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4 years ago

Can someone help me with this one

Mathematics

2 answers:

zaharov [31]4 years ago

4 0

There's this app where you get a picture of the math problem and it gives you the answer

Arada [10]4 years ago

3 0

To find the x coordinate of the vertex, use -b/2a.
b is 6 and a is 3 in this equation.

So you'd get -6/2(3) which equals -6/6 = -1.

Now that we have -1 as our x coordinate, plug it in to the equation to find y.

so y is equal to 3(-1^2) + 6(-1) - 11.

3 - 6 - 11 = -14.

Put it together and your vertex is (-1, -14)

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Ook

Leya [2.2K]

The cars speed in kilometres per hour is = 87.2 Km/hr

<h3>Calculation of speed of a car:</h3>

The speed of the car = d/t

The distance(d) travelled by the car = 41 kilometers

The time(t) taken to cover the distance = 28 minutes

To solve, convert minutes to hours

But 60 mins = 1 hour

28mims = X

cross multiply,

X = 28/60

X = 0.47hour

To calculate the speed substitute d and t with it's variables provided.

speed = 41/0.47

=87.2 Km/hr ( which is to the nearest tenth)

Therefore, The cars speed in kilometres per hour is = 87.2 Km/hr

Learn more about speed here:

brainly.com/question/24872445

7 0

3 years ago

The first column contains five pairs of relationships. Determine if each pair forms

3241004551 [841]

1. Proportional
2. Not Proportional
3. Not Proportional
4. Proportional
5. Proportional

8 0

3 years ago

|x |-8= -5 <br>what's the answer

ladessa [460]

Answer:

|x |-7 = -5 ( x = -3 )

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Just wondering is 71 a PRIME number? ≡≡≡

Sergeu [11.5K]

Yes 71 is a Prime Number

7 0

3 years ago

Read 2 more answers

Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=x+4y subject to the constraint x2+y2=9, if such values

Vesnalui [34]

The Lagrangian is

$L(x,y,\lambda)=x+4y+\lambda(x^2+y^2-9)$

with critical points where the partial derivatives vanish.

$L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}$

$L_y=4+2\lambda y=0\implies y=-\dfrac2\lambda$

$L_\lambda=x^2+y^2-9=0$

Substitute $x,y$ into the last equation and solve for $\lambda$ :

$\left(-\dfrac1{2\lambda}\right)^2+\left(-\dfrac2\lambda\right)^2=9\implies\lambda=\pm\dfrac{\sqrt{17}}6$

Then we get two critical points,

$(x,y)=\left(-\dfrac3{\sqrt{17}},-\dfrac{12}{\sqrt{17}}\right)\text{ and }(x,y)=\left(\dfrac3{\sqrt{17}},\dfrac{12}{\sqrt{17}}\right)$

We get an absolute maximum of $3\sqrt{17}\approx12.369$ at the second point, and an absolute minimum of $-3\sqrt{17}\approx-12.369$ at the first point.

4 0

3 years ago