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3 years ago

Error analysis

2 answers:

6 0

In the second to last part, there should have been a 4 added to the right side instead of subtracted, making it x=5 and not x=-3

You put it in a little weird in that part too, watch out for that next time. The 4 that should've been in the second to last part looked like it was being multiplied by the x in the last part

professor190 [17]3 years ago

5 0

Answer:

it should be x=5 and not x=-3

Step-by-step explanation:

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PLEASE HELP URGENTLY

STALIN [3.7K]

Answer:

V= 120

Step-by-step explanation:

due to the formula, V=1/3 * S * H = 1/3 * (9*4) * 10= 1/3*360= 360/3= 120

3 0

3 years ago

Which is equivalent to the expression below?

OleMash [197]

To find the answer, combine like terms:

3p + p = 4p

4 + 12 = 16

3q has no like terms so it stays the same.

Now put them all together:

4p + 3q + 16 = Option A).

8 0

4 years ago

Identify the slope and y-intercept of the equation 2x + 5y = 20

DIA [1.3K]

Answer:

Slope = - ⅖

Y-intercept = (0, 4), where b = 4

Step-by-step explanation:

Given the linear equation in standard form, 2x + 5y = 20, where A = 2, B = 5, and C = 20:

Start by transforming the standard equation into its slope-intercept form, y = mx + b, where m = slope, and b = y-intercept.

Subtract 2x from both sides:

2x -2x + 5y = - 2x + 20

5y = -2x + 20

Divide both sides by 5 to isolate y:

$\frac{5y}{5} = \frac{-2x + 20}{5}$

y = - ⅖x + 4 ⇒ This is the slope-intercept form where the slope, m = -⅖, and the y-intercept, b = 4. The y-intercept is the point on the graph where it crosses the y-axis, and has coordinates of (0, b). The y-coordinate is the value of b in the slope-intercept form. Therefore, the y-intercept is (0, 4).

3 0

3 years ago

Triangle not drawn to scale

Liono4ka [1.6K]

Answer:

D. scalene triangle

Step-by-step explanation:

This triangle has sides that all have different lengths. Therefore, this triangle is a scalene triangle.

Tips:

Isosceles triangle - two sides have equal lengths

Equilateral triangle - all three sides have equal lengths

I hope this helped! :)

5 0

3 years ago

How do i solve that question?

yawa3891 [41]

a) The solution of this ordinary differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ .

b) The integrating factor for the ordinary differential equation is $-\frac{1}{x}$ .

The particular solution of the ordinary differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ .

<h3>How to solve ordinary differential equations</h3>

a) In this case we need to separate each variable ( $y, t$ ) in each side of the identity:

$6\cdot \frac{dy}{dt} = y^{4}\cdot \sin^{4} t$ (1)

$6\int {\frac{dy}{y^{4}} } = \int {\sin^{4}t} \, dt + C$

Where $C$ is the integration constant.

By table of integrals we find the solution for each integral:

$-\frac{2}{y^{3}} = \frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32} + C$

If we know that $x = 0$ and $y = 1$ , then the integration constant is $C = -2$ .

The solution of this ordinary differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ . $\blacksquare$

b) In this case we need to solve a first order ordinary differential equation of the following form:

$\frac{dy}{dx} + p(x) \cdot y = q(x)$ (2)

Where:

$p(x)$ - Integrating factor
$q(x)$ - Particular function

Hence, the ordinary differential equation is equivalent to this form:

$\frac{dy}{dx} -\frac{1}{x}\cdot y = x^{2}+\frac{1}{x}$ (3)

The integrating factor for the ordinary differential equation is $-\frac{1}{x}$ . $\blacksquare$

The solution for (2) is presented below:

$y = e^{-\int {p(x)} \, dx }\cdot \int {e^{\int {p(x)} \, dx }}\cdot q(x) \, dx + C$ (4)

Where $C$ is the integration constant.

If we know that $p(x) = -\frac{1}{x}$ and $q(x) = x^{2} + \frac{1}{x}$ , then the solution of the ordinary differential equation is:

$y = x \int {x^{-1}\cdot \left(x^{2}+\frac{1}{x} \right)} \, dx + C$

$y = x\int {x} \, dx + x\int\, dx + C$

$y = \frac{x^{3}}{2}+x^{2}+C$

If we know that $x = 1$ and $y = -1$ , then the particular solution is:

$y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$

The particular solution of the ordinary differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ . $\blacksquare$

To learn more on ordinary differential equations, we kindly invite to check this verified question: brainly.com/question/25731911

3 0

3 years ago