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3 years ago

Form a sequence that has two arithmetic means between -1 and 59.

2 answers:

7 0

The formula for solving the problem is as follow:
an = a1 + (n - 1)d
Where:
n = number of figure in the sequence = 4
d = difference between successive number =?
a1 = -1
a4 = 59
Insert the given values into the formula,
59 = -1 + (4 - 1)d
59 = -1 + 3d
59 + 1 = 3d
60 = 3d
d = 60/3 = 20
Therefore, d = 20. This implies that, there is a difference of 20 between successive numbers.
The number sequence is as follow:
-1, 19, 39, 59.

Stels [109]3 years ago

7 0

<span><span>two arithmetic means between -1 and 59.
</span>b.-1, 19, 39, 59</span>

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Answer:

4) $\frac{x}{7\cdot x +x^{2}}$ is equivalent to $\frac{1}{7+x}$ for all $x \ne -7$ . (Answer: A)

5) $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$ is equivalent to $-\frac{14}{1-5\cdot x}$ for all $x \ne \frac{1}{5}$ . (Answer: B)

6) $\frac{x+7}{x^{2}+4\cdot x - 21}$ is equivalent to $\frac{1}{x-3}$ for all $x \ne 3$ . (Answer: None)

7) $\frac{x^{2}+3\cdot x -4}{x+4}$ is equivalent to $x - 1$ . (Answer: None)

8) $\frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}$ is equivalent to $\frac{4}{3\cdot a^{3}}$ for all $a\ne 0$ . (Answer: A)

Step-by-step explanation:

We proceed to simplify each expression below:

4) $\frac{x}{7\cdot x +x^{2}}$

(i) $\frac{x}{7\cdot x +x^{2}}$ Given

(ii) $\frac{x}{x\cdot (7+x)}$ Distributive property

(iii) $\frac{1}{7+x} \cdot \frac{x}{x}$ Distributive property

(iv) $\frac{1}{7+x}$ Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

$7+x = 0$

$x = -7$

Hence, we conclude that $\frac{x}{7\cdot x +x^{2}}$ is equivalent to $\frac{1}{7+x}$ for all $x \ne -7$ . (Answer: A)

5) $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$

(i) $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$ Given

(ii) $\frac{x^{3}\cdot (-14)}{x^{3}\cdot (1-5\cdot x)}$ Distributive property

(iii) $\frac{x^{3}}{x^{3}} \cdot \left(-\frac{14}{1-5\cdot x} \right)$ Distributive property

(iv) $-\frac{14}{1-5\cdot x}$ Commutative property/Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

$1-5\cdot x = 0$

$5\cdot x = 1$

$x = \frac{1}{5}$

Hence, we conclude that $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$ is equivalent to $-\frac{14}{1-5\cdot x}$ for all $x \ne \frac{1}{5}$ . (Answer: B)

6) $\frac{x+7}{x^{2}+4\cdot x - 21}$

(i) $\frac{x+7}{x^{2}+4\cdot x - 21}$ Given

(ii) $\frac{x+7}{(x+7)\cdot (x-3)}$ $x^{2} -(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2} = (x-r_{1})\cdot (x-r_{2})$