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2 years ago

What are the solutions of 3x? - x +4- 0 ?

Mathematics

2 answers:

ss7ja [257]2 years ago

8 0

Answer: B

Hope this helps :)

Karolina [17]2 years ago

6 0

Answer:

We can use quadratic formula

a=3, b= -1, c=4

Step-by-step explanation:

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Please help! I'm getting timed.

Molodets [167]

Hi there! The answer is A.

$t = 2(s + 2)$
To get a proper idea of the type of function this formula represents we work out the parenthesis.

Working out the parenthesis can for instance be done using rainbow technique.
$t = 2s + 4$

We can now see sinilarities between this function an the general function of a line in slope intercept form
$y = mx + n$

The function t therefore describes a line with slope 2 and y-intercept 4. Therefore the answer is A.

3 0

2 years ago

What is 10 times as many as

Talja [164]

Answer:

To answer this question, more information needs to be disclosed.

3 0

2 years ago

I'm so confused!!!!! Can somebody plz help me!!!plz

ZanzabumX [31]

I would suggest looking up a converter chart, I don't know all of them but I can help with some.
For yards to feet you have to divide by 3 because there are 3 feet in a yard.
For lbs, it's 16 oz in a pound, so divide by 16.
For miles to yards you divide by 1760 because there are 1760 yards in a mile.
I hope this helps a little bit.

8 0

2 years ago

There are 48 students in a speech contest. Yesterday, 2/3 of them gave their speeches. Today, 3/4 of the remaining students gave

Dmitry [639]

Answer:

24 :))))))))))))))

Step-by-step explanation:

2/3 of 48 is 32, so 3/4 of 32 is 24.

8 0

2 years ago

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

geniusboy [140]

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$ , so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$ , we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$ . The critical point with the negative $x$ -coordinates gives the maximum value, $9+\sqrt{185}$ .

8 0

3 years ago