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2 years ago

8

More points also what is 2x3+4x6x8+9

2 answers:

-Dominant- [34]2 years ago

7 0

Answer:

the answer is 207

Step-by-step explanation:

using bodmas

ryzh [129]2 years ago

4 0

Answer:

The answer is 207

Step-by-step explanation:

you just do what it says to find the answer.

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Explain why the oppisite of a positive number will always be a negitive number

Aloiza [94]

Answer:

because a positive is over zero and a negative is below zero so from that you can tell that the opposite of any positive will be a negative.

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

P³ = 1/8 please help me if anybody can .

ivanzaharov [21]

Answer:

$p= \dfrac{1}{2}$

Step-by-step explanation:

Given equation:

$p^3=\dfrac{1}{8}$

Cube root both sides:

$\implies \sqrt[3]{p^3}= \sqrt[3]{\dfrac{1}{8}}$

$\implies p= \sqrt[3]{\dfrac{1}{8}}$

$\textsf{Apply exponent rule} \quad \sqrt[n]{a}=a^{\frac{1}{n}}:$

$\implies p= \left(\dfrac{1}{8}\right)^{\frac{1}{3}}$

$\textsf{Apply exponent rule} \quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}:$

$\implies p= \dfrac{1^{\frac{1}{3}}}{8^{\frac{1}{3}}}$

$\textsf{Apply exponent rule} \quad 1^a=1:$

$\implies p= \dfrac{1}{8^{\frac{1}{3}}}$

Rewrite 8 as 2³:

$\implies p= \dfrac{1}{(2^3)^{\frac{1}{3}}}$

$\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\implies p= \dfrac{1}{2^{(3 \cdot \frac{1}{3})}}$

Simplify:

$\implies p= \dfrac{1}{2^{\frac{3}{3}}}$

$\implies p= \dfrac{1}{2^{1}}$

$\implies p= \dfrac{1}{2}$

3 0

1 year ago

Read 2 more answers

Algebric expressions formula

Arada [10]

Answer:

a2 – b2 = (a – b)(a + b)

(a + b)2 = a2 + 2ab + b

a2 + b2 = (a + b)2 – 2ab

(a – b)2 = a2 – 2ab + b

That's all i know- rip TvT

6 0

2 years ago

Find the global maximum and global minimum values (if they exist) of x 2 + y 2 in the region x + y = 1. If there is no global ma

Grace [21]

Given that $x+y=1$ , we have $y=1-x$ , so that

$f(x,y)=x^2+y^2\iff g(x)=x^2+(1-x)^2$

Take the derivative and find the critical points of $g$ :

$g'(x)=2x-2(1-x)=4x-2=0\implies x=\dfrac12$

Take the second derivative and evaluate it at the critical point:

$g''(x)=4>0$

Since $g''$ is positive for all $x$ , the critical point is a minimum.

At the critical point, we get the minimum value $g\left(\frac12\right)=f\left(\frac12,\frac12\right)=\frac12$ .

5 0

3 years ago

The manager of a music store has kept records of

Marysya12 [62]

Answer:

(a) $P(x\le 3) = 0.75$

(b) $P(x\le 3) = 0.75$

<em>(b) is the same as (a)</em>

(c) $P(x \ge 5) = 0.10$

(d) $P(x = 1\ or\ 2) = 0.55$

(e) $P(x > 2) = 0.45$

Step-by-step explanation:

Given

$\begin{array}{ccccccc}{CDs} & {1} & {2} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.30} & {0.25} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}$

Solving (a): Probability of 3 or fewer CDs

Here, we consider:

$\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}$

This probability is calculated as:

$P(x\le 3) = P(1) + P(2) + P(3)$

This gives:

$P(x\le 3) = 0.30 + 0.25 + 0.20$

$P(x\le 3) = 0.75$

Solving (b): Probability of at most 3 CDs

Here, we consider:

$\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}$

This probability is calculated as:

$P(x\le 3) = P(1) + P(2) + P(3)$

This gives:

$P(x\le 3) = 0.30 + 0.25 + 0.20$

$P(x\le 3) = 0.75$

<em>(b) is the same as (a)</em>

<em />

Solving (c): Probability of 5 or more CDs

Here, we consider:

$\begin{array}{ccc}{CDs} & {5} & {6\ or\ more}\ \\ {Prob} & {0.05} & {0.05}\ \ \end{array}$

This probability is calculated as:

$P(x \ge 5) = P(5) + P(6\ or\ more)$

This gives:

$P(x\ge 5) = 0.05 + 0.05$

$P(x \ge 5) = 0.10$

Solving (d): Probability of 1 or 2 CDs

Here, we consider:

$\begin{array}{ccc}{CDs} & {1} & {2} \ \\ {Prob} & {0.30} & {0.25} \ \ \end{array}$

This probability is calculated as:

$P(x = 1\ or\ 2) = P(1) + P(2)$

This gives:

$P(x = 1\ or\ 2) = 0.30 + 0.25$

$P(x = 1\ or\ 2) = 0.55$

Solving (e): Probability of more than 2 CDs

Here, we consider:

$\begin{array}{ccccc}{CDs} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}$

This probability is calculated as:

$P(x > 2) = P(3) + P(4) + P(5) + P(6\ or\ more)$

This gives:

$P(x > 2) = 0.20+ 0.15 + 0.05 + 0.05$

$P(x > 2) = 0.45$

3 0

3 years ago