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3 years ago

The square of the difference of two numbers

2 answers:

4 0

Answer: The difference between consecutive square numbers is always odd. The difference is the sum of the two numbers that are squared. ... The difference between alternate square numbers is always even; it is twice the sum of the two numbers that are squared.

Step-by-step explanation:

Dmitriy789 [7]3 years ago

4 0

Let’s consider 2 numbers be ‘a’ and ‘b’
Then
Square of difference of two numbers means square of difference of a-b which equals:
=(a-b)^2
=a^2+2ab+b^2

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the equation of a line is y=-1/2x-2. what is the equation of the line that is perpendicular to the first line and passes through

Alexxandr [17]

Perpendicular means that the slopes are negartive inverses, aka
slope1 times slope2=-1

y=mx+b
slope=m

y=-1/2x-2
slope=-1/2

we need to find negative inverse
-1/2 times what=-1
answer is 2

y=2x+b
input the point (2,-2) to find b
(x,y)
(2,-2)
-2=2(2)+b
-2=4+b
minus 4 from both sides
-6=b
y=2x-6

the equtaion is y=2x-6

8 0

3 years ago

A sum of money that is set aside to be spent for a specific purpose is a

Vadim26 [7]

Answer:

A. budget

Step-by-step explanation:

3 0

4 years ago

Help me please<br>help.me

levacccp [35]

Answer:

The answers of the questions are given below :

a) = 4096
b) = 1.25
3) = m²
4) = r⁴s³
5) = a⁸/b¹²

Step-by-step explanation:

$\large{\tt{\underline{\underline{\red{QUESTION}}}}}$

$\begin{gathered}\footnotesize\boxed{\begin{array}{c|c|c}\bf\underline{Given}&\bf\underline{Solution}&\bf{\underline{Simple\: Form}}\\\\\rule{60pt}{0.5pt} &\rule{70pt}{0.5pt}& \rule{70pt}{0.5pt}\\\\ 1.\: {4}^{6} & & \\\\ 2.\: \bigg(\dfrac{2^6}{5^3} \bigg)& &\\\\ 3. \: \Big({m}^{\frac{2}{3}}\Big)\bull \Big({m}^{\frac{4}{3}}\Big) & &\\\\4. \: \big({r}^{12} {s}^{9}\big)^{ - \frac{1}{3}} &&\\\\ 5.\bigg(\dfrac{a^4}{a^6}\bigg)^{2}& &\end{array}}\end{gathered}$

$\begin{gathered}\end{gathered}$

$\large{\tt{\underline{\underline{\red{SOLUTION}}}}}$

Question. 1

>> 4⁶

$\begin{gathered}\qquad{= 4 \times 4 \times 4 \times 4 \times 4 \times 4} \\ \qquad{= 16 \times 4 \times 4 \times 4 \times 4} \\ \qquad{= 64 \times 4 \times 4 \times 4} \\ \qquad{= 256\times 4 \times 4} \\ \qquad{= 1024 \times 4} \\ \qquad{= 4096} \end{gathered}$

Hence, the answer is 4096.

$\begin{gathered}\end{gathered}$

Question. 2

>> (2⁶/5³)^-⅓

$\begin{gathered} \qquad\implies{\bigg(\frac{2^6}{5^3}\bigg)^{ - \frac{1}{3}}}\\ \\ \qquad\implies{\bigg(\frac{64}{125}\bigg)^{ - \frac{1}{3}}}\\ \\\qquad\implies{\bigg( \frac{1}{\frac{64}{125}}\bigg)^{ \frac{1}{3}}} \\ \\ \qquad\implies{\bigg( 1 \times \frac{125}{64} \bigg)^{ \frac{1}{3}}} \\ \\ \qquad\implies{\bigg( \frac{125}{64} \bigg)^{ \frac{1}{3}}} \\ \\\qquad\implies{\bigg( \sqrt[3]{ \frac{125}{64}}\bigg)} \\ \\ \qquad\implies{\bigg( \dfrac{5}{4} \bigg)} \\ \\ \qquad\implies{\Big( 1.25\Big)}\end{gathered}$

Hence, the answer is 1.25.

$\begin{gathered}\end{gathered}$

Question. 3

>> (m^2/3)•(m^4/3)

$\begin{gathered} \qquad{= \Big({m}^{\frac{2}{3}}\Big) \bull \Big({m}^{ \frac{4}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{\frac{2}{3} + \frac{4}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{\frac{2 + 4}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{\frac{6}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{2}\Big)}\end{gathered}$

Hence, the answer is m².

$\begin{gathered}\end{gathered}$

Question. 4

>> (r¹² s⁹)^⅓

$\begin{gathered} \qquad\implies{\Big( {r}^{12} \: {s}^{9}\Big)^{\frac{1}{3}}}\\\\ \qquad\implies{\Big({r}^{\frac{12}{3} } \: {s}^{\frac{9}{3}}\Big)} \\ \\ \qquad\implies{\Big({r}^{\cancel{\frac{12}{3}}} \: {s}^{\cancel{\frac{9}{3}}}\Big)} \\ \\ \qquad\implies{\Big({r}^{4} \: {s}^{3}\Big)} \end{gathered}$

Hence, the answer is r⁴s³.

$\begin{gathered}\end{gathered}$

Question. 5

>> (a⁴/b⁶)^2

$\begin{gathered} \qquad{ = \Big(\frac{a^4}{b^6}\Big)^{2}} \\ \\ \qquad{ = \Big(\frac{a^{4 \times 2}}{b^{6 \times 2}}\Big)} \\ \\ \qquad{ = \Big(\frac{a^{8}}{b^{12}}\Big)} \end{gathered}$

Hence, the answer is a⁸/b¹².

$\underline{\rule{220pt}{3pt}}$

4 0

2 years ago

Last summer, Karl went to the beach every 5 days. Antonia went to the beach every 33 days. How often did they see each other at

vfiekz [6]

Asuming they started summer at the same time. The logical answer will be none, because it will take at least 165 days for the lcd(5) to overlap and there is not 165 days in a summer.

so it will be a rate of 1 for every 5 times Antonia went to the beach.

3 0

3 years ago

Read 2 more answers

A single die is rolled twice. The set of 36 equally likely outcomes is {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1),

lana [24]

Answer:

(9) $\frac{1}{12}$ (10) $\frac{1}{12}$ (11) $\frac{5}{12}$ (12) $\frac{1}{4}$ (13) $\frac{1}{6}$ 14) $\frac{5}{36}$ (15) $\frac{1}{12}$ (16)0

Step-by-step explanation:

The sample Space of the single die rolled twice is presented below:

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6),}.

n(S)=36

(9)Probability of getting two numbers whose sum is 9.

The possible outcomes are: (3, 6), (4, 5), (5, 4)

$P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}$

10) Probability of getting two numbers whose sum is 4.

The possible outcomes are: (1, 3),(2, 2),(3, 1),

$P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}$

11.)Find the probability of getting two numbers whose sum is less than 7.

The possible outcomes are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)

$P(\text{two numbers whose sum is less than 7})=\frac{15}{36}=\frac{5}{12}$

12.Probability of getting two numbers whose sum is greater than 8

The possible outcomes are:(4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)

$P(\text{two numbers whose sum is greater than 8})=\frac{9}{36}=\frac{1}{4}$

(13)Probability of getting two numbers that are the same (doubles).

The possible outcomes are:(1, 1)(2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

$P(\text{two numbers that are the same})=\frac{6}{36}=\frac{1}{6}$

14.Probability of getting a sum of 7 given that one of the numbers is odd.

The possible outcomes are: (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

$P(\text{getting a sum of 7 given that one of the numbers is odd.})=\frac{5}{36}$

(15)Probability of getting a sum of eight given that both numbers are even numbers.

The possible outcomes are: (2, 6), (4, 4), (6, 2)

$P(\text{getting a sum of eight given that both numbers are even numbers.})=\frac{3}{36}\\=\frac{1}{12}$

16.Probability of getting two numbers with a sum of 14.

$P(\text{getting two numbers with a sum of 14.})=\frac{0}{36}=0$

4 0

4 years ago