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3 years ago

Is each line parallel, perpendicular, or neither parallel nor perpendicular to a line whose slope is −34?

Mathematics

1 answer:

Zigmanuir [339]3 years ago

8 0

Given:

The slope of a line is $-\dfrac{3}{4}$ .

To find:

The lines in the options are parallel, perpendicular or neither parallel nor perpendicular to the given line.

Solution:

We know that the slopes of parallel lines are equal.

The slope of line Q and the slope of given line are same, i.e., $-\dfrac{3}{4}$ . So, the line Q is parallel to the given line.

The slope of a perpendicular line is the opposite reciprocal of the slope of the line because the product of slopes of two perpendicular lines is -1.

The slope of a line is $-\dfrac{3}{4}$ . It means the slope of the perpendicular line must be $\dfrac{4}{3}$ . So, the line N is perpendicular to the given line.

The slopes of line M and P are neither equal to the slope of the given line nor opposite reciprocal of the slope of the line.

Therefore, the lines M and P are neither parallel nor perpendicular.

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You purchased a rare painting for $150 that is increasing in value by 3% annually. How many years will it take until it is doubl

Ahat [919]

Answer:

23 years.

Step-by-step explanation:

It is given that the initial price of painting is $150 and its values increasing by 3% annually.

We need to find how many years will it take until it is doubled in value.

The value of painting after t years is given by

$y=150(1+0.03)^t$

$y=150(1.03)^t$

The value of painting after double is 300. Substitute y=300.

$300=150(1.03)^t$

Divide both sides by 150.

$2=(1.03)^t$

Taking log both sides.

$\log 2=\log (1.03)^t$

$\log 2=t\log (1.03)$

$t=\dfrac{\log 2}{\log (1.03)}$

$t=23.44977$

$t\approx 23$

Therefore, the required number of years is 23.

6 0

3 years ago

Suppose that the overall probability of living to age 70 is 0.62 and that the overall probability of living to age 80 is 0.23. I

velikii [3]

Given :

P( X = 70 ) = 0.62 .

P( X = 80 ) = 0.23 .

To Find :

We need to find If a person reaches their 70th birthday, what is the probability that they will live to age 80 .

Solution :

We need to find :

$P(X=70 | X=80 )$

We know ,

$P(x,y ) = \dfrac{P(x)}{P(y)}$

So ,

$P(X=70 | X=80 )=\dfrac{P(X=80)}{P(X=70)}\\\\=\dfrac{0.62}{0.23}=0.371$

Hence , this is the required solution .

8 0

3 years ago

Recall that the primes fall into three categories: Let Pi be the set of

a_sh-v [17]

Answer:

Check the explanation

Step-by-step explanation:

(a)Let p be the smallest prime divisor of (n!)^2+1 if p<=n then p|n! Hence p can not divide (n!)^2+1. Hence p>n

(b) (n!)^2=-1 mod p now by format theorem (n!)^(p-1)= 1 mod p ( as p doesn't divide (n!)^2)

Hence (-1)^(p-1)/2= 1 mod p hence [ as p-1/2 is an integer] and hence( p-1)/2 is even number hence p is of the form 4k+1

(C) now let p be the largest prime of the form 4k+1 consider x= (p!)^2+1 . Let q be the smallest prime dividing x . By the previous exercises q> p and q is also of the form 4k+1 hence contradiction. Hence P_1 is infinite

4 0

3 years ago

I need help on how to get the answer and how ?????

aksik [14]

Answer: For your work try to just find the X or missing sides. once you find that match it with the degrees connecting to the next question. repeat until you get to finish.

7 0

3 years ago

A gift box shown below is packed with small cubic 1/2 inch blocks. The blocks are packed tightly with no spaces between them.

svetlana [45]

Step-by-step explanation:

the box is

9 in × 3.5 in × 4.5 in = 141.75 in³

each block is

0.5 in × 0.5 in × 0.5 in = 0.125 in³ (1/8 in³)

so, we can fit

141.75 / 0.125 = 1134

blocks into the box.

and they fit neatly, as the dimensions of the box and the cubes allow a tight packing without any empty left over space in the box.

6 0

3 years ago