All categories

3 years ago

In ΔOPQ, o = 500 cm, p = 600 cm and q=380 cm. Find the measure of ∠P to the nearest 10th of a degree.

Mathematics

1 answer:

TEA [102]3 years ago

4 0

Answer: 84.8°

Step-by-step explanation:

a²+b²+c²-2bc(cosA)

cosA = b² + c² - a² / 2bc

That is the formula we will be using.

Then, plug in our numbers.

cosP= 500² + 380² - 600² / 2(500)(380)

cosP= 34400/380000

cosP≈ 0.09052631

cos⁻¹(0.09052631) ≈ 84.81

∠P = 84.8°

You might be interested in

Using long division to find the quotient and remainder 4x^4+2x^2-2 divided by x^2+2

emmasim [6.3K]

4x² -6
__________________
x²+2 l 4x^4 +2x² -2
l 4x^4 +8x²
--------------------
-6x² -2
-6x² -12
---------------
10

so the answer is $(4x^2-6) \frac{10}{x^2+2}$

8 0

3 years ago

1. During a field trip, 60 students are put into equal sized groups.

loris [4]

If you use for fingers and count bye five you should eventually reach 60 and how ever many fingers you have up is your answer. Witch =7

6 0

2 years ago

If you have a demand function of Qd=48-9P and a supply function of Qs=-12+6P,

jenyasd209 [6]

Answer:

The equilibrium quantity is 26.4

Step-by-step explanation:

Given

$Q_d = 48 - 9P$

$Q_s = 12 + 6P$

Required

Determine the equilibrium quantity

First, we need to determine the equilibrium by equating Qd to Qs

i.e.

$Q_d = Q_s$

This gives:

$48 -9P = 12 + 6P$

Collect Like Terms

$-9P - 6P = 12 - 48$

$-15P = -36$

Solve for P

$P = \frac{-36}{-15}$

$P =2.4$

This is the equilibrium price.

Substitute 2.4 for P in any of the quantity functions to give the equilibrium quantity:

$Q_d = 48 - 9P$

$Q_d = 48 - 9 * 2.4$

$Q_d = 26.4$

<em>Hence, the equilibrium quantity is 26.4</em>

7 0

3 years ago

What‘s 1+1? ;);) have a good day :)

galben [10]

Answer: 2 :))

Step-by-step explanation: hope you also have a good day!!

4 0

3 years ago

Read 2 more answers

ln(y+1)-lny=1+3lnx . Express y in terms of x, in a form not involving logarithms. (4 marks, Maths CIE A level)

Delicious77 [7]

$remember$
$ln(x)=log_e(x)$
$and$
$log_x(a)-log_x(b)= log_x( \frac{a}{b} )$
$and$
$log_a(b)=c$ means $a^c=b$
$and$
$blog(a)=log(a^b)$

$so$

$ln(y+1)-ln(y)=1+3ln(x)$
$ln( \frac{y+1}{y} )=1+ln(x^3)$

$minus$ $ln(x^3)$ $from$ $both$ $sides$

$ln( \frac{y+1}{y}-ln(x^3) )=1$
$ln( \frac{y+1}{yx^3})=1$
$log_e( \frac{y+1}{yx^3})=1$

$translate$

$e^1=\frac{y+1}{yx^3}$
$e=\frac{y+1}{yx^3}$

$times$ $both$ $sides$ $by$ $yx^3$

$eyx^3=y+1$

$minus$ $y$ $from$ $both$ $sides$

$eyx^3-y=1$
$y(ex^3-1)=1$

$divide$ $both$ $sides$ $by$ $(ex³-1)$

$y= \frac{1}{ex^3-1}$

6 0

3 years ago

Read 2 more answers