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3 years ago

PQ is parallel to RS. PR and QS are perpendicular to PQ and RS the ratio of the lengths of PR and QS is......

Mathematics

1 answer:

Lady bird [3.3K]3 years ago

6 0

$\frac{PQ}{PR} = \frac{RS}{QS}$ → $\frac{PR}{PQ} = \frac{QS}{RS}$

the ratio of PR to QS is 1:1 because they are congruent (equal)

If you draw a picture, you will see that PQRS is a parallelogram.

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The weight of 72 books is 9 kg.what is the weight of 80 such books? How many such books would weigh 6kg?

Alex17521 [72]

9 divided by 72 = 0.125
0.125 is the weight of one book
0.125 times 80 = 10kg.
So 80 books weight 10kg.
72 divided by 9 = 8 books make up 1 kg.
8 times 6 =48

4 0

3 years ago

Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

mario62 [17]

Answer: The required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$

From equation (ii), we get

$B=\dfrac{7}{3}.$

Thus, the required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

7 0

3 years ago

-3x - 3y - z= -6

kifflom [539]

Answer:

Ammm, I think the answer you have is wrong, but what I really think is that the equations you put in are not the right ones because the answer for these is "ugly" (decimals etc...). Want to check your equations?

Step-by-step explanation:

Here is a step by step explanation to your equations (I assumed +3z for the second but you can change it). You can edit it for your equations if you want to.

https://simplisico.com/share/q/QiWKOyN5uYgkSG4iBXOmdG0i

5 0

3 years ago

Solve by factoring: x^4-12x^2= 64

USPshnik [31]

The final solution is all the values that make (x+4)(x−4)(x2+4)=0x+4⁢x-4⁢x2+4=0 true.x=−4,4,2i,−2i

7 0

3 years ago

Read 2 more answers

Y varies directly as x. Y=42 when x=6. Find the value of y when x =15.

pentagon [3]

Answer:

y=120

Step-by-step explanation:

Y varies directly as x is written as

$y \alpha x$

introducing a constant

y=kx

from the question, when y=48, x=6

substitute it in the formula

48=6k

making k the subject by dividing through by 6

$\frac{48}{6} = \frac{6k}{6}$

k=48/6

k=8

put the value of k in the expression y=kx

y=8x

from the question,the value of y when x=15 is given by

y=8×15

y=120.

4 0

3 years ago