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2 years ago

Each cement block weighs 4 3/5 pounds. How much do 3 cement blocks weigh in total?

Mathematics

1 answer:

natali 33 [55]2 years ago

3 0

Answer:

13 4/5 pounds

Step-by-step explanation:

You multiply 4 3/5 by 3, which equals 13.8, which equals 13 4/5 pounds.

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There can be more than one option answer it fast in two minutes

san4es73 [151]

all of them except the last efg and bcd

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3 years ago

Can some one help me on this pls :(

Julli [10]

Answer:

Step-by-step explanation:

1= 2/3

2=5/6

3=21/40

4= 2 11/20

5= 1 1/3

6=5/6

7=1 1/2

8=9/10

9= 5 1/6

10= 2 1/2

6 0

2 years ago

Simplify the expression: 2/3 divided by -4 minus (1/6 - 8/6)

guapka [62]

Answer:

Step-by-step explanation:

$\frac{2}{3} \div - 4 - ( \frac{1}{6} - \frac{8}{6} )$

$- \frac{2}{3} \times \frac{1}{4} - ( - \frac{7}{6} )$

$- \frac{1}{3} \times \frac{1}{2} + \frac{7}{6}$

$- \frac{1}{6} + \frac{7}{6}$

$\boxed{\green{= 1}}$

5 0

2 years ago

boyakko [2]

Answer:

hey ok

Step-by-step explanation:

3 0

3 years ago

Apply Gaussian quadrature with n = 4 to approximate integrate sin x^2dx from 1 to 5

maks197457 [2]

First, recall that Gaussian quadrature is based around integrating a function over the interval [-1,1], so transform the function argument accordingly to change the integral over [1,5] to an equivalent one over [-1,1].

$x=2t+3\iff t=\dfrac x2-\dfrac32\implies2\mathrm dt=\mathrm dx$
$x=1\implies t=\dfrac{2-6}4=-1$
$x=5\implies t=\dfrac{10-6}4=1$

So,

$\displaystyle\int_{x=1}^{x=5}\sin x^2\,\mathrm dx=\displaystyle2\int_{t=-1}^{t=1}\sin(2t+3)^2\,\mathrm dt$

Let $f(t)=2\sin(2t+3)^2$ . With $n=4$ , we're looking for coefficients $c_i$ and nodes $x_i$ , with $1\le i\le4$ , such that

$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx c_1f(x_1)+\cdots+c_4f(x_4)$

You can either try solving for each with the help of a calculator, or look up the values of the weights and nodes (they're extensively tabulated, and I'll include a link to one such reference).

Using the quadrature, we then have

$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.3749f(-0.8611)+0.6521f(-0.3400)+0.6521f(0.3400)+0.3749f-0.8611)$
$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.5790$

4 0

3 years ago