By satisfying the equation of a function 'f' by each coordinates given in the options we can get the point which lies on the graph of f(x) = $2\times (5)^x$

Option (A). (1, 10)

f(1) = $2\times (5)^1$

10 = 10

True.

Therefore, point (1, 10) lies on the graph.

Option (B). (0, 10)

f(0) = $2\times 5^0$

10 = 2

Not true.

Therefore, point (0, 10) doesn't lie on the graph.

Option (C). (10, 1)

f(10) = $2\times 5^{10}$

1 = 19531250

Not true.

Therefore, point (10, 1) doesn't lie on the graph.

Option (D). (0, 0)

f(0) = $2\times 5^0$

0 = 2

Not True.

Point (0, 0) doesn't lie on the graph.

Option (A) will be the answer.

4 0

3 years ago

5.5 divided by 83 in quotient

sammy [17]

the answer is 15.090090

7 0

3 years ago

Solve for x:a(a²+b²)x²+b²x-a

m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}$

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b² = (b² + 2a²)²

2. Solve for x

$\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}$

$\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}$

5 0

3 years ago

How many feet are in 2000 meters?

11Alexandr11 [23.1K]

First, note that 1 meter is equal to 3.28084 feet.

Next, multiply
2000 m • 3.28084 = 6561.68 ft

6 0

4 years ago

Read 2 more answers