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3 years ago

Approximately how many strides does it take to complete a marathon? Choose 1 answer:

Mathematics

2 answers:

Hunter-Best [27]3 years ago

5 0

Answer:

Step-by-step explanation:

Viefleur [7K]3 years ago

3 0

Male steps: 55,375
Female steps:62,926

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I need help please and thx you

NARA [144]

Answer:

1- 3/4

2- 5/6

3- 2/3

4- 7/10

Step-by-step explanation:

1= a

2=b

3=c

4=d

__________

4 0

3 years ago

Read 2 more answers

G(x)=8x²+5x+6 como se resuelve está operasion

Debora [2.8K]

I don't know exactly what you want to know about this operation so I will put multiple answers.

No sé exactamente qué quiere saber sobre esta operación, así que pondré varias respuestas.

x-intercept = Nothing

y-intercept = ( 0 , 6 )

Axis of symmetry = $-\frac{5}{16}$

Vertex $(-\frac{5}{16}, \frac{167}{32})$

Parabola opens up and the Vertex is the minimum

6 0

3 years ago

Point S is located at (-3,2) on the coordinate plane. Point S is reflected over the y-

grigory [225]

Answer:

(3,2)

Step-by-step explanation:

3 0

3 years ago

Solve.

Eva8 [605]

Answer:

The answer is 154.8 ml

Step-by-step explanation:

In this question, the amount of pure acid is 18% of the total solution, that is, 18% of 860 milliliters. So

0.18*860 = 154.8 ml

The answer is 154.8 ml

5 0

3 years ago

For what values of x does the curve y^2x^3-15x^2=4 have horizontal tangent lines

enyata [817]

We have the following curve:<span>

$y^2x^3-15x^2=4$

To find the tangent lines to the curve we need to use the concept of derivative, but we can’t solve this problem for $y$ , thus, let's apply implicit differentiation, so:

$\frac{d}{dx}(y^2x^3-15x^2=4) \\ \\
\frac{d}{dx}(y^2x^3)-\frac{d}{dx}(15x^2)=\frac{d}{dx}(4) \\ \\
(y^{2})'x^3+(x^3)'y^2-15(x^2)'=0 \\ \\ 2yy'x^3+3x^2y^2-30x=0 \\ \\
y'=\frac{dy}{dx}=\frac{30x-3x^2y^2}{2x^3y}=\frac{x(30-3xy^2)}{2x^3y} \\ \\
y'=\frac{30-3xy^2}{2x^2y}$

Therefore the horizontal lines occurs when $y'=0$ , then:

$\frac{30-3xy^2}{2x^2y}=0 \\ \\ That \ is, \
when: \\ \\ 30-3xy^2=0 \\ \\ \therefore xy^2=10 \rightarrow y^2=\frac{10}{x}$

If we substitute this in the original equation we have:

$\frac{10}{x}x^3-15x^2=4 \\ \\ \therefore
10x^2-15x^2=4 \\ \\ \therefore x^2=-\frac{4}{5}$

This is an absurd result because it is impossible for a squared number to get a negative number. So the conclusion is that there is no any value of $x$ in which the curve has horizontal tangent lines.</span>

8 0

3 years ago