Tenisha solved the equation below by graphing a system of equations. log35x = log (2x+8)
Which point approximates the solution for Tenisha's system of equations?
0 (0.9, 0.8)
O (1.0, 1.4)
O (2.3, 1.1)
O (2.7, 13.3)

Option B:(1.0, 1.4)

Step-by-step explanation:

log3 5x = log5 (2x + 8) implies log3 5x = 2log5 . x + 8log5

log3 5x - 2log5 . x = 8log5 and (5log3 - 2log5)x = 8log5

Finally, x = 8log5 / (5log3 - 2log5) = a let y = log5 (2x + 8), so y(a) = log5 (2a + 8)   The point approximating the solution is  P( a, y(a))  or P ( 8log5 / (5log3 - 2log5,    log5[2 (8log5 / (5log3 - 2log5)) + 8])

Therefore, by graphing a system of equations, the points that approximate the solution for Tenisha’s system of equations are points (1.0, 1.4).

Option B.)

Step-by-step explanation:

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(1.0, 1.4)

Step-by-step explanation:

x = 17

step-by-step explanation:

in the given figure, two triangles are inscribed in a circle where ∠bae ≅ ∠dae which makes the side be and de equal so we can write it as:

3x - 24 = x + 10

solving this equation for x to get:

3x - x = 10 + 24

2x = 34

x = 34/2

x = 17

therefore, the value of x = 17.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: