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2 years ago

I need help anyone plss :)

Mathematics

1 answer:

Brilliant_brown [7]2 years ago

7 0

Answer:

Step-by-step explanation:

the following are true

1) yes

2)No

3)yes

4)yes

5)yes

6)no

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Factor by grouping. 9c^3-12c^2+18c-24

xeze [42]

Answer:

$9c^3 - 12c^2 - 18c - 24= [3c^2 - 6][3c - 4]$

Step-by-step explanation:

Given

$9c^3 - 12c^2 - 18c - 24$

Required

Factor

Group into 2

$[9c^3 - 12c^2] - [18c + 24]$

Factorize each group

$3c^2[3c - 4] - 6[3c - 4]$

Factor out 3c - 4

$[3c^2 - 6][3c - 4]$

Hence:

$9c^3 - 12c^2 - 18c - 24= [3c^2 - 6][3c - 4]$

8 0

2 years ago

Keiko sold 3 less than three-fourths of his sister’s sales. Which expression represents what Keiko sold?

Valentin [98]

Let’s represent his sister’s sales as x

3/4ths of his sister’s sales would be:

3/4x

3 less than 3/4ths of his sister’s sales would be:

3/4x - 3

3 0

3 years ago

The braking distance, in feet of a car a Travling at v miles per hour is given.

irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

The braking distance at 25 mph, is approximately 63.7 ft.

The braking distance at 55 mph, is approximately 298.35 ft.

The braking distance at 85 mph, is approximately 708.92 ft.

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

$D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.$

At 25 mph, the braking distance is approximately 63.7 ft.

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945 ft.s

$D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.$

At 55 mph, the braking distance is approximately 298.35 ft.

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

$D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.$

At 85 mph, the braking distance is approximately 708.92 ft.

b. The speed of the car when the braking distance is 450 feet is given as follows;

$450 = 2.6 + \dfrac{v^2}{22}$

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ 98.211 ft./s

Learn more here:

brainly.com/question/18591940

8 0

2 years ago

Name the theorem or postulate that lets you immediately conclude ∆ABD ≈ ∆CBD.

enot [183]

I think that the correct answer is D. SAS
But I’m no sure tho
Hope you get it right

7 0

2 years ago

Read 2 more answers

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago