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3 years ago

Solve 3x2 + 4x = −5.

Mathematics

2 answers:

erastova [34]3 years ago

3 0

Answer:

Step-by-step exExplanation:

Solve:

−

is a quadratic equation in standard form:

where:

−

Quadratic formula

−

√

−

Plug in the known values.

−

(

−

)

√

(

−

)

−

⋅

−

⋅

Simplify.

√

Prime factorize

√

Apply rule:

√

Simplify.

√

Solutions for

√

−

√

Related questions

How do you know how many solutions

−

has?

What is the Quadratic Formula?

planation:

baherus [9]3 years ago

3 0

Answer:

x= -11/4

Step-by-step explanation:

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What are the three steps you should use when using algebra to solve word problems?

KonstantinChe [14]

Answer: understand , strategize and implement

Hoped it helped , if it didn’t I apologize
Good luck

3 0

2 years ago

Gourmet Pretzels bakes soft pretzels on an assembly line. It currently bakes 800 pretzels each eight-hour shift. If the producti

Delvig [45]

Answer:

Change in productivity will be 50%

So option (b) will be correct option

Step-by-step explanation:

It is given that it takes 8 hour to bake 800 pretzels

And there is a shift of 8 hours

So hourly production = $\frac{800}{8} =100$

Now the production is increases to 1200

Now hourly production = $\frac{1200}{8}=150$

So change in productivity = $\frac{150-100}{100}=50$ %

So option (b) will be correct answer

6 0

3 years ago

raph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the ci

andreyandreev [35.5K]

Answer:

Equation:

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

The point (0,-5), (0,7), (5,0) and (-7,0)also lie on this circle.

Step-by-step explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

The center of this circle is the midpoint of (-3, 4) and (5, -2).

We use the midpoint formula:

$( \frac{x_1+x_2}{2}, \frac{y_1+y_2,}{2} )$

Plug in the points to get:

$( \frac{ - 3+5}{2}, \frac{ - 2+4}{2} )$

$( \frac{ -2}{2}, \frac{ 2}{2} )$

$( - 1, 1)$

We find the radius of the circle using the center (-1,1) and the point (5,-2) on the circle using the distance formula:

$r = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$r = \sqrt{ {(5 - - 1)}^{2} + {( - 2- - 1)}^{2} }$

$r = \sqrt{ {(6)}^{2} + {( - 1)}^{2} }$

$r = \sqrt{ 36+ 1 } = \sqrt{37}$

The equation of the circle is given by:

$(x-h)^2 + (y-k)^2 = {r}^{2}$

Where (h,k)=(-1,1) and r=√37 is the radius

We plug in the values to get:

$(x- - 1)^2 + (y-1)^2 = {( \sqrt{37}) }^{2}$

$(x + 1)^2 + (y - 1)^2 = 37$

We expand to get:

${x}^{2} + 2x + 1 + {y}^{2} - 2y + 1 = 37$

${x}^{2} + {y}^{2} + 2x - 2y +2 - 37= 0$

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

We want to find at least four points on this circle.

We can choose any point for x and solve for y or vice-versa

When y=0,

${x}^{2} + {0}^{2} + 2x - 2(0) - 35= 0$

${x}^{2} +2x - 35= 0$

$(x - 5)(x + 7) = 0$

$x = 5 \: or \: x = - 7$

The point (5,0) and (-7,0) lies on the circle.

When x=0

${0}^{2} + {y}^{2} + 2(0) - 2y - 35= 0$

${y}^{2} - 2y - 35= 0$

$(y - 7)(y + 5) = 0$

$y = 7 \: or \: y = - 5$

The point (0,-5) and (0,7) lie on this circle.

3 0

2 years ago

Write a polynomial in factored form.

enyata [817]

Answer:

f(x)= x^3

Step-by-step explanation:

5 0

1 year ago

Alyssa spent 9.20 for 16 packs of gum at this rate how much would it cost alyssa to buy 30 packs of gum.

zhuklara [117]

$17.25. Divide 9.20 by 16 and mulitply by 30.

6 0

3 years ago

Read 2 more answers