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3 years ago

Plz help im so confused plzzzzzzzzzz

Mathematics

1 answer:

horsena [70]3 years ago

5 0

Your Want Help With All The Questions

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Which of the following functions has the largest value when x = 3? c(x) = 3x2 + 5x + 22 j(x) = 12x a(x) = 9x

Olin [163]

To determine the function which has the largest value at x=3,

We will calculate the value of each function by substituting the value of x=3 in each of the given function.

Let us consider the first function,

$c(x)=3x^{2}+5x+22$

$c(3)=3\times 3^{2}+5 \times 3+22$

$c(3)=27+15+22$

$c(3)=64$

Let us consider the second function,

$j(x)=12x$

$j(x)=12 \times 3 =36$

Let us consider the third function,

$a(x)=9x$

$a(x)=9 \times 3=27$

Therefore, the function c(x) has the largest value at x=3.

3 0

3 years ago

156÷34 partial quotients

AveGali [126]

The answer is 4. 156/34 = 4

6 0

3 years ago

Sia’s leggings weigh 6 grams. How many milligrams do they weigh?_

777dan777 [17]

Answer:

they weight 6000 gram .............

7 0

4 years ago

Find the length of X

nasty-shy [4]

Answer:

$x = 8\sqrt{2}$

Step-by-step explanation:

Leg = x

Hypotenuse = 16

$x\sqrt{2} = 16$

$x = \dfrac{16}{\sqrt{2}}$

$x = \dfrac{16}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}}$

$x = \dfrac{16\sqrt{2}}{2}$

$x = 8\sqrt{2}$

8 0

3 years ago

What are the zeros of the quadratic function f(x) = 6x2 + 12x - 7?<br>

mina [271]

Answer:

$\frac{-6+\sqrt{78} }{6}$ and $\frac{-6-\sqrt{78} }{6}$

Step-by-step explanation:

We are asked that what are the zeros of the quadratic function f(x) = 6x² +12x -7.

So, we have to find the roots of the equation, f(x) = 6x² +12x -7 =0 ...... (1)

Since the quadratic function can not be factorized, so we have to apply Sridhar Acharya's formula.

This formula gives if, ax² +bx +c =0, the the two roots of the equation are

$\frac{-b+\sqrt{b^{2}-4ac } }{2a} , \frac{-b-\sqrt{b^{2}-4ac } }{2a}$

Therefore, in our case 'a' being 6, 'b' being 12 and 'c' being -7, the two roots of the equation (1) will be

$\frac{-12+\sqrt{12^{2}-4*6*(-7) } }{2*6}, \frac{-12-\sqrt{12^{2}-4*6*(-7) } }{2*6}$

= $\frac{-12+\sqrt{312} }{12},\frac{-12-\sqrt{312} }{12}$

= $\frac{-6+\sqrt{78} }{6}$ and $\frac{-6-\sqrt{78} }{6}$

Hence, x= $\frac{-6+\sqrt{78} }{6}$

and x= $\frac{-6-\sqrt{78} }{6}$

(Answer)

9 0

4 years ago