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Naddika [18.5K]
3 years ago
7

Rounded to the nearest thousands 6,842

Mathematics
2 answers:
mr_godi [17]3 years ago
4 0
7,000 is the answer to this. Your welcome
ValentinkaMS [17]3 years ago
4 0
7,000.
1 thousands place, rounded by the 100 place.
Below 5 round down, 5 or above round up.
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If f(x)=3x^2-2 and g(x)=4x+2, what is the value of (f+g)(2)
elena-14-01-66 [18.8K]

(f+g)(x)=3x^2-2+4x+2=3x^2+4x\\\\(f+g)(2)=3\cdot2^2+4\cdot2=12+8=20

5 0
3 years ago
Index and laws<br><br> evaluate <br> A) 55^0<br> B)100^1
lisabon 2012 [21]

Answer:

Step-by-step explanation:

Hey there!

so answer for (A is 1)

Answer for (B is 100)

Thank you have fun!

5 0
3 years ago
Read 2 more answers
(n^3 +67² + 4n - 2)/(n + 1)
Alla [95]

Answer:

(n^3 + 4 n + 4487)/(n + 1)

Step-by-step explanation:

Simplify the following:

(n^3 + 4 n - 2 + 67^2)/(n + 1)

| | 6 | 7

× | | 6 | 7

| 4 | 6 | 9

4 | 0 | 2 | 0

4 | 4 | 8 | 9:

(n^3 + 4 n - 2 + 4489)/(n + 1)

Grouping like terms, n^3 + 4 n - 2 + 4489 = n^3 + 4 n + (4489 - 2):

(n^3 + 4 n + (4489 - 2))/(n + 1)

4489 - 2 = 4487:

Answer: (n^3 + 4 n + 4487)/(n + 1)

5 0
3 years ago
Simplify the left side of equation so it looks like the right side. cos(x) + sin(x) tan(x) = sec (x)
uranmaximum [27]

Step-by-step explanation:

Consider LHS

\cos(x)  +  \sin(x)  \tan(x)  =  \sec(x)

Apply quotient identies

\cos(x)  +   \sin(x) \times  \frac{ \sin(x) }{ \cos(x) }  =  \sec(x)

Multiply the fraction and sine.

\cos(x)  +  \frac{ \sin {}^{2} (x) }{ \cos(x) }  =  \sec(x)

Make cos x a fraction with cos x as it denominator.

\cos(x)  \times  \cos(x)  =  \cos {}^{2} (x)

so

\frac{ \cos {}^{2} (x) }{ \cos(x) }  +  \frac{ \sin {}^{2} (x) }{ \cos(x) }  =  \sec(x)

Pythagorean Identity tells us sin squared and cos squared equals 1 so

\frac{1}{ \cos(x) }  =  \sec(x)

Apply reciprocal identity.

\sec(x)  =  \sec(x)

7 0
3 years ago
Used use euler's method with step size 0.1 to estimate y(0.5), where y(x) is the solution of the initial-value problem y ' = y +
brilliants [131]
We want to solve the Initial Value Problem y' = y + 4xy, with y(0) = 1.

To use Euler's method, define
y(i+1) = y(i) + hy'(i),  for i=0,1,2, ..., 
where
h = 0.1, the step size.,
x(i) = i*h

1st step.
y(0) = 1 (given) and x(0) = 0.
y(1) ≡ y(0.1) = y(0) + h*[4*x(0)*y(0)] = 1 

2nd step.
x(1) = 0.1
y(2) ≡ y(0.2) = y(1) + h*[4*x(1)*y(1)] = 1 + 0.1*(4*0.1*1) = 1.04

3rd step.
x(2) = 0.2
y(3) ≡ y(0.3) = y(2) + h*[4*x(2)*y(2)] = 1.04 + 0.1*(4*0.2*1.04) = 1.1232

4th step.
x(3) = 0.3
y(4) ≡ y(0.4) = y(3) + h*[4*x(3)*y(3)] = 1.1232 + 0.1*(4*0.3*1.1232) = 1.258

5th step.
x(4) = 0.4
y(5) ≡ y(0.5) = y(4) + h*[4*x(4)*y(4)] = 1.258 + 0.1*(4*0.4*1.258) = 1.4593

Answer:  y(0.5) = 1.4593
4 0
3 years ago
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