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4 years ago

Can someone teach me how to do this?

Mathematics

1 answer:

Tanya [424]4 years ago

7 0

Step-by-step explanation:

$pythagorean \: theorem \\ {hypothenes}^{2} = { {x}^{2} + {y}^{2} } \\ {17}^{2} = {x}^{2} + {15}^{2} \\ 289 = {x}^{2} + 225 \\ {x}^{2} = 64 \\ x = 8cm$

You might be interested in

Which statement describes the relationship between labor cost and time for a car repair?

Nimfa-mama [501]

Answer:

Option A is correct.

Step-by-step explanation:

As we see the graph, we can say that the correct statement is :

A.)All repairs requiring 1 hour or less have the same labor cost. We can see that the coat from 0 hours to 1 hour is $50. So, the number of hours falling in this range has the same repairing cost.

B.) Labor costs the same no matter how many hours are used for a repair. This is wrong as the graph is increasing after 1 hour.

C.) Labor costs for a repair are more expensive as the number of hours increases. This is wrong as the hours are increasing from 0.25 to 0.5 then to 0.75 but they all have the same cost.

D.)There is no cost of labor for a repair requiring less than 1 hour. This is also wrong. The cost is $50.

7 0

3 years ago

In 2000 the population of a country reached 1 billion, and in 2025 it is projected to be 1.2 billion. (a) Find values for C an

Mice21 [21]

Answer:

(a) The value of C is 1.

(b) In 2010, the population would be 1.07555 billions.

(c) In 2047, the population would be 1.4 billions.

Step-by-step explanation:

(a) Here, the given function that shows the population(in billions) of the country in year x,

$P(x)=Ca^{x-2000}$

So, the population in 2000,

$P(2000)=Ca^{2000-2000}$

$=Ca^{0}$

$=C$

According to the question,

$P(2000)=1$

$\implies C=1$

(b) Similarly,

The population in 2025,

$P(2025)=Ca^{2025-2000}$

$=Ca^{25}$

$=a^{25}$ (∵ C = 1)

Again according to the question,

$P(2025)=1.2$

$a^{25}=1.2$

Taking ln both sides,

$\ln a^{25}=\ln 1.2$

$25\ln a = \ln 1.2$

$\ln a = \frac{\ln 1.2}{25}\approx 0.00729$

$a=e^{0.00729}=1.00731$

Thus, the function that shows the population in year x,

$P(x)=(1.00731)^{x-2000}$ ...... (1)

The population in 2010,

$P(2010)=(1.00731)^{2010-2000}=(1.00731)^{10}=1.07555$

Hence, the population in 2010 would be 1.07555 billions.

(c) If population P(x) = 1.4 billion,

Then, from equation (1),

$1.4=(1.00731)^{x-2000}$

$\ln 1.4=(x-2000)\ln 1.00731$

$0.33647 = (x-2000)0.00728$

$0.33647 = 0.00728x-14.56682$

$0.33647 + 14.56682 = 0.00728x$

$14.90329 = 0.00728x$

$\implies x=\frac{14.90329}{0.00728}\approx 2047$

Therefore, the country's population might reach 1.4 billion in 2047.

3 0

3 years ago

Ayudaaaaaaaaaaaaaaaaaa

Zina [86]

Answer:

A (-5, -6)

B (3, 2)

C (-8, 0)

D (-3, 7)

E (7, -1)

F (0,10)

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

ASAP HELP<br> 10.Simplify the following expression: 2n+4-0.3n-3

Oksana_A [137]

1.7n +1
I think the is the answer
I hope this helps :)

7 0

3 years ago

PLEASE HELP ON THIS AND EXPLAIN<br> THX

Wewaii [24]

N=1→f(1)=-5.25

f(n)=f(n-1)+1.75, for n=2, 3, 4, ...
n=2→f(2)=f(2-1)+1.75=f(1)+1.75=-5.25+1.75→f(2)=-3.5
n=3→f(3)=f(3-1)+1.75=f(2)+1.75=-3.5+1.75→f(3)=-1.75
n=4→f(4)=f(4-1)+1.75=f(3)+1.75=-1.75+1.75→f(4)=0
n=5→f(5)=f(5-1)+1.75=f(4)+1.75=0+1.75→f(5)=1.75

Answer:
Sequence
-5.25, -3.5, -1.75, 0, 1.75

7 0

3 years ago