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3 years ago

I need healpp pleaseeeee

Mathematics

2 answers:

solong [7]3 years ago

7 0

I don’t really understand

DENIUS [597]3 years ago

5 0

Answer:

I don't get it...

Step-by-step explanation:

Please elaborate on the question...

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Please help me prove that ED is congruent to BA! I thought I had it right by proving right and vertical angles, but I’m missing

Vikki [24]

Step-by-step explanation:

$\overline{DA} \text{ bisects } \overline{EB}$ is given

$\overline{BC} \cong \overline{EC}$ definition of bisect

$\overline{EB} \perp \overline{ED}, \, \overline{EB} \perp \overline{BA}$ are both given

$\angle B, \, \angle E \text{ are right angles}$ definition of perpendicular

$\angle B \cong \angle E$ because all right angles are congruent

$\angle{ACB} \cong \angle{DCE}$ vertical angles are congruent

$\triangle{ACB} \cong \triangle{DCE}$ ASA (angle-side-angle)

$\overline{ED} \cong \overline{BA}$ CPCTC (corresponding parts of congruent triangles are congruent)

5 0

3 years ago

A 450-gallon tank full of water is draining at a rate of 20 gallons per minute.

Stels [109]

Answer:

The answer is below

Step-by-step explanation:

A function show the relationship between an independent variable and a dependent variable. The independent variable (input) does not depend on other variables while the dependent variable (output) depend on other variables.

In this question, amount of water in tank is dependent variable and the time taken is the independent variable. Let y represent amount of water in tank and x represent the time it has been draining. Therefore:

a) This can be represented by the equation:

y = 450 - 20x

At 7 minutes (x = 7)

y = 450 - 20(7)

y = 310 gallons

b) For 200 gallons (y = 200), the time taken is:

200 = 450 - 20x

200 - 450 = -20x

-20x = -250

x = -250/-20

x = 12.5 minutes

c) y = 450 - 20x

d) The graph was plotted using geogebra onling graphing calculator.

e) When the tank is empty, y = 0, hence

0 = 450 - 20x

20x = 450

x = 450 / 20

x = 22.5 minutes

6 0

3 years ago

Does 4=-3/2(-4)-2 equal out to (-4,4)

Marizza181 [45]

Answer:

yes

Step-by-step explanation:

well im going to assume that the original equation is y = -3/2x - 2

if that is the case, then if you plug in the numbers

4 = -3/2(-4) - 2

4 = 6 - 2

4 = 4

so yes when x is -4, y is 4; (x, y) (-4, 4)

5 0

4 years ago

Find the 15th term of the arithmetic sequence whose common difference is d=9 and whose first term is a, = 2.

Alenkinab [10]

Nose rkrbtje de donde salió la luz de

6 0

3 years ago

For the following integral, find the approximate value of the integral with 4 subdivisions using midpoint, trapezoid, and Simpso

PIT_PIT [208]

Answer:

$\textsf{Midpoint rule}: \quad \dfrac{2\pi}{\sqrt[3]{2}}$

$\textsf{Trapezium rule}: \quad \pi$

$\textsf{Simpson's rule}: \quad \dfrac{4 \pi}{3}$

Step-by-step explanation:

Midpoint rule

$\displaystyle \int_{a}^{b} f(x) \:\:\text{d}x \approx h\left[f(x_{\frac{1}{2}})+f(x_{\frac{3}{2}})+...+f(x_{n-\frac{3}{2}})+f(x_{n-\frac{1}{2}})\right]\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}$

Trapezium rule

$\displaystyle \int_{a}^{b} y\: \:\text{d}x \approx \dfrac{1}{2}h\left[(y_0+y_n)+2(y_1+y_2+...+y_{n-1})\right] \quad \textsf{where }h=\dfrac{b-a}{n}$

Simpson's rule

$\displaystyle \int_{a}^{b} y \:\:\text{d}x \approx \dfrac{1}{3}h\left(y_0+4y_1+2y_2+4y_3+2y_4+...+2y_{n-2}+4y_{n-1}+y_n\right)\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}$

Given definite integral:

$\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x$

Therefore:

a = 0
b = 2π

Calculate the subdivisions:

$\implies h=\dfrac{2 \pi - 0}{4}=\dfrac{1}{2}\pi$

Midpoint rule

Sub-intervals are:

$\left[0, \dfrac{1}{2}\pi \right], \left[\dfrac{1}{2}\pi, \pi \right], \left[\pi , \dfrac{3}{2}\pi \right], \left[\dfrac{3}{2}\pi, 2 \pi \right]$

The midpoints of these sub-intervals are:

$\dfrac{1}{4} \pi, \dfrac{3}{4} \pi, \dfrac{5}{4} \pi, \dfrac{7}{4} \pi$

Therefore:

$\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2}\pi \left[f \left(\dfrac{1}{4} \pi \right)+f \left(\dfrac{3}{4} \pi \right)+f \left(\dfrac{5}{4} \pi \right)+f \left(\dfrac{7}{4} \pi \right)\right]\\\\& = \dfrac{1}{2}\pi \left[\sqrt[3]{\dfrac{1}{2}} +\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}\right]\\\\ & = \dfrac{2\pi}{\sqrt[3]{2}}\\\\& = 4.986967483...\end{aligned}$

Trapezium rule

$\begin{array}{| c | c | c | c | c | c |}\cline{1-6} &&&&&\\ x & 0 & \dfrac{1}{2}\pi & \pi & \dfrac{3}{2} \pi & 2 \pi \\ &&&&&\\\cline{1-6} &&&&& \\y & 0 & 1 & 0 & 1 & 0\\ &&&&&\\\cline{1-6}\end{array}$

$\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2} \cdot \dfrac{1}{2} \pi \left[(0+0)+2(1+0+1)\right]\\\\& = \dfrac{1}{4} \pi \left[4\right]\\\\& = \pi\end{aligned}$

Simpson's rule

$\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(0+4(1)+2(0)+4(1)+0\right)\\\\& = \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(8\right)\\\\& = \dfrac{4}{3} \pi\end{aligned}$

6 0

2 years ago