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sdas [7]
3 years ago
12

You your driving from Austin to Houston with a constant speed of 120km/hr (75 mph). After 1.5 hr you stop for 0.5 hr. Then you c

ontinue driving with a new constant speed v2 and reach Houston 45 mins later. If the total distance from Austin to Houston is 265km, what is the speed v2?
Mathematics
1 answer:
Aneli [31]3 years ago
7 0

Answer:

69

Step-by-step explanation:

i'm just hear to get points sorry

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What value of a makes the statement below true?
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The answer would be D, 75.

Step-by-step explanation:

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artcher [175]
The answer is A because the square root function needs to be greater than or equal to zero bc the circle is filled and you can’t have a negative radical. It’s also choice A because the -1 is outside of the radical which means it’s being translated downwards from the parent function. Hope that helps ;-;
5 0
3 years ago
Express it in slope-intercept form.
natka813 [3]

Hey there! :)

Answer:

y = 1/4x - 3.

Step-by-step explanation:

Use the slope-formula to find the slope of the line:

m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}

Plug in two points from the line. Use the points (-4, -4) and (0, 3):

m  = \frac{-3 - (-4)}{0 - (-4)}

Simplify:

m = 1/4.

Slope-intercept form is y = mx + b.

Find the 'b' value by finding the y-value at which the graph intersects the y-axis. This is at y = -3. Therefore, the equation is:

y = 1/4x - 3.

7 0
2 years ago
Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y
ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; (x^2-4)^2y'' + (x + 2)y' + 7y = 0

The above equation can be better expressed as:

y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

p(x) = \dfrac{(x+2)}{(x^2-4)^2} \

p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \

p(x) = \dfrac{1}{(x+2)(x-2)^2}

Also;

q(x) = \dfrac{7}{(x^2-4)^2}

q(x) = \dfrac{7}{(x+2)^2(x-2)^2}

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

\lim \limits_{x \to-2} (x+ 2) p(x) =  \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}

\implies  \lim \limits_{x \to2}  \dfrac{1}{(x-2)^2}

\implies \dfrac{1}{16}

\lim \limits_{x \to-2} (x+ 2)^2 q(x) =  \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}

\implies  \lim \limits_{x \to2}  \dfrac{7}{(x-2)^2}

\implies \dfrac{7}{16}

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

5 0
3 years ago
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