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2 years ago

WHO EVER ANSWERS CORRECTLY FIRSRT GETS BRAINLIEST

Mathematics

2 answers:

NARA [144]2 years ago

4 0

Answer:

Total length of the road = 6 3/4 miles.

Each markers are spaced at a distance = 1/8 of a mile.

Step-by-step explanation:

Total length of the road ÷ each marker space =

a) 6 3/4 ÷ 1/8.

b) In order to solve it, we need to convert that fraction into improper fraction:

6 3/4 = 27/4

6 3/4 ÷ 1/8 = 27/4 ÷ 1/8

We need to convert division sign into multiplication sign and flip the second fraction.

27/4 × 8/1

then simplify to get

27 × 2 = 54.

There would be total number of 54 markers of lengths (1/8) of a mile.

Rainbow [258]2 years ago

3 0

I got 54, I might be wrong. I divided 6 3/4 by 1/8

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Find the key features of the equation and graph it. (Can you explain how to find them and not just the answers please)

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2 years ago

A rectangle is 51.6 cm tall and 45.6 cm wide. If it is reduced to a width of 3.8 cm, then how tall will it be?

MA_775_DIABLO [31]

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2 years ago

How do i solve that question?

yawa3891 [41]

a) The solution of this <em>ordinary</em> 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 <em>ordinary</em> differential equation is $-\frac{1}{x}$ .

The <em>particular</em> solution of the <em>ordinary</em> 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 $y = 1$ <em>, </em>then the integration constant is $C = -2$ .

The solution of this <em>ordinary</em> 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 <em>ordinary</em> 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 <em>particular</em> solution of the <em>ordinary</em> 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

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2 years ago

Multiple choice, need help with this ASAP; I'll give BRANIEST to whoever gets it right and no links

Sunny_sXe [5.5K]

Answer:

so it should be a,c,e, and b

Step-by-step explanation:

49s square root is 7 and 7 is an integer so check.

7 is a whole number check

7 is rational check

and 7 is a real number check.

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2 years ago

Consider the polynomial function.

svp [43]

The number of positive zeros is 1. The number of negative zeros is either 3 or 1.

<h3>Rule of Descartes</h3>

This states that the number of real positive zeros of a polynomial are equal to or less than by an even number the number of sign changes of the coefficients of the polynomial, f(x). Also, the number of real negative zeros of a polynomial are equal to or less than by an even number the number of sign changes of the coefficients of the polynomial, f(-x).

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Since f(x) = x⁴ + 2x³ - 11x² - 5x - 6

The cofficients are +1, + 2, -11, -5, -6

There is no sign change from + 1 to + 2.

There is a sign change from + 2 to - 11.

There is no sign change from - 11 to - 5.

There is no sign change from -5 to - 6.

Since there is only one sign change, there is 1 positive zero.

<h3>The negative zero</h3>

f(-x) = x⁴ - 2x³ - 11x² + 5x - 6

The cofficients are +1, - 2, -11, +5, -6

There is a sign change from + 1 to - 2.

There is no sign change from - 2 to - 11.

There is a sign change from - 11 to + 5.

There is a sign change from +5 to - 6.

Since there are 3 sign changes, we have 3 or 3 - 2 = 1 negative zeros.

So, the number of positive zeros is 1. The number of negative zeros is either 3 or 1.

Learn more about rule of descartes here:

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