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DIA [1.3K]
3 years ago
15

Find x on this triangle

Mathematics
1 answer:
blsea [12.9K]3 years ago
7 0

Answer:

3 sqrt(3) =x

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

cos theta = adj / hyp

cos 30 = x/6

6 cos 30 = x

6 ( sqrt(3)/2) = x

3 sqrt(3) =x

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What is 35 cents times 500 ​
kvasek [131]

Answer:

17500

Step-by-step explanation:

you just multiply it by 35 if it is pennies but you have to tell me if it is dimes or whatever

7 0
4 years ago
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Find the smallest 4 digit number such that when divided by 35, 42 or 63 remainder is always 5
alex41 [277]

The smallest such number is 1055.

We want to find x such that

\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}

Taking everything together, we end up with the system

\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

Now the moduli are coprime and we can apply the CRT.

We start with

x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2

which means the first term is fine and doesn't require adjustment.

Taken modulo 3, we have

x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since 2\cdot4\equiv8\equiv1\pmod7, the inverse of 2 is 4.

So, we have to adjust x to

x=3\cdot5\cdot7+2^2\cdot5\cdot7+0+2^3\cdot3\cdot5^2=845

and from the CRT we find

x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}

so that the general solution x=210n+5 for all integers n.

We want a 4 digit solution, so we want

210n+5\ge1000\implies210n\ge995\implies n\ge\dfrac{995}{210}\approx4.7\implies n=5

which gives x=210\cdot5+5=1055.

5 0
3 years ago
Whitch of the following can not be classified as a parallagrams
Sindrei [870]
Can you show the choices you have
3 0
4 years ago
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Raman mixed juice and water in the ratio 2 : 3 and Meera mixed juice and water in the ratio 1 : 4. They then mixed 150 ml of Ram
Kay [80]

Answer:

80

Step-by-step explanation:

ramans ratio is 2:3

so in 150ml there are 2 parts of juice and 3 parts of water

2x + 3x = 150

x = 30

so, juice in 150 ml of raman's mixture = 2x = 2* 30 = 60ml

meera ratio is 1:4

similarly,

1y + 4y = 100

y = 20

so, juice in 100 ml of meera's mixture is y = 20ml

so, quanitity of juice in final mixture is 60 + 20 = 80

4 0
3 years ago
If m and n are whole numbers such that m^n = 121, then find the value of (m-1)^n+1.​
stiv31 [10]

Because m and n are whole numbers, we only have 3 values for m, and 3 values for n :

1^{121} ; 11^{11} ;121^{1} (or you can say because 121 only have the factors of 1 ; 11 ; 121, those are the only possible values for m and n, then conclude the possible values)

Now, we will substitute them in :

If m = 1 ; n = 121

(1-1)^{121+1} = 0

If m = 11 ; n = 11

(11-1)^{11+1} = 1000000000000

If m = 121 ; n = 1

(121-1)^{1+1} = 14400

3 0
2 years ago
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