All categories

3 years ago

I’m kinda confused. Can someone help.

Mathematics

1 answer:

horsena [70]3 years ago

5 0

Answer:

PQ = $PM\cos MPQ + MQ\cos MQP=2PM\cos MQP$

Step-by-step explanation:

given MQR = 60, PQR = 75

MQP = 75 -60= 15 $QR = ((18(1+\cos 30))^{2}-(2\cos 15)^{2})^{\frac{1}{2}}$

WE KNOW THAT MQR +QRM +QMR = 180

MPQ = MQP AS ANGLE OPPOSITE TO EQUAL SIDES ARE EQUAL

THEREFORE QMR = 90 - 60 = 30

therefore PQM = 75 - 60 = 15

PM = PQ because M is the mid point

therefore PR = PM + $MQ\cos QMR$

PR = $18(1+\cos 30)$

$QR = (PR^{2}-PQ^{2}))^{\frac{1}{2}}$

PQ = $PM\cos MPQ + MQ\cos MQP=2PM\cos MQP$

You might be interested in

Two quadratic functions are shown.

slava [35]

Answer:

Option (1) is the correct option.

Step-by-step explanation:

Function 1,

f(x) = 4x² + 8x + 1

= 4(x² + 2x) + 1

= 4(x² + 2x + 1 - 1) + 1

= 4(x² + 2x + 1) - 4 + 1

f(x) = 4(x + 1)² - 3

This graph opens up with the vertex or minimum point at (-1, -3)

So, the minimum value of the function is (-3) at x = -1.

Function (2)

From the given table minimum value of the function is 0 at x = -1 or minimum point as (-1, 0)

Therefore, Function 1 has the least minimum value and its coordinates are (-1, -3)

Option (1) is the correct option.

7 0

2 years ago

A company requires a password to access a data storage vault. The password must be 4 characters long and may

lisov135 [29]

Answer:

There are 36 possibilities per character (26 lower case letters + 10 numbers), so the total number of permutations are 36^4 = 1,679,616

Step-by-step explanation:

5 0

3 years ago

If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

Alisiya [41]

If $k$ is odd, then

$\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor$

while if $k$ is even, then the sum would be

$\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4$

The latter case is easier to solve:

$\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0$

which means $k=6$ .

In the odd case, instead of considering the above equation we can consider the partial sums. If $k$ is odd, then the sum of the even integers between 1 and $k$ would be

$S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)$

Now consider the partial sum up to the second-to-last term,

$S^*=2+4+6+\cdots+(k-5)+(k-3)$

Subtracting this from the previous partial sum, we have

$S-S^*=k-1$

We're given that the sums must add to $2k$ , which means

$S=2k$
$S^*=2(k-2)$

But taking the differences now yields

$S-S^*=2k-2(k-2)=4$

and there is only one $k$ for which $k-1=4$ ; namely, $k=5$ . However, the sum of the even integers between 1 and 5 is $2+4=6$ , whereas $2k=10\neq6$ . So there are no solutions to this over the odd integers.