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3 years ago

At a school trivia competition, contestants can answer two kinds of questions: easy questions and hard

Mathematics

1 answer:

lorasvet [3.4K]3 years ago

5 0

Answer:

Step-by-step explanation:

so 8*5 is 40 which leaves 39 points to be accounted for.

39/3 is 13 (easy questions)

8+13=21 which is the amount of questions Chantrea answered correctly so yee

You might be interested in

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

3 years ago

Pleasehelp, im inda stuck

padilas [110]

Answer:

I think the answer is C, sorry if I'm wrong

4 0

2 years ago

Read 2 more answers

What is the value of n? <br> n/-3 - 2 = -18

SCORPION-xisa [38]

Answer:

Step-by-step explanation:

n/-3=-16

n=-16*-3

n=48

4 0

3 years ago

Read 2 more answers

A professional baseball team won 84 games this season. The team won 14 more games than it lost. There were no ties. How many gam

siniylev [52]

14+x=84

x=70

The team lost 70 games.

84+70=x

154=x

The team played a total of 154 games.

5 0

3 years ago

Read 2 more answers

The curved urface area of a cone i 140π cm2. What will be the radiu of a cone whoe lant height i 5 cm

VashaNatasha [74]

The radius of a cone with a curved surface area of 140π cm² and a slant height of 5 cm will be 28 cm.

<h3>What is curved surface area?</h3>

The region with just curved surfaces, leaving the circular top and base, is referred to as the curved surface area. Total Surface Area is the combined area of the bases and the curved surface. The measurement of a solid's curved surface area is its outer area, which excludes the top and bottom extensions. Surface area of the cylinder that is curved: A right circular cylinder is the solid that results when a rectangle circles around one side and makes a full revolution. The curved surface area of a cylinder (CSA) is also known as the lateral surface area and is defined as the area of the curved surface of any given cylinder having a base radius "r" and height "h".

Here,

Curved surface area of cone=πrl

=140π

l=5 cm

140π=πr*5

r=140/5

r=28 cm

The radius of cone that has 5 cm as slant height and 140π cm² as the curved surface area will be 28 cm.

To know more about curved surface area,

brainly.com/question/1037971

#SPJ4

7 0

2 years ago