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3 years ago

O SYSTEMS OF EQUATIONS AND MATRICES

Mathematics

1 answer:

Hoochie [10]3 years ago

8 0

Answer:

Find the midpoint of the line segment joining the points R(,) and S(,).

Step-by-step explanation:

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PLEASE ANSWER THIS!!! IF DO, YOU ARE SMART!

Ratling [72]

Answer:

33.64

Step-by-step explanation:

4 0

3 years ago

It is impossible to slice a triangular pyramid with a plane and get a quadrilateral cross section.

Ksivusya [100]

Answer:

false

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

(d) Let p represent the probability of a successful long hit. What values of p will make the long hit better than the short hit

seropon [69]

Answer:

The probability of successful long hit is greater than 0.7083

Step-by-step explanation:

Solution

The expected value for long hit is given below

P (successful hit) * 4.2 + P (Unsuccessful hit) *5.4

so,

= P * 4.2 + (1-p) * 5.4

= 5.4 - 1.2p

Now,

If the expected value of long hit is less than that of the short it

Then,

5.4 - 1.2p <4.55

P> (5.4 - 4.55)/1.2

P> 0.7083

Therefore the probability of successful long hit is greater than 0.7083

Note:

Due to my research and findings to this example, the complete question for is attached below, i solved or Question 2(d) only

8 0

3 years ago

Choose the correct answer to 1,889/36. 42 R17 51 R53 52 R15 52 R17

ozzi

Answer:

52 R17

Step-by-step explanation:

52 * 36 = 1872

1889 - 1872 = 17

--> 52 R17

3 0

3 years ago

Read 2 more answers

Question -

Sauron [17]

$\rule{200}4$

Answer : The required ratio is (14m-6):(8m+23) .

$\rule{200}4$

Here we are given that the ratio of sum of first n terms of two AP's is (7n + 1):(4n + 27) .

That is.

$\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{7n +1}{4n +27} \\$

As , we know that the sum of n terms of an AP is given by ,

$\small\sf\longrightarrow \pink{ S_n =\dfrac{n}{2}[2a +(n-1)d]} \\$

Assume that ,

First term of 1st AP = a
First term of 2nd AP = a'
Common difference of 1st AP = d
Common difference of 2nd AP = d'

Using this we have ,

$\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{\dfrac{n}{2}[2a + (n-1)d]}{\dfrac{n}{2}[2a' +(n-1)d'] } \\$

$\small\sf\longrightarrow \dfrac{7n+1}{4n+27}=\dfrac{2a + (n -1)d}{2a' + (n -1)d' } . . . . . (i) \\$

Now also we know that the nth term of an AP is given by ,

$\longrightarrow\sf\small \pink{ T_n = a + (n-1)d}\\$

Therefore,

$\longrightarrow\sf\small \dfrac{T_{m_1}}{T_{m_2}}= \dfrac{ a + (n-1)d }{a'+(n-1)d'}. . . . . (ii)\\$

$\longrightarrow\sf\small \dfrac{T_1}{T_2}=\dfrac{2a + (2n-2)d}{2a'+(2n-2)d'} . . . . . (iii)\\$

From equation (i) and (iii) ,

$\longrightarrow\sf\small n-1 = 2m-2\\$

$\longrightarrow\sf\small n = 2m -2+1 \\$

$\longrightarrow\sf\small n = 2m -1 \\$

Substitute this value in equation (i) ,

$\longrightarrow \sf\small \dfrac{2a+ (2m-1-1)d}{2a' +(2m-1-1)d'}=\dfrac{7(2m-1)+1}{4(2m-1) +27}\\$

Simplify,

$\longrightarrow\sf\small \dfrac{ 2a + (2m-2)d}{2a' +(2m-2)d'}=\dfrac{14m-7+1}{8m-4+27}\\$

$\longrightarrow\sf\small \dfrac{2[a + (m-1)d]}{2[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\$

$\longrightarrow\sf\small \dfrac{[a + (m-1)d]}{[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\$

From equation (ii) ,

$\longrightarrow\sf\small \underline{\underline{\blue{ \dfrac{T_{m_1}}{T_{m_2}}=\dfrac{ 14m-6}{8m+23}}}}\\$

$\rule{200}4$

8 0

3 years ago