All categories

3 years ago

-5x^2-6=-4x using the Quadratic Formula.

Mathematics

1 answer:

a_sh-v [17]3 years ago

5 0

Answer:

Step-by-step explanation:

-5x² - 6 = -4x

-5x² + 4x - 6 = 0

a = -5 ; b = 4 ; c = -6

Discriminant = b² - 4ac

= 4² - 4*(-5)*(-6)

= 16 - 120

= -104

roots = $\dfrac{-b+\sqrt{D}}{2a} \ ; \ \dfrac{-b-\sqrt{D}}{2a}$

$=\dfrac{-4+\sqrt{-104}}{2*(-5)};\dfrac{-4-\sqrt{-104}}{2*(-5)}\\\\=\dfrac{-4+2i\sqrt{26}}{-10} ; \dfrac{-4-2i\sqrt{26}}{-10}\\\\=\dfrac{(-2)[2-i\sqrt{26}]}{-10} \ ; \ \dfrac{(-2)[2+i\sqrt{26}]}{-10}\\\\=\dfrac{2-i\sqrt{26}}{5} \ ; \ \dfrac{2+i\sqrt{26}}{5}$

You might be interested in

Which of the following represents a negative correlation?

babunello [35]

Answer:

The answer is letter B, The water level in a bathtub as it drains over time.

Step-by-step explanation:

It is best to learn the meaning of "negative correlation" first.

Correlation- In Mathematics, "correlation" means a measure of the relationship between two variables.

Positive Correlation- when the values of the variables increase together.

Negative Correlation- when one value of a variable increases while the other variable's value decreases.

*The situation above is asking for a negative correlation, therefore the answer is letter B (The water level in a bathtub as it drains over time).

The variables are: water level and draining time

As the draining time increases, the water level in the bathtub decreases. This is clearly an example of a negative correlation.

3 0

3 years ago

Robert weighs 20 pounds more than his friend Genesis. The sum of their weights is more than 220 lbs. What is are the smallest po

Vesnalui [34]

Answer:robert=120lbs,genesis=100

Step-by-step explanation:take genesis as (x) and since robert is 20 pounds heavier he would be (X+20) so:

x+(x+20) =220lbs

2x+20=220

2x=220-20

2x=200

x=100 which we took as genesis

which leaves robert's weight 100 +20 =120

5 0

3 years ago

In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

7 0

3 years ago

I need help with the last part

Lyrx [107]

Answer:

ok gimmie a min i will put answr

Step-by-step explanation:

4 0

3 years ago

Simplify 4 whole 2/7 + 5/7

tia_tia [17]

Answer:

Step-by-step explanation:

2/7+5/7=7/7=1 because 7/7 would be a 1 whole

4 0

3 years ago

Read 2 more answers