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2 years ago

What is the height of the tree show all work

Mathematics

2 answers:

ryzh [129]2 years ago

4 0

Step-by-step explanation:

hello,

see attached and have a beautiful day!

poizon [28]2 years ago

3 0

Answer:

x=32.91

Step-by-step explanation:

Given: Triangle

Find: x

_______

Tan(35°)=x/47

x=tan(35)×47

x=0.7×47

x=32.91

Hope it helps!

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What is the value of x ? (Make sure you give the value of x !!!!)

DochEvi [55]

Answer:

The value of x is 20 degrees.

Step-by-step explanation:

The three angles of a triangle must equal 180 degrees.

100 + 40 + 2x = 140 + 2x

180 - 140 = 40 (goal)

2(x) = 40

2(20) = 40

A) 20 degrees

6 0

2 years ago

Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?

likoan [24]

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=15,\:\quad a_n=2$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5$

$=\left(x+3\right)\left(2x^2-11x+5\right)$

$Factor: 2x^2-11x+5$

$2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)$

$=x\left(2x-1\right)-5\left(2x-1\right)$

$2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)$

$\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:}$

thus zeros of f(x) is

$x=-3,\:x=5,\:x=\frac{1}{2}$

5 0

2 years ago

Read 2 more answers

How to factor a quadratic equation and find the vertex?

Jet001 [13]

To convert a quadratic from y = ax2 + bx + c form to vertex form, y = a(x - h)2+ k, you use the process of completing the square. Let's see an example. Convert y = 2x2- 4x + 5 into vertex form, and state the vertex.

6 0

3 years ago

In which situation is the distance traveled proportional to time? A car moving in stop-and-go traffic A person biking on a trail

Sindrei [870]

For this case, what you must do is to see in which scenario the speed of keeping constant during a certain time.
"A person biking on a trail at 1212 miles per hour for 2020 minutes"
We observe that the distance in this case is proportional to the time and the constant of proportionality is the speed.
In other words:
d = v * t
Answer:
the distance traveled is proportional to time in:
"A person biking on a trail at 1212 miles per hour for 2020 minutes"

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:

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