Ask question

Ask question

All categories

valentina_108 [34]

2 years ago

11

Triangle A B C is shown. Angle A B C is a right angle. The length of hypotenuse A C is 9, the length of B C is 2, and the length

of A B is 8.77. Law of cosines: a2 = b2 + c2 – 2bccos(A) What is the measure of AngleC to the nearest whole degree? 70° 77° 80° 85°

2 answers:

Paha777 [63]2 years ago

6 0

Answer:

The correct answer is 77 degrees

Step-by-step explanation:

Its right on Edge.

IrinaVladis [17]2 years ago

5 0

85 is the answer for this measure

You might be interested in

You just graduated college (age 23) and now have your dream job of a computer animator. Your gross salary is 43,000 a year. Answ

Kipish [7]

Answer:

$93.72

Step-by-step explanation:

To be able to find how much is deducted from our paycheck monthly, we have to compute for the total amount left to pay for the health insurance.

Insurance = $4500

Company % = 75% or 0.75

Total amount left = $4500- $4500x0.75

Total amount left = $1125

Now that we know the total amount left, we simply divide the total by 12 to find the monthly payments.

Monthly Payment = $1125/12

Monthly Payment = $93.72

So a total of $93.72 will be deducted from our monthly paycheck for the health insurance.

5 0

2 years ago

Read 2 more answers

Which number is least 145. 155 or 144.809

GREYUIT [131]

145.555 is the least

7 0

2 years ago

Read 2 more answers

Any 10th grader solve it <br>for 50 points

kkurt [141]

Answer:

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$ is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., $a_{1}=A$ and common difference=D

The nth term can be written as

$a_{n}=A+(n-1)D$

pth term of given arithmetic progression is a

$a_{p}=A+(p-1)D=a$

qth term of given arithmetic progression is b

$a_{q}=A+(q-1)D=b$ and

rth term of given arithmetic progression is c

$a_{r}=A+(r-1)D=c$

We have to prove that

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0$

Now to prove LHS=RHS

Now take LHS

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)$

$=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)$

$=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)$

$=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}$

$=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}$

$=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$\neq 0$

ie., $RHS\neq 0$

Therefore $LHS\neq RHS$

ie., $\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$

Hence proved

5 0

2 years ago

Rajesh invested $30,000 into an account that compounds interest monthly at a rate of 2.16%. He has made arrangements to withdraw

Bond [772]

By Evaluating the Compound Interest, we come to know that Rajesh will have enough money in the account to cover all of the required loan payments.

The Principal Amount(P) = $30,000

Rate of Interest (r) = 2.16 %

Time(t) = 10 years

Number of Times it is Compounded in a year(n) = 12

Now, we have

$A =P(1+\frac{r}{100n}) ^{nt}$

Putting all the values, we evaluate the amount,

$A =30,000(1+\frac{2.16}{100*12}) ^{12*10}\\\\A = 30,000 * 1.240\\A = 37,225.84$

Hence, the Amount after Compound Interest = $37,225.87

Now, The loan amount that he pays = 300 *12*10 = $ 36,000

Yes, he will have enough money in the account to cover all of the required loan payments.

To read more about Compound Interest, visit brainly.com/question/29335425

#SPJ1

5 0

9 months ago

One afternoon in January, the temperature outside was -4*F. By midnight, the temperature had dropped by 10*F. what was the tempe

loris [4]

The temperature at midnight would be -14*F. Subtract 10 from -4 and you get -14*F

3 0

2 years ago