1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kirill115 [55]
3 years ago
11

A building casts a shadow that is 18 ft long. A woman standing near the building is 5.75 ft tall and 4 feet long. What is the he

ight of the building in feet?
Mathematics
1 answer:
Lynna [10]3 years ago
5 0

Answer:

25 feet 6 inches

Step-by-step explanation:

Assuming that the woman is standing up straight, you have a situation where you have two similar right triangles. The legs of the smaller triangle are the woman's height and the length of her shadow, and the legs of the larger triangle are the building's height and the length of its shadow. We know that the sides of similar triangles are proportional, so we can write:

5ft8in =2ft=x%2Aft%2F18ft

Now you need to cross-multiply and solve for x, but multiplying feet times feet and inches is a messy calculation. Changing 5 ft 8 inches to 5 and 2/3 feet is messy also. My recommendation is to convert everything to inches, and then convert back to feet and inches when you are done.

5 ft 8 inches = (5 *12) + 8 = 68 inches

4 ft = 4 * 12 = 48 inches, and

18 ft = 18 * 12 = 216 inches.

Now the proportion looks like:

68%2F48=x%2F216, so

48x=14688 after cross-multiplying, and

x=306 after dividing by 48. So the building is 306 inches tall. But we want to know the height of the building in feet and inches. To convert back to feet and inches from inches, just do integer division by 12 where your answer is expressed as a quotient and a remainder. The quotient is the number of feet, and the remainder is the number of inches.

306 = 25 remainder 6, so the answer is 25 feet 6 inches.

Note that this answer makes sense. 5'8" is nearly 6'. 6' is one and a half times 4 feet, so the woman is a little less than one and a half times taller than her shadow is long. One and a half times the building's 18 ft shadow is 27 feet, and that is a little bit more than the 25'6' answer that we

You might be interested in
Help PLEASE ive been stuck on this forever!
AlekseyPX

Answer:

1/5

Step-by-step explanation:

2/10=1/5

4 0
3 years ago
HELP IVE GOT 5 MINS!!
faust18 [17]

Answer:

< A = 148°

Step-by-step explanation:

<A = <B → 6x - 2° = 4x + 48° → 2x = 50° → x = 25°

<A = 6(25°) - 2° = 150° - 2° = 148°

8 0
3 years ago
What is the best first step in solving the equation 3x2 + 8x = -5?
Rashid [163]
I would assume it would be dividing both sides by three although if someone were to actually solve this I would say multiply 2 and 3 together first.

Hope this helps :)
5 0
3 years ago
44. Express each of these system specifications using predicates, quantifiers, and logical connectives. a) Every user has access
DENIUS [597]

Answer:

a. ∀x (User(x) → (∃y (Mailbox(y) ∧ Access(x, y))))

b. FileSystemLocked → ∀x Access(x, SystemMailbox)

c. ∀x ∀y ((Firewall(x) ∧ Diagnostic(x)) → (ProxyServer(y) → Diagnostic(y))

d. ∀x (ThroughputNormal ∧(ProxyServer(x)∧ ¬Diagnostic(x))) → (∃y Router(y)∧Functioning(y))

Step-by-step explanation:

a)  

Let the domain be users and mailboxes. Let User(x) be “x is a user”, let Mailbox(y) be “y is a mailbox”, and let Access(x, y) be “x has access to y”.  

∀x (User(x) → (∃y (Mailbox(y) ∧ Access(x, y))))  

(b)

Let the domain be people in the group. Let Access(x, y) be “x has access to y”. Let FileSystemLocked be the proposition “the file system is locked.” Let System Mailbox be the constant that is the system mailbox.  

FileSystemLocked → ∀x Access(x, SystemMailbox)  

(c)  

Let the domain be all applications. Let Firewall(x) be “x is the firewall”, and let ProxyServer(x) be “x is the proxy server.” Let Diagnostic(x) be “x is in a diagnostic state”.  

∀x ∀y ((Firewall(x) ∧ Diagnostic(x)) → (ProxyServer(y) → Diagnostic(y))  

(d)

Let the domain be all applications and routers. Let Router(x) be “x is a router”, and let ProxyServer(x) be “x is the proxy server.” Let Diagnostic(x) be “x is in a diagnostic state”. Let ThroughputNormal be “the throughput is between 100kbps and 500 kbps”. Let Functioning(y) be “y is functioning normally”.  

∀x (ThroughputNormal ∧(ProxyServer(x)∧ ¬Diagnostic(x))) → (∃y Router(y)∧Functioning(y))

4 0
3 years ago
PLEASE HELP ASAP! I don’t recall how to do this!
MakcuM [25]

Answer:

Step-by-step explanation:

For a. we start by dividing both sides by 200:

(1.05)^x=1.885

In order to solve for x, we have to get it out from its position of an exponent.  Do that by taking the natural log of both sides:

ln(1.05)^x=ln(1.885)

Applying the power rule for logs lets us now bring down the x in front of the ln:

x * ln(1.05) = ln(1.885)

Now we can divide both sides by ln(1.05) to solve for x:

x=\frac{ln(1.885)}{ln(1.05)}

Do this on your calculator to find that

x = 12.99294297

For b. we will first apply the rule for "undoing" the addition of logs by multipllying:

ln(x*x^2)=5

Simplifying gives you

ln(x^3)=5

Applying the power rule allows us to bring down the 3 in front of the ln:

3 * ln(x) = 5

Now we can divide both sides by 3 to get

ln(x)=\frac{5}{3}

Take the inverse ln by raising each side to e:

e^{ln(x)}=e^{\frac{5}{3}}

The "e" and the ln on the left undo each other, leaving you with just x; and raising e to the power or 5/3 gives you that

x = 5.29449005

For c. begin by dividing both sides by 20 to get:

\frac{1}{2}=e^{.1x}

"Undo" that e by taking the ln of both sides:

ln(.5)=ln(e^{.1x})

When the ln and the e undo each other on the right you're left with just .1x; on the left we have, from our calculators:

-.6931471806 = .1x

x = -6.931471806

Question d. is a bit more complicated than the others.  Begin by turning the base of 4 into a base of 2 so they are "like" in a sense:

(2^2)^x-6(2)^x=-8

Now we will bring over the -8 by adding:

(2^2)^x-6(2)^x+8=0

We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution.  Let's let u=2^x

When we do that, we can rewrite the polynomial as

u^2-6u+8=0

This factors very nicely into u = 4 and u = 2

But don't forget the substitution that we made earlier to make this easy to factor.  Now we have to put it back in:

2^x=4,2^x=2

For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:

2^2=2^x

Now that the bases are the same, we can say that

x = 2

For the second solution, we will raise the 2 on the right to a power of 1 to get:

2^x=2^1

Now that the bases are the same, we can say that

x = 1

5 0
3 years ago
Other questions:
  • Find the discriminant<br> 9x^2+14x+13=8x^2
    12·1 answer
  • 20 POINTS AND WILL MARK BRAINLIEST
    8·2 answers
  • Are there less than 1 million exactly 1 million or greater than 1 million miligrams in a kilogram?
    9·2 answers
  • What points names the ordered -4 -2
    9·1 answer
  • John knows that his first 4 tests grades were 84, 79, 82, and 88. Find John's grade on the test if his average was 83.8. ​
    9·1 answer
  • Simplify the expression 2[13-1)^2/2]
    15·1 answer
  • In ΔOPQ, m∠O = (3x+12)^{\circ}(3x+12)
    9·1 answer
  • Find the unit rate of 3÷12​
    9·2 answers
  • How do you simplify square roots.​
    10·1 answer
  • Solve 3a<br> 17<br> 7<br> 8<br> 9<br> - (2² + 1)² + a<br> when a = 8
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!