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4 years ago

G(6) + f(4) I need help with this ASAP can you pls help me

Mathematics

1 answer:

Karolina [17]4 years ago

7 0

Answer:

2(3g+2f)

Step-by-step explanation:

g(6) + f(4)

rewrite 4 as 2 . 2

rewrite 6 as 3 . 2

3 . 2g + 2 . 2f

factor out the common term 2

= 2(3g+2f)

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Grace [21]

Answer:

Step-by-step explanation:

Rearrange the terms and take out a common factor of 4

4(8x^2 + 2x + 1)

I think this is as far down as you want to take it. It gives an imaginary root which you may not be familiar with.

4 0

3 years ago

The numbers of students in the four

Ludmilka [50]

The total number of students in the four classes is 100 students.

<h3>How to compute the value?</h3>

From the information, the numbers of students in the four sixth-grade classes at Northside School are 26, 19, 34, and 21.

Therefore, the total number of students will be:

= 26 + 19 + 34 + 21

= 100

Therefore there are 100 students.

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8 0

2 years ago

Evaluate <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%7Bm%7D%5E%7B2%7D%20%20%2B%20%20%7Bm%7D%5E%7B%20-%202%7D%20%2B%20

zysi [14]

Answer:

9.5

Step-by-step explanation:

(0.5^2 + 0.5^-2 + 0.5) / 0.5 = 9.5

Hope this help! :)

4 0

4 years ago

Consider the expression 6x6^0x6^-3

Montano1993 [528]

Answer:

6*(6^-3)=0.027777 recurring

Step-by-step explanation:

6^0=1

6^-3=0.004629629 recuuring

0.004629629 recurring *6=0.0.02777recurring

6 0

3 years ago

Which relationships have the same constant of proportionality between y and x as in the equation y=1/2x Choose 3 answers:

Andrew [12]

Answer:

A and B has the same constant of proportionality

Step-by-step explanation:

$y \propto x$

$y = kx ----1$

Where k is the constant of proportionality

We are supposed to find Which relationships have the same constant of proportionality between y and x as in the equation $y=\frac{1}{2}x$

On comparing with 1

$k = \frac{1}{2}$

A)6y = 3x

$y = \frac{3}{6}x\\y = \frac{1}{2}x$

So, this equation has the same constant of proportionality

B) $(x_1,y_1)=(2,1)\\(x_2,y_2)=(4,2)$

To find the equation :

Formula : $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

So, $y - 1=\frac{2-1}{4-2}(x-2)\\y-1=\frac{1}{2}(x-2)\\y-1=\frac{1}{2}x-1\\y=\frac{1}{2}x$

So, this equation has the same constant of proportionality

$(x_1,y_1)=(1,2)\\(x_2,y_2)=(2,4)$

To find the equation :

Formula : $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

So, $y - 2=\frac{4-2}{2-1}(x-1)\\y - 2=2(x-1)\\y - 2=2x-2\\y=2x$

So, this equation do not has the same constant of proportionality

$(x_1,y_1)=(2,1)\\(x_2,y_2)=(3,2.5)$

To find the equation :

Formula : $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

So, $y - 1=\frac{2.5-1}{3-2}(x-2)$

$y-1=1.5(x-2)\\y-1=1.5x-3\\y=1.5x-2$

So, this equation do not has the same constant of proportionality

Hence A and B has the same constant of proportionality

3 0

4 years ago

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