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2 years ago

A ___ equation is a second-degree equation of the form ax^2+bx+c=0 where a ≠ 0

Mathematics

1 answer:

klasskru [66]2 years ago

6 0

Answer:

A quadratic equation is second degree equation with the standard form ax^2+bx+c=0 as the highest power is 2.

__________________.

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A car driving on a straight track accelerates at a rate of 3.2 m/s2 for 14 s. If the initial velocity of the car was 5.1 m/s, an

liraira [26]

Answer:

The data we have is:

The acceleration is 3.2 m/s^2 for 14 seconds

Initial velocity = 5.1 m/s

initial position = 0m

Then:

A(t) = 3.2m/s^2

To have the velocity, we integrate over time, and the constant of integration will be equal to the initial velocity.

V(t) = (3.2m/s^2)*t + 5.1 m/s

To have the position equation, we integrate again over time, and now the constant of integration will be the initial position (that is zero)

P(t) = (1/2)*(3.2 m/s^2)*t^2 + 5.1m/s*t

Now, the final position refers to the position when the car stops accelerating, this is at t = 14s.

P(14s) = (1/2)*(3.2 m/s^2)*(14s)^2 + 5.1m/s*14s = 385m

So the final position is 385 meters ahead the initial position.

6 0

3 years ago

N/-3 - 2 = 8 n=___ help

lys-0071 [83]

Answer:

-30

Step-by-step explanation:

$\frac{N}{-3} -2 =8$ ( Add 2 on both sides)

$\frac{N}{-3} = 10$ ( Multiply by -3 on both sides)

$N = -30$

Verify:

$\frac{-30}{-3} =10$ (Correct)

7 0

3 years ago

Read 2 more answers

Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3>Let's understand the concept:-</h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3>Required Answer:-</h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

3 0

2 years ago

Which of the following terms is described as “the ratio of two polynomial

Sergio [31]

Answer:

rational algebraic expression

8 0

2 years ago

HELP!! PROBABILITIES!!

denis23 [38]

2×3×4 She has 24 options of cars

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3 years ago

A ___ equation is a second-degree equation of the form ax^2+bx+c=0 where a ≠ 0​

A ___ equation is a second-degree equation of the form ax^2+bx+c=0 where a ≠ 0