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3 years ago

HELP WITH MATH!! RLLY COULD USE IT

Mathematics

2 answers:

san4es73 [151]3 years ago

7 0

Answer:

120 square units

Step-by-step explanation:To find the area you do length times hight or l x h. If you want you can also make the triangle and move it to the other side to create a square if thats helps :) but you always do LxH for area :)

Alborosie3 years ago

7 0

Answer:

120

Step-by-step explanation:

Since it's a parallelogram, it has to be A = BH.

Base in this problem: 10

Height in this problem: 12

10x12=120

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F(x)=3^x+2 domain and range

Irina-Kira [14]

Answer:

domain: (-∞ ,∞)

range: (2,∞ )

Step-by-step explanation:

6 0

3 years ago

Solve for x in the following triangle.

Pavlova-9 [17]

Answer:

x = 12

Step-by-step explanation:

The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

8 + 6x is an exterior angle of the triangle, thus

8 + 6x = 30 + 4x + 2 = 4x + 32 ( subtract 4x from both sides )

8 + 2x = 32 ( subtract 8 from both sides )

2x = 24 ( divide both sides by 2 )

x = 12

3 0

3 years ago

Use the form of the definition of the integral given in the theorem to evaluate the integral. ∫ 0 − 2 ( 7 x 2 + 7 x ) d x

Murrr4er [49]

Answer:

$\int _{-2}^07x^2+7xdx=\frac{14}{3}$

Step-by-step explanation:

The definite integral of a continuous function <em>f</em> over the interval [a,b] denoted by $\int\limits^b_a {f(x)} \, dx$ , is the limit of a Riemann sum as the number of subdivisions approaches infinity. That is,

$\int\limits^b_a {f(x)} \, dx=\lim_{n \to \infty} \sum_{i=1}^{n}\Delta x \cdot f(x_i)$

where $\Delta x = \frac{b-a}{n}$ and $x_i=a+\Delta x\cdot i$

To evaluate the integral

$\int\limits^{0}_{-2} {7x^{2}+7x } \, dx$

you must:

Find $\Delta x$

$\Delta x = \frac{b-a}{n}=\frac{0+2}{n}=\frac{2}{n}$

Find $x_i$

$x_i=a+\Delta x\cdot i\\x_i=-2+\frac{2i}{n}$

Therefore,

$\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} f(-2+\frac{2i}{n})$

$\int\limits^{0}_{-2} {7x^{2}+7x } \, dx=\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7(-2+\frac{2i}{n})^{2} +7(-2+\frac{2i}{n})$

$\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7(-2+\frac{2i}{n})^{2} +7(-2+\frac{2i}{n})\\\\\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7[(-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})]\\\\\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} (-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})$ $\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} (-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})\\\\\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} 4-\frac{8i}{n}+\frac{4i^2}{n^2} -2+\frac{2i}{n}\\\\\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} \frac{4i^2}{n^2}-\frac{6i}{n}+2$

$\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} \frac{4i^2}{n^2}-\frac{6i}{n}+2\\\\\lim_{n \to \infty}\frac{14}{n}[ \sum_{i=1}^{n} \frac{4i^2}{n^2}-\sum_{i=1}^{n}\frac{6i}{n}+\sum_{i=1}^{n}2]\\\\\lim_{n \to \infty}\frac{14}{n}[ \frac{4}{n^2}\sum_{i=1}^{n}i^2 -\frac{6}{n}\sum_{i=1}^{n}i+\sum_{i=1}^{n}2]$

We can use the facts that

$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$

$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$

$\lim_{n \to \infty}\frac{14}{n}[ \frac{4}{n^2}\cdot \frac{n(n+1)(2n+1)}{6}-\frac{6}{n}\cdot \frac{n(n+1)}{2}+2n]\\\\\lim_{n \to \infty}\frac{14}{n}[-n+\frac{2\left(n+1\right)\left(2n+1\right)}{3n}-3]\\\\\lim_{n \to \infty}\frac{14\left(n^2-3n+2\right)}{3n^2}$

$\frac{14}{3}\cdot \lim _{n\to \infty \:}\left(\frac{n^2-3n+2}{n^2}\right)\\\\\mathrm{Divide\:by\:highest\:denominator\:power:}\:1-\frac{3}{n}+\frac{2}{n^2}\\\\\frac{14}{3}\cdot \lim _{n\to \infty \:}\left(1-\frac{3}{n}+\frac{2}{n^2}\right)\\\\\frac{14}{3}\left(\lim _{n\to \infty \:}\left(1\right)-\lim _{n\to \infty \:}\left(\frac{3}{n}\right)+\lim _{n\to \infty \:}\left(\frac{2}{n^2}\right)\right)\\\\\frac{14}{3}\left(1-0+0\right)\\\\\frac{14}{3}$