Mr. Chu's cherry tree grew 12 inches in

Mathematics

1 answer:

DanielleElmas [232]3 years ago

7 0

Answer:

16 inches after one year

Step-by-step explanation:

in 3/4 a year tree grew 12 inches

divide by 3

multiply by 4

16 inches

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The side lengths of a triangle are given by the expressions 5x+3,5x+5,and3x-2.write and simplify a linear expression for the per

serg [7]

The perimeter of the triangle is: 13x+6

Step-by-step explanation:

The perimeter is the length of the outer boundary of any closed geometrical shape.

So,

Given

$Side\ 1 = s_1 = 5x+3\\Side\ 2 = s_2 = 5x+5\\Side\ 3 = s_3 = 3x-2$

The perimeter will be:

$P = s_1+s_2+s_3\\= (5x+3)+(5x+5)+(3x-2)\\=5x+5x+3x+3+5-2\\=13x+6$

The perimeter of the triangle is: 13x+6

Keywords: Perimeter, geometry

Learn more about perimeter at:

#LearnwithBrainly

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3 years ago

In the diagram, two parallel lines are intersected by a transversal. What are the measurements of angles 1, 2, 4, 5, 6, 7, and 8

larisa86 [58]

Answer:

Step-by-step explanation:

1 6 8 55 degrees

2 4 5 7 125 degrees

8 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

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