The equation y = log b^x is equivalent to the exponential form

Mumz [18]

The formula that is useful for solving both of these problems is ...

$(v_2)^2-(v_1)^2=2ad\qquad\text{a=acceleration, d=distance}\\\\ \text{where $v_1$ and $v_2$ are initial and final velocities}$

9. Given v₁=15, a=9.8, d=10, find v₂.

... (v₂)² = 15² + 2·9.8·10

... v₂ = √421 ≈ 20.5 . . . . m/s

10. Given d=12 m when a=-9.8 m/s² and v₂=0, find d when a=0.17·(-9.8 m/s²).

The formula tells us that d=(v₁)²/(2a), which is to say that the distance is inversely proportional to the acceleration. If acceleration is 0.17 times that on earth, distance will be 1/0.17 ≈ 5.88 times that on earth.

(12 m)/0.17 ≈ 70.6 m

7 0

2 years ago

Un edificio proyecta una sombra de 7.5m.mientras que un arbol que mide 1.6 m de altura proyecta una sombra 1.85 M¿Cual es la alt

sattari [20]

Answer: The height of the building is 6.49 meters.

Step-by-step explanation:

This can be translated to:

"A building projects a 7.5 m shadow, while a tree with a height of 1.6 m projects a shadow of 1.85 m.

Which is the height of the building?"

We can conclude that the ratio between the projected shadow is and the actual height is constant for both objects, this means that if H is the height of the building, we need to have:

(height of the building)/(shadow of the building) = (height of the tree)/(shadow of the tree)

H/7.5m = 1.6m/1.85m

H = (1.6m/1.85m)*7.5m = 6.49m

The height of the building is 6.49 meters.

5 0

2 years ago

{ vowels in the word ALGEBRA}

daser333 [38]

Answer:

2As and 1E

Step-by-step explanation:

Please mark as brainliest

7 0

2 years ago

25x + 4x = x + 3x +7

Musya8 [376]

For this equation its the same as simplifying any other equation -Simplify both sides of the equation then isolate the variable- Easy then once you've do so the answer should be

x = $\frac{7}{26\\}$