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3 years ago

PLZ HELP ASAP SIMILAR AND CONGRUENT TRIANGLES

Mathematics

1 answer:

Stells [14]3 years ago

7 0

We know that

in the triangle ROA
tan (∡ROA)=12/5
(∡ROA)=arc tan(12/5)---------> 67.38°

in the triangle OPA
(∡POA)=(∡ROA)-------> 67.38°
sin (∡POA)=PA/5-------> PA=5*sin (67.38°)-----> 4.62 units

the answer is
PA=4.62 units

alternative method

in the triangle ROA
RO²=12²+5²-------> RO=13 units
sin (∡ROA)=12/13

in the triangle OPA
(∡POA)=(∡ROA)
sin (∡POA)=12/13
sin (∡POA)=PA/5
12/13=PA/5-------> PA=12*5/13-------> PA=4.62 units

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Please help guys!! this is arithmetic sequence

weqwewe [10]

ap + 1993 = 3983.

i.e ap 1990, add 1993 its then 3983.

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3 years ago

Brian invests $10,000 in an account earning 4% interest, compounded annually for 10 years. Five years after Brian's initial inve

pentagon [3]

Answer:

Brian has $776 more account in his account than Chris.

Step-by-step explanation:

Compound interest Formula:

$A=P(1+r)^t$

$I$ = A-P

A= Amount after t years

P= Initial amount

r= Rate of interest

t= Time in year

Given that,

Brian invests $10,000 in an account earning 4% interest, compounded annually for 10 years.

Here P = $10,000 , r= 4%=0.04, t=10 years

The amount in his account after 10 years is

$A=10000(1+0.04)^{10}$

=$14802.44

≈$14802

Five years after Brian's investment,Chris invests $10,000 in an account earning 7% interest, compounded annually for 5 years.

Here P = $10,000 , r= 7%=0.07, t=5 years

The amount in his account after 5 years is

$A=10000(1+0.07)^{5}$

=$14025.51

≈$14026

From the it is cleared that Brian has $(14802-14026)=$776 more account in his account than Chris.

6 0

3 years ago

5. A tank initially contains 20 lb of salt dissolved in 200-gallon of water. Starting at time t = 0, a solution containing 0.5 l

sladkih [1.3K]

Answer:

(a) Amount of salt as a function of time

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt in the tank reaches 50 lb is 23.5 minutes.

Step-by-step explanation:

The rate of change of the amount of salt can be written as

$\frac{dA}{dt} =rate\,in\,-\,rate\,out\\\\\frac{dA}{dt}=C_i*q_i-C_o*q_o=C_i*q_i-\frac{A(t)}{V}*q_o\\\\\frac{dA}{dt}=0.5*4-\frac{A(t)}{200}*4\\\\\frac{dA}{dt} =\frac{100-A(t)}{50}=-(\frac{A(t)-100}{50})$

Then we can rearrange and integrate

$\frac{dA}{dt}= -(\frac{A(t)-100}{50})\\\\\int \frac{dA}{A-100}=-\frac{1}{50} \int dt\\\\ln(A-100)=-\frac{t}{50}+C_0\\\\ A-100=Ce^{-0.02*t}\\\\\\A(0)=20 \rightarrow 20-100=Ce^0=C\\C=-80\\\\A=100-80e^{-0.02t}$

Then we have the model of A(t) like

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt reaches 50 lb is

$A(t)=100-80e^{-0.02t}=50\\\\e^{-0.02t}=(50-100)/(-80)=0.625\\\\-0.02*t=ln(0.625)=-0.47\\\\t=(-0.47)/(-0.02)=23.5$

(c) When t approaches to +infinit, the term e^(-0.02t) approaches to zero, so the amount of salt in the solution approaches to 100 lb.

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3 years ago

Rate of change for 71, 62,53,44

Vladimir79 [104]

Answer:-9

Step-by-step explanation:

71-62=9

62-53=9

53-44=9

It is negative because the numbers are decreasing

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3 years ago

In a home derby competition, there were 32 home runs hit out of 124 hits. Approximately what percent of hits were home runs?

marta [7]

Answer:

26%

Step-by-step explanation:

In home run derby competition there were 32 home rinse hit out of 124 hits approximately

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3 years ago