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2 years ago

Quanto é o resultado e como ae resolve a seguinte operação: 1.205 × 32 Me ajudem preciso para hj.

Mathematics

1 answer:

Elenna [48]2 years ago

5 0

Answer:

la respuesta es 38560

Step-by-step explanation:

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If P(x) = 0.35, find the complement

katovenus [111]

Hello :
P(x) + P(x') =1 ...( x' is the <span>complement of x)
0.3 +</span> P(x') =1
P(x') = 1 - 0.3 = 0.7

4 0

3 years ago

PLEASE HELP

Usimov [2.4K]

Answer:

$r\approx0.03\text{ or about $3\%$}$

Step-by-step explanation:

The standard compound interest formula is given by:

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

Where A is the amount afterwards, P is the principal, r is the rate, n is the times compounded per year, and t is the number of years.

Since we are compounding annually, n=1. Therefore:

$\displaystyle A = P (1+r)^{t}$

Lester wants to invest $10,000. So, P=10,000.

He wants to earn $1000 interest. Therefore, our final amount should be 11000. So, A=11000.

And our timeframe is 3.3 years. So, t=3.3. Substituting these values, we get:

$11000=10000(1+r)^{3.3}$

Let’s solve for our rate r.

Divide both sides by 10000:

$1.1=(1+r)^{3.3}$

We can raise both sides to 1/3.3. So:

$\displaystyle (1.1)^\frac{1}{3.3}=((1+r)^{3.3})^\frac{1}{3.3}$

The right side will cancel:

$\displaystyle r+1=(1.1)^\frac{1}{3.3}$

So:

$\displaystyle r=(1.1)^\frac{1}{3.3}-1$

Use a calculator:

$r\approx0.03$

So, the annual rate of interest needs to be about 0.03 or 3% in order for Lester to earn his interest.

5 0

2 years ago

HELP DUE IN 10 MINS! Use the Pythagorean Theorem to solve for x.

irakobra [83]

Answer:

B. 5sqrt(2)

Step-by-step explanation:

x² = 4² + (√34)²
x² = 16 + 34
x² = 50
x = √50
x = 5√2

8 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

Explain how finding 7x20 is similar to finding 7 x 2,000. Then find each product

Sphinxa [80]

7×20 and 7×2000 are similar. The amount of zeros are the dependent factor.

7×20= 140
7×2000= 14000

The base equation is 7×2=14
Any additions of zeroes in this formula will be added at the end result.

Ex.
7×2=14
7×20=140
7×200=1400
70×200=14000
700×200=140000

5 0

2 years ago