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3 years ago

Write the first four terms in the multiplication pattern given by the formula:10 · 5n

1 answer:

7 0

We have that
an=10 * 5n---------------> an=50*n

then

for n=1
a1=50*1=50

for n=2
a2=50*2=100

for n=3
a3=50*3=150

for n=4
a4=50*4=200

the answer is
the first four terms are [50,100,150,200]

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

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2 years ago

Read 2 more answers

3x^2+kx=-3 What is the value of K will result in exactly one solution to the equation?

Scorpion4ik [409]

Answer:

For k = 6 or k = -6, the equation will have exactly one solution.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

If $\bigtriangleup = 0$ , the equation has only one solution.

In this problem, we have that:

$3x^{2} + kx + 3 = 0$

$a = 3, b = k, c = 3$

$\bigtriangleup = b^{2} - 4ac$

$\bigtriangleup = k^{2} - 4*3*3$

$\bigtriangleup = k^{2} - 36$

We will only have one solution if $\bigtriangleup = 0$ . So

$\bigtriangleup = 0$

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$k^{2} = 36$

$k = \pm \sqrt{36}$

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2 years ago

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Answer:

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2 years ago