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3 years ago

Round 5649 to the nearst ten hundred thousand

Mathematics

1 answer:

elixir [45]3 years ago

4 0

5,649 rounded to the nearest thousand is:

6,000 because 6 is larger than 5

Hope I Helped!!! :-)

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Among the career home run leaders for Major League Baseball, hank aaron has 181 fewer than twice the number of chipper Jones. Ha

adoni [48]

Answer: He has 287 home runs

Step-by-step explanation:

Let's define:

H = home runs of Hank

J = home runs of Jones.

We know that:

H = 2*J - 181

H = 755

Then we can replace the second equation into the first one:

755 = 2*J - 181

(755 - 181)/2 = J

287 = J

6 0

2 years ago

Select all the expressions that are solutions to 5=2/3x. A. 5 2/3 B. 5 2/3 D. 15/2 E. 10/3

Leona [35]

Answer:

C. 5 divided by 2/3

D. 15/2

E. 3/2 X 5

hope this helps

Step-by-step explanation:

6 0

2 years ago

What is 7/8+4/7 please help

Alinara [238K]

The answer is 1 25/56

I hope this helps!

8 0

3 years ago

Read 2 more answers

Evaluate the Riemann sum for f(x) = 3 - 1/2 times x between 2 and 14 where the endpoints are included with six subintervals taki

Digiron [165]

Answer:

-6

Step-by-step explanation:

Given that :

we are to evaluate the Riemann sum for $f(x) = 3 - \dfrac{1}{2}x$ from 2 ≤ x ≤ 14

where the endpoints are included with six subintervals, taking the sample points to be the left endpoints.

The Riemann sum can be computed as follows:

$L_6 = \int ^{14}_{2}3- \dfrac{1}{2}x \dx = \lim_{n \to \infty} \sum \limits ^6 _{i=1} \ f (x_i -1) \Delta x$

where:

$\Delta x = \dfrac{b-a}{a}$

a = 2

b =14

n = 6

∴

$\Delta x = \dfrac{14-2}{6}$

$\Delta x = \dfrac{12}{6}$

$\Delta x =2$

Hence;

$x_0 = 2 \\ \\ x_1 = 2+2 =4\\ \\ x_2 = 2 + 2(2) \\ \\ x_i = 2 + 2i$

Here, we are using left end-points, then:

$x_i-1 = 2+ 2(i-1)$

Replacing it into Riemann equation;

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} \begin {pmatrix}3 - \dfrac{1}{2} \begin {pmatrix} 2+2 (i-1) \end {pmatrix} \end {pmatrix}2$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - (2+2(i-1))$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - (2+2i-2)$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 -2i$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 2i$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - 2 \lim_{n \to \infty} \sum \imits ^{6}_{i=1} i$

Estimating the integrals, we have :

$= 6n - 2 ( \dfrac{n(n-1)}{2})$

= 6n - n(n+1)

replacing thevalue of n = 6 (i.e the sub interval number), we have:

= 6(6) - 6(6+1)

= 36 - 36 -6

= -6

5 0

3 years ago

Evaluate the expression a2+6ab for a=−3 and b=4.

inn [45]

First you plug in the variables:

(-3)(2)+(6)(-3)(4)

Multiply:
-6+(-18)(4)
-6+-72

Add and your answer comes out to be:
-78

5 0

3 years ago