64\\y = \pm 8

But since the terminal point is above the initial point, thus,

y > 0\\\Rightarrow y = 4

Thus, the terminal point is Q(0,4)

4 0
3 years ago
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