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4 years ago

3.(4+5) is how much more than 3.4+5?

Mathematics

2 answers:

Vesna [10]4 years ago

7 0

Answer:

Step-by-step explanation:

3(4+5) = 12

3.4+5 = 8.5

12 - 8.5 = 7.5

hope this helps you

kifflom [539]4 years ago

5 0

Answer:10 more

Step-by-step explanation:

distributive property so 3*4+3*5 so 12 an 15 or 27. 3*4+5 is 12+5 so 17 27-17 is 10

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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

A mountain climber starts a climb at an elevation of 453 feet above sea level. At his first rest stop he has climbed 162 feet, a

Sedaia [141]

162
+207
= 369

Okay, so it says he descends 285 feet after he's climbed 369 feet. That means he needs to descend a bit more to reach 453 feet, which we'll just call 0 for now. First, we need to do some subtraction:

369
- 285
= 84

Okay, we know he needs to descend 84 feet to get to his starting point.
So your answer is C

8 0

3 years ago

What are the first three terms of a geometric sequence in which (picture below) and the common ratio is 5?

Marta_Voda [28]

Answer:

5, 10, 15

Step-by-step explanation:

the increase in the sequence is 5 so since it starts from 0 which you can find out because 25 is the 5th sequence and that means that +2*5 =25 . x=0

0+ 5 = sequence 1

0+ 5+5 = seq 2

0+5+5+5= seq 3

5 0

2 years ago

Can someone help me fast pls

kykrilka [37]

Your answer would be C.) 18oz

5 0

3 years ago

Lindsey owes $5,381 on a credit card with a 24.9% interest rate compounded monthly. What is the monthly payment she should make

Digiron [165]

If you multiply .249*5381= 1339.87

3 0

3 years ago