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2 years ago

8

What constant acceleration is required to increase the speed of a car from 30 mi/h to 52 mi/h in 5 seconds? (Round your answer t

o two decimal places.)

1 answer:

Alenkasestr [34]2 years ago

6 0

Answer:

$a = 9.84 m/s^2$

Step-by-step explanation:

Given:

Initial speed (Vi) = 30 mi/h = 67.1 m/s

Final speed (Vf) = 52 mi/h = 116.32

Time period(t) = 5 seconds

We know the formula of acceleration in terms of initial speed, final speed and time:

$a = \frac{Vf -Vi}{t}$

$a = \frac{116.32-67.10}{5}$

$a = 9.84 m/s^2$

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Find all the missing elements. NEED HELP PLZZZZ

IgorC [24]

Answer:

B=42°

C=98°

c=10.7 units

Step-by-step explanation:

Given

Angle A = Ф = 40°
a=8
b=7

We know that trigonometric ratio as

sin Ф = opposite/hypotenuse

cos Ф = adjacent/hypotenuse

tan Ф = opposite/adjacent

as

sin A = sin 40 = opposite/hypotenuse

From the diagram, it is clear that the opposite of angle A is a=8 and the hypotenuse is 'c'.

also the angle

sin A = sin 40 = opposite/hypotenuse = a/c = 8/c

sin 40 = 8/c

c = 8/sin 40

c = 8/0.745

c = 10.7

so the value of c will be: 10.7 units

so

tan B = opposite/adjacent

= b/a

= 0.9

$B=\tan ^{-1}\left(0.9\right)$

$=42^{\circ \:}$

Therefore, the value of B = $42^{\circ \:}$

As the sum of the angles of triangle ABC is 180.

i.e.

A+B+C = 180

(40)° + (42)° + C = 180°

C = 180° - 40° - 42°

= 98°

Therefore, the value of C = 98°

Hence,

B=42°

C=98°

c=10.7 units

7 0

3 years ago

A triangle has measurements of 39, 52, and 65 units. Is it a right triangle?

Ivanshal [37]

Hi!

We can use the Pythagorean Theorem to see if it is a right triangle.

The Pythagorean Theorem is a² + b² = c².

c = the longest side, or the hypotenuse. In this case, the hypotenuse should be 65.
39 can be a, and 52 could be b, or 52 could be a, and 39 could be b. It doesn't matter.

39² + 52² = 65²

Is this equation true?

Yes! This equation is true.

The answer is Yes, this is a right triangle.

Hope this helps! :)

8 0

3 years ago

Read 2 more answers

Need help plz Compare using <, >, or =.

jek_recluse [69]

Meanings of inequalities :

< less than
> more than
= equal

convert both fractions to their common denominators , which is 40.

7/2/5 = 7/16/40
7/3/8 = 7/15/40

as you can see, 7/2/5 is bigger. thus the answer is > .

3 0

2 years ago

1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat

brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

$x = 2$ is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) $f(x) = 6x + 6/x$

$f\prime(x) = 6 - 6/x^2 = 0$ and $x \neq 0$

So, the roots are $x = -1$ and $x = 1$

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum.

2) $f(x)=6-4/x+2/x^2$

$f\prime(x)=4/x^2-4/x^3=0$ and $x \neq 0$

So the root is $x = 1$

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum.

3) $f(x) = 8x^2/(x-4)$

$f\prime(x) = (8x^2-64x)/(x-4)^2=0$ and $x \neq 4$

So the roots are $x = 0$ and $x = 8$

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum.

4) $f(x)=6(x-2)^{2/3} +4=0$

$f\prime(x) = 4/(x-2)^{1/3}$ has no solution and $x = 2$ is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

$x = 2$ is absolute minimum.

5) $f(x)=8\sqrt x - 6x$ for $x>0$

$f\prime(x) = (4/\sqrt x)-6 = 0$

So the root is $x = 4/9$

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum.

5 0

2 years ago

S=4LW+2WH;S+=136,L6,W=4 WHAT IS H

Mekhanik [1.2K]

Answer:

H=5

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers