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3 years ago

What is the slope of the linear function? A) -4/3 B) -3/4 C) 3/4 D) 1

Mathematics

2 answers:

Romashka-Z-Leto [24]3 years ago

7 0

The slope is equal to rise over run.

Each time the equation travels 3 units up on the y-axis, it travels 4 units across on the x-axis.

Therefore your slope is m = 3/4.

iragen [17]3 years ago

7 0

You count the distance between one point and another, in this case from one point to the next is three up and four to the right, therefore your slope is 3/4 which means 3 up 4 to the right this positive numbers mean positive direction.

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Ada makes sparkling juice by mixing 2 cups of sparkling water with every 4 cups of apple juice. Let s represent the number of cu

Dafna11 [192]

Answer:

s = 2j

Step-by-step explanation:

Given - Ada makes sparkling juice by mixing 2 cups of sparkling water

with every 4 cups of apple juice. Let s represent the number of

cups of sparkling water and j represent the number of cups of

apple juice.

To find - Write an equation that shows how s and j are related.

Proof -

Let total number of cups of sparkling water make = s

Total number of cups of apple juice make = j

Now,

Given that, Ada makes sparkling juice by mixing 2 cups of sparkling water with every 4 cups of apple juice.

⇒2s = 4j

⇒s = 2j

∴ we get

Ada can makes sparking water by mixing 2j cups of sparking water with every j cups of apple juice.

6 0

3 years ago

2x + 2x2 + 2x3 + ... + 2x O arithmetic O geometric O both O neither

Mkey [24]

Answer:

16 ALB

Step-by-step explanation:

ayudame wey andale

6 0

3 years ago

Em 1 Graph the system of equations. { x−y=6 ,4x+y=4

muminat

Graphing the system of equations is shown in figure attached.

Solution set is (2,-4). The lines will intersect at (2,-4)

Step-by-step explanation:

We need to graph the system of equations. $x-y=6 ,4x+y=4$

First we will find value of x and y

Let:

$x-y=6\,\,eq(1)\\4x+y=4\,\,eq(2)$

Add eq(1) and eq(2)

$x-y=6\,\,\\4x+y=4\,\,\\------\\5x=10\\x=10/5\\x=2$

Putting value of x in eq(1) and finding y

$x-y=6\\2-y=6\\-y=6-2\\-y=4\\y=-4$

So, y=-4

Graphing the system of equations is shown in figure attached.

Solution set is (2,-4). The lines will intersect at (2,-4)

Keywords: System of equations

Learn more about system of equations at:

#learnwithBrainly

8 0

3 years ago

How do I expand the following equation with the binomial theorem?

Vsevolod [243]

Answer:

16x^4+32x^3+24x^2+8x+1

Step-by-step explanation:

(2x+1)^4

(2x+1)*(2x+1)*(2x+1)*(2x+1)

4 0

2 years ago

Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))

NNADVOKAT [17]

I believe the given limit is

$\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)$

Let

$a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1$

Now rewrite the expression as a difference of cubes:

$a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}$

Then

$a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2$

The limit is then equivalent to

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}$

From each remaining cube root expression, remove the cubic terms:

$a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}$

$(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}$

$b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}$

Now that we see each term in the denominator has a factor of x ², we can eliminate it :

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}$

$=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}$

As x goes to infinity, each of the 1/x ⁿ terms converge to 0, leaving us with the overall limit,

$\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}$

8 0

3 years ago