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2 years ago

Find the formula for the nth term of these sequences:

1 answer:

8 0

A child makes a pitcher of fruit punch by completely dissolving fruit- flavoured powder in water. The child drinks a glass and decides the punch needs to be sweeter, so she stirs in more powder. This time the powder does not all dissolve, but settles on the bottom of the pitcher. A) Is the fruit-flavoured powder a pure substance or a mixture? * helpme anyone?

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When light arrives and leaves a reflecting surface the angle with the line perpendicular to the surface is what?

sashaice [31]

Answer:

the angle of reflection is the angle between the reflected wave and the "normal" (the perpendicular line to the reflecting surface).

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

HELP PLEASE <br><br>i dont understand at all...

bija089 [108]

An exponent signifies repeated multiplication.

$x\cdot x\cdot x=x^{3}$ the factor x is repeated 3 times

Exponents can be added and subtracted to express the effects of multiplication and division.

$\dfrac{x\cdot x\cdot x\cdot x\cdot x}{x\cdot x\cdot x}=\dfrac{x^{5}}{x^{3}}\\\\=\dfrac{x\cdot x\cdot x}{x\cdot x\cdot x}\cdot x\cdot x=x\cdot x\\\\=x^{(5-3)}=x^{2}$

The addition and subtraction of exponents works the same even when there are more denominator factors than numerator factors.

$\dfrac{x\cdot x\cdot x}{x\cdot x\cdot x\cdot x\cdot x}=\dfrac{x^{3}}{x^{5}}=\dfrac{1}{x^{2}}\\\\=x^{(3-5)}=x^{-2}$

That is, a negative numerator exponent is the same as a positive denominator exponent and vice versa. You can move a factor with an exponent from denominator to numerator and change the sign of the exponent, and vice versa.

Your expression has 3 in the denominator with a negative exponent. It can be moved to the numerator and the exponent changed to positive:

$\dfrac{1}{3^{-2}}=3^{2}\\\\=3\cdot 3=\bf{9}$

8 0

2 years ago

Solve (x + 1)2 – 4(x + 1) + 2 = 0 using substitution.

solniwko [45]

For this case we have to:

Let $u = x + 1$

So:

$u ^ 2-4u + 2 = 0$

We have the solution will be given by:

$u = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

Where:

$a = 1\\b = -4\\c = 2$

Substituting:

$u = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (1) (2)}} {2 (1)}\\u = \frac {4 \pm \sqrt {16-8}} {2}\\u = \frac {4 \pm \sqrt {8}} {2}\\u = \frac {4 \pm \sqrt {2 ^ 2 * 2}} {2}\\u = \frac {4 \pm2 \sqrt {2}} {2}$

The solutions are:

$u_ {1} = \frac {4 + 2 \sqrt {2}} {2} = 2 + \sqrt {2}\\u_ {2} = \frac {4-2 \sqrt {2}} {2} = 2- \sqrt {2}$

Returning the change:

$2+ \sqrt {2} = x_ {1} +1\\x_ {1} = 1 + \sqrt {2}\\2- \sqrt {2} = x_ {2} +1\\x_ {2} = 1- \sqrt {2}$

Answer:

$x_ {1} = 1 + \sqrt {2}\\x_ {2} = 1- \sqrt {2}$

3 0

2 years ago

Read 2 more answers

If x = -4, calculate 2x squared - 5

Varvara68 [4.7K]

You just substitute the value of x in the given equation:

f(x) = 2x^2 - 5

f(-4) = 2 (-4)^2 -5 = 27

f(x) : a name given to the equation
f(-4): refer to the equation when the value of the independent variable x has the value between the brackets.

5 0

2 years ago

Write down a fraction whose value lies between 3 and 4; whose denominator is a multiple of 5; whose numerator is a multiple of 1

slamgirl [31]

Answer:

33/10 and 55/15

Step-by-step explanation:

Possible denominators that are multiples of 5 less than 17 are 5, 10, 15.

Corresponding numerator ranges are [15, 20], [30, 40], and [45, 60]. In only two of these ranges are there any multiples of 11.

33 in [30, 40]

55 in [45, 60]

So, there are only two possible fractions meeting your requirement:

33/10 and 55/15

6 0

2 years ago