~please only answer if you know for sure~
-Thank you-

Answers

Answer 1

Answer: ZW and DA are similar sides
WX and AB are similar sides.

Make a proprtion of the sides.

$\sf{ (WX)/(AB) = (ZW)/(DA) }$

Plug in the numbers

$\sf{ (6)/(24) = (ZW)/(18) }$

Cross multiply

$\sf{ 6* 18=ZW * 24$

And then solve for ZW

$\sf{ZW = 4.5}$

So our final answer is

$\boxed{\bf{4.5~centimeters}}$

Hope that helps :P

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What is the effect on the graph of the linear parent function, f(x) = x, when f(x) is replaced by f(x + 5)?A) The graph of the line becomes steeper.B) The graph of the line shifts horizontally right 5 units.C) The graph of the line shifts vertically down 5 units.D) The graph of the line shifts horizontally left 5 units.
you earn $3 an hour as a waitress. After working 3 hours, you earn $12, $5, and $7 in tips. How much money did you earn in total? Explain how you found your answer.

If F(x) = x+ 6 and G(x) = x^4, what is G(F(x))?O
A. A +x+6
B. (X+6)^4
C.x^4+6
D.x^4(x+6)

Answers

ANSWER

B. $(x + 6)^(4)$

EXPLANATION

The given functions are:

f(x)=x+6

and

$g(x)= {x}^(4)$

We want to find

$g(f(x)) = g(x + 6)$

This is the composition of functions, where one function becomes the input of the second function.

We plug in x+6 into the expression for g(x) to obtain,

$g(f(x)) = (x + 6)^(4)$

The correct answer is B.

The high school band and chorus are putting on a spring concert. The two groups together want to perform no more than 15 songs and the concert has to be 60 minutes or less in length. Each piece the band performs is about 5 minutes in length. Each piece the chorus sings is about 3 minutes in length. Which system of inequalities best represents the number of songs the band performs (x) and the number of longs the chorus performs (y) for the concert?

Answers

Answer:

Step-by-step explanation:

x = band songs

y = chorus songs

x + y < = 15 (thats less then or equal to)

5x + 3y < = 60 (thats less then or equal)

Explain how you can use mental math to compare 7 pounds to 120 ounces.

Answers

16 oz=1 pound
we either convert 7 pounds to oz or 120 ounces to pound
I would ocnvert ounces to pounds since it is bigger to make it smaller
120/16=60/8=30/4=15/2=7.5

120oz=7.5pounds

so we compare 7 to 7.5 (pounds)

You could use mental math if you know how many ounces = 1 pound, which is 16, so you could easily multiply 16 times 7 in your head which will equal 112, and you could see dat 120 ounces > 7 pounds, (120 ounces is greater than 7 pounds), and DAT'S how you could use mental math! =D
I hope I helped! =D

Tell whether the ratio form a proportion
14/21 =26/39

Answers

if the ratios form a proportion, the ratios will be equal. first we'll simplify the ratios:

14/21
divide top and bottom by 7
2/3

26/39
divide top and bottom by 13
2/3

ss you can see now, the ratios are indeed equal, so they do form a proportion

an item was selling for $72 is reduced to $60 find the precent decrease in price round your answer to the nearest tenth

Answers

Remember this, 5 or higher goes up, 4 or lower the number stays the same. soooo.... lets see how i can start this off...
well,... you could....60 divided by 72 equals.... 0.8333333333333
then, starting from the decimal.. you go two spaces to the right... so thats 83.3333333333
so, since 3 is lower than 4, it would be the same.. and take off the extra three after the decimal. and we're basically... rounding off right??
so your answer should be 80%
Hope i helped!!! :D

A bus goes from town A to B in an exact time. If the bus goes at the rate of 50km/h, then it will arrive B 42min later than it had to and if it increases its speed 5.5/9 m/sec., it will arrive B 30min earlier. Find:A) The distance between the two towns;
B) The exact time that it takes to arrive town B
C) The speed of the bus(by schedule) for the exact time.

Answers

Let the speed of bus for the exact time = x km/h
the distance between the cities = y km
then the exact time would be, t hours = (y/x) hours

If the bus goes at the rate of 50km/h, then it will arrive B 42min later,
speed = 50 km/h
42 minutes = 42/60 hours = 7/10 hours = 0.7 hours
time taken = t+0.7
distance = speed×time
⇒ y = 50×(t+0.7)
⇒ y = 50t + 35 ---------------------(1)

it increases its speed 5.5/9 m/sec, it will arrive B 30min earlier.
5.5/9 m/s = (5.5/9)×(18/5) km/h = 2.2 km/h
30 minutes = 30/60 = 0.5 hour
speed = (x+2.2) km/h
time = (t - 0.5) hours

distance = speed×time
⇒ y = (x+2.2)×(t-0.5)
⇒ y = ((y/t) +2.2)×(t-0.5) (t = y/x)
⇒ y = y - 0.5 (y/t) + 2.2t - 1.1
⇒ 0.5 (y/t) - 2.2t + 1.1 = 0   (subtracting y from both sides)
⇒ (y/t) - 4.4t - 2.2 = 0    (dividing both sides by 0.5)
⇒ y - 4.4t² - 2.2t = 0    (multiplying both sides by t)
⇒ 50t + 35 - 4.4t² - 2.2 t = 0 (from equation 1)
⇒ -4.4t² + 35 + 47.8t = 0
⇒ 4.4t² - 47.8t - 35 = 0

solving the quadratic equation, we get t = 11.55 hours
y = 50t + 35 = 612.5 km
x = 612.5/11.55 = 53 km/h

A) 612.5 km
B) 11.55 hours
C) 53 km/h

The Logic Defined:

1 Minute=t, (a unit of time)

Time (By schedule)=nt, (n>0), nt=number of minutes

Metre(s)=m

Speed=s (in metres per minute), s=[distance in metres]/[time in minutes]

Distance=d (in metres), d=[speed in metres per minute]*[time in minutes]

---------------------------------------------

Statement (1):

"If the bus goes at the rate of 50km/h, then it will arrive B 42min later than it had to."

Conclusion 1:

$\frac { 50km }{ h } =\frac { 50,000m }{ 60t } =\frac { 2,500m }{ 3t } \n \n \therefore \quad \frac { 2,500m }{ 3t } =nt+42t\n \n \frac { 2,500m }{ 3t } =t\left( n+42 \right)$

$\n \n 2,500m=3{ t }^( 2 )\left( n+42 \right) \n \n m=\frac { 3{ t }^( 2 )\left( n+42 \right) }{ 2,500 }$

Statement (2):

"if it increases its speed 5.5/9 m/sec., it will arrive B 30min earlier."

Conclusion 2:

$\frac { 5.5m }{ 9\quad seconds } =\frac { 5.5m }{ \frac { 9 }{ 60 } t } =\frac { 110m }{ 3t }$

$\n \n \therefore \quad \frac { 110m }{ 3t } =nt-30t\n \n \frac { 110m }{ 3t } =t\left( n-30 \right) \n \n 110m=3{ t }^( 2 )\left( n-30 \right)$

$\n \n m=\frac { 3{ t }^( 2 )\left( n-30 \right) }{ 110 }$

Conclusion 3, because of conclusion 1 and 2:

$\frac { 3{ t }^( 2 )\left( n-30 \right) }{ 110 } =\frac { 3{ t }^( 2 )\left( n+42 \right) }{ 2,500 } \n \n 7,500{ t }^( 2 )\left( n-30 \right) =330{ t }^( 2 )\left( n+42 \right) \n \n 7,500\left( n-30 \right) =330\left( n+42 \right)$

$\n \n 7,500n-225,000=330n+13,860\n \n 7,500n-330n=13,860+225,000\n \n 7,170n=238,860\n \n n=\frac { 238,860 }{ 7,170 } \n \n \therefore \quad n=\frac { 7962 }{ 239 }$

Therefore,

$Time\quad by\quad schedule=\frac { 7962 }{ 239 } t\n \n Approx:\quad 33.3\quad mins$

Now we want to find the distance between the two towns, so we say that:

$d=\frac { 2,500m }{ 3t } \cdot \left( \frac { 7962 }{ 239 } t+42t \right) \n \n =\frac { 2,500m }{ 3t } \cdot \frac { 18,000 }{ 239 } t$

$\n \n =\frac { 45,000,000 }{ 717 } m\n \n Approx:\quad 62,761.5\quad metres\n \n In\quad km\quad (approx):\quad 62.761\quad km$

So now you want to know how fast the bus has to travel to get to its destination on time...

Use the formula: s=d/t

Therefore:

$s=\frac { \frac { 45,000,000 }{ 717 } m }{ \frac { 7962 }{ 239 } t } \n \n Approx:\quad 1,883.9\quad metres\quad per\quad minute$

~please only answer if you know for sure~-Thank you-

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~please only answer if you know for sure~
-Thank you-