All categories

2 years ago

WILL MARK BRAINLIEST! PLEASE HELP

1 answer:

4 0

Part 1) In triangle PQR, mP= 43, PQ=7.5 and PR= 8.4. What is mR to the nearest degree?

we know that
the law of cosines establishes that
c²=a²+b²-2*a*b*cos C

in this problem

c=QR
a=PQ=7.5
b=PR=8.4
C=angle P=43°
c²=a²+b²-2*a*b*cos C----> c²=7.5²+8.4²-2*7.5*8.4*cos 43°
c²=34.66--------> c=5.89
QR=5.9

applying the law of sines
7.5/sin R=5.9/sin 43-----> 7.5*sin 43=5.9*sin R---> sin R=7.5*sin 43/5.9
sin R=0.8669---------> R=arc sin (0.8669)----> R=60.11°----> R=60°

the answer Part 1) is the option
D. 60

Part 2) In triangle ABC mA=43, mB=62 and BC=22 in. What is AB to the nearest tenth of an inch?

we know that
∠C=180-(62+43)-----> ∠C=75°

Applying the law of sines
22/sin 43=AB/sin 75------> AB=22*sin 75/ sin 43-----> AB=31.16 in
AB=31.2 in

the answer Part 2) is
D.31.2

You might be interested in

How many square inches in a foot

Cloud [144]

Answer:

There are 144 square inches in a square foot.

Step-by-step explanation:

A square foot is a square that is 1 foot by 1 foot. Since there are 12 inches in 1 foot, and to get the area you need to multiple 2 adjacent sides together, you do 12 times 12 which is 144 square inches. Many people have the misconception that a square foot has 12 square inches, but this is wrong since a square foot has dimensions of 12 inches by 12 inches, but when you multiply them together, you can see that the answer is actually 144.

5 0

2 years ago

2) AY RA O positive zero O negative O undefined

bezimeni [28]

Answer:

negative

hope it helps;)

8 0

3 years ago

Read 2 more answers

the number of members in a club dropped from 80 to 52. what is the percent of decrease in the number of members

Illusion [34]

Percent of decrease :
Original - New / Original
so...
80 - 52 / 80 ---->
28/80
.35 multiply by 100 to get percent
35% increase Hope I helped if you have any questions let me know! :)

7 0

2 years ago

Find the integral using substitution or a formula.

Nadusha1986 [10]

$\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx$

Derivative of the denominator:
$\rm (x^2+2x+5)'=2x+2$

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

$\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx$

and then split the fraction,

$\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx$

Based on our previous test, we know that a simple substitution will work for the first integral: $\rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx$

So the first integral changes,

$\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx$

integrating to a log,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx$

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

$\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2$

So we have this going on,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx$

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx$

Bringing all that stuff together as a single square,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx$

Making the substitution: $\rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx$

giving us,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du$

simplying a lil bit,

$\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du$

and hopefully from this point you recognize your arctangent integral,

$\rm ln|x^2+2x+5|+\frac52arctan(u)$

undo your substitution as a final step,
and include a constant of integration,

$\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c$

Hope that helps!
Lemme know if any steps were too confusing.

8 0

3 years ago

Of the 120 rooms in a hotel, 55% are single bed rooms, 30% are double bed rooms and the rest are deluxe rooms. How many deluxe r

Nat2105 [25]

Hey there !

Check the attachment.
Hope it helps you :)

6 0

3 years ago