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2 years ago

11

What is the least common multiple of 7 and 10?

1 answer:

Lina20 [59]2 years ago

8 0

Answer:

70

Step-by-step explanation:

plz give brainliest (the crown)

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CAn someone give me the answers pls it would really help

nasty-shy [4]

Do you need answer for all paper

3 0

2 years ago

Express your answer as a decimal. 7÷25=

Over [174]

Answer:

0.28

Step-by-step explanation:

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2 years ago

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For what value of k is there one solution to the given quadratic equation (k+1)x²+4kx+2=0.

andriy [413]

Answer:

If $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

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:

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

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

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

The signal of $\bigtriangleup$ determines how many real roots an equation has:

$\bigtriangleup > 0$ : Two real and different solutions

$\bigtriangleup = 0$ : One real solution

$\bigtriangleup < 0$ : No real solutions

In this problem, we have the following second order polynomial:

$(k+1)x^{2} + 4kx + 2 = 0$ .

This means that $a = k+1; b = 4k; c = 2$

It has one solution if

$\bigtriangleup = 0$

$b^{2} - 4ac = 0$

$16k^{2} -8(k+1) = 0$

$16k^{2} - 8k - 8 = 0$

We can simplify by 8

$2k^{2} - k - 1 = 0$

The solution is:

$k = 1$ or $k = -\frac{1}{2}$

So, if $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

4 0

3 years ago

I need help, Please hurry, It's timed.

kenny6666 [7]

The answer is
C) 22x^2-x+17

5 0

2 years ago

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Does that look right

Tju [1.3M]

looks right to me! :)

7 0

2 years ago

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