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3 years ago

Help marking Brainly!!!

Mathematics

2 answers:

lorasvet [3.4K]3 years ago

7 0

Use the app Socratic

loris [4]3 years ago

4 0

Socratic sucks don’t get it trust

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(6.04)The equation of the line of best fit of a scatter plot is y = 6x − 9. What is the slope of the equation?

Ghella [55]

Y=mx+b
m=slope

given
y=6x-9
m=6

the slope is 6

5 0

3 years ago

Fred is making a bouquet of carnations and roses. The carnations cost $5.25 in all. The roses cost $1.68 each. How many roses di

olga nikolaevna [1]

Answer:

8 roses

Step-by-step explanation:

First start by subtracting $18.69 by $5.25 because you already know how much the carnations cost and how much the bouquet cost in all. When you subtract 18.69 by 5.25 you should get $13.44. If each rose cost $1.68 then divide 13.44 by 1.68 to get the amount of roses in the bouquet.

8 0

3 years ago

What is the product of (3a + 2)(4a2 - 2a + 9)?

natka813 [3]

Answer:

12a³+2a²+23a+18

Step-by-step explanation:

(3a+2)(4a²-2a+9)=

12a³-6a²+27a+8a²-4a+18=

12a³+2a²+23a+18

If you need more explanation, reply to this answer.

4 0

4 years ago

A winery has a vat with two pipes leading to it. The inlet pipe can fill the vat in 5 hours, while the outlet pipe can empty it

Rashid [163]

Answer:

$time=40/3hours\approx 13.3hours$

Step-by-step explanation:

The rates are additive: you can calculate the inlet rate and the outlet rate and add them algebraically, i.e. the inlet rate will be positive and the outlet rate will be negative.

1. Inlet rate:

$1vat/5hours$

2. Outlet rate:

$1vat/8hours$

3. Net rate:

$\text{Inlet rate - outlet rate}=1vat/5hours-1vat/8hours\\\\ \text{Net rate}=(8-5)vat/40hour=3vat/40hour=(3/40)vat/hour$

4. Time to fill the vat

$rate=amount/time\implies time=amount/rate\\ \\ time=1vat/(3vat/40hour)$

$time=40/3hours\approx 13.3hours$

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5x%2F8y%7D" id="TexFormula1" title="\sqrt[4]{5x/8y}" alt="\sqrt[4]{5x/8y}" al

Furkat [3]

Answer: $\frac{\sqrt[4]{10xy^3}}{2y}$

where y is positive.

The 2y in the denominator is not inside the fourth root

==================================================

Work Shown:

$\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\$

The idea is to get something of the form $a^4$ in the denominator. In this case, $a = 2y$

To be able to reach the $16y^4$ , your teacher gave the hint to multiply top and bottom by $2y^3$

For more examples, search out "rationalizing the denominator".

Keep in mind that $\sqrt[4]{(2y)^4} = 2y$ only works if y isn't negative.

If y could be negative, then we'd have to say $\sqrt[4]{(2y)^4} = |2y|$ . The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

3 0

3 years ago