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3 years ago

1. Use the quadratic formula to solve the equation. -2x^2+3x+5=0

Mathematics

2 answers:

hram777 [196]3 years ago

8 0

The answer is x= 5/2 or -1

artcher [175]3 years ago

4 0

The formula for the solution is:
$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In your case: a=-2, b=3, c=5

Replace these in the formula and you'll get:
$x_1=\frac{-b + \sqrt{b^2 - 4ac}}{2a}=\frac{-3 + \sqrt{3^2 - 4 \cdot (-2) \cdot 5}}{2 \cdot (-2)}=-1$
$x_2=\frac{-b - \sqrt{b^2 - 4ac}}{2a}=\frac{-3 - \sqrt{3^2 - 4 \cdot (-2) \cdot 5}}{2 \cdot (-2)}= \frac{5}{2}$

So you have two roots:
x1 = -1
x2 = 5/2

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If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each o

Ludmilka [50]

Answer:

16/33

Step-by-step explanation:

12C4(total possible outcomes)=495

Since there are 6 married couples there must be 6*[2C2×10C2]=270

must subtracted 15 since when there are 2 couples in a group would have doubled 270-15=255

the no of groupings with no married couples 495-255 =240

probability=no of favourable outcomes /no of possible outcome

probability=240/495= 16/33

6 0

3 years ago

-2/3x+7(3/7-3/7x) -12 divided by 6*3

Anna71 [15]

Answer:

x = (-27)/11

Step-by-step explanation:

Solve for x:

(-11 x)/54 - 1/2 = 0

Put each term in (-11 x)/54 - 1/2 over the common denominator 54: (-11 x)/54 - 1/2 = (-27)/54 - (11 x)/54:

(-27)/54 - (11 x)/54 = 0

(-27)/54 - (11 x)/54 = (-11 x - 27)/54:

(-11 x - 27)/54 = 0

Multiply both sides of (-11 x - 27)/54 = 0 by 54:

(54 (-11 x - 27))/54 = 54×0

(54 (-11 x - 27))/54 = 54/54×(-11 x - 27) = -11 x - 27:

-11 x - 27 = 54×0

0×54 = 0:

-11 x - 27 = 0

Add 27 to both sides:

(27 - 27) - 11 x = 27

27 - 27 = 0:

-11 x = 27

Divide both sides of -11 x = 27 by -11:

(-11 x)/(-11) = 27/(-11)

(-11)/(-11) = 1:

x = 27/(-11)

Multiply numerator and denominator of 27/(-11) by -1:

Answer:x = (-27)/11

5 0

3 years ago

A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tan

Artyom0805 [142]

Answer:

1) $\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) $y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) 98.23lbs

4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let $y$ represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: $\frac{dy}{dt}=rate\:in-rate\:out$

The amount coming in is $0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}$

The rate going out depends on the concentration of salt in the tank at time t.

If there is $y(t)$ pounds of salt and there are $100+2t$ gallons at time t, then the concentration is: $\frac{y(t)}{2t+100}$

The rate of liquid leaving is is 3gal\min, so rate out is $=\frac{3y(t)}{2t+100}$

The required differential equation becomes:

$\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) We rewrite to obtain:

$\frac{dy}{dt}+\frac{3}{2t+100}y=2.5$

We multiply through by the integrating factor: $e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }$

to get:

$(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }$

This gives us:

$((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }$

We integrate both sides with respect to t to get:

$(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C$

Multiply through by: $(50+t)^{-\frac{3}{2}}$ to get:

$y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }$

$y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}$

We apply the initial condition: $y(0)=0$

$0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}$

$C=-12500\sqrt{2}$

The amount of salt in the tank at time t is:

$y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

$y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}$

$y(50)=98.23$

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

$\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})$

As $t\to \infty$ , $50+t\to \infty$ and $\frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0$

This implies that:

$\lim_{t \to \infty}y(t)=\infty- 0=\infty$

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

6 0

3 years ago

HURRY PLZ! also please find tan as well!!! will mark brainliest!!

vazorg [7]

I honestly don’t know man but don’t worry, if you get a bad score it won’t matter in 30 years :)

3 0

3 years ago

10th grade geometry <br> Select statements that are true about angles on diagram

Goshia [24]

Answer:

B, C

Step-by-step explanation:

The only statements that must be true are ones that describe the markings on the diagram:

NO as an angle bisector

N is the vertex of a pair of congruent angles

_____

The only other fact that must be true is that segments on the left side of the diagram are proportional to those (corresponding) on the right. (This is a property of the angle bisector.) There is nothing that says the sides must be equal, or that the triangle is isosceles.

3 0

2 years ago