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3 years ago

Plz, Help me with this!!!! Plz, Help me with this!!!!

Mathematics

2 answers:

Kazeer [188]3 years ago

6 0

Answer:

-3x-4

Step-by-step explanation:

first you add 5x and -8x to get -3x, then you get -4.

Maslowich3 years ago

3 0

Answer:

-3x+2

Step-by-step explanation:

Since it's addition you just combine all like terms so,

5x+4

-8x-2

= -3x+2

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Hong drove 348 miles using 15 gallons of gas. At this rate, how many gallons of gas would he need to drive 232 miles?

SCORPION-xisa [38]

348 miles——>15 gallons of gas
232 miles——-> x
X=(232)(15)/348=10 gallons of gas

8 0

3 years ago

What is the rate of change of the graph shown below? PLS HELP!!!! ^^^

const2013 [10]

Answer:

$m=\frac{-1}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Find points from graph.

Point (3, 0)

Point (0, 1)

Step 2: Find slope m

Plug in points to find the rate of change of the graph.

Substitute [SF]: $m=\frac{1-0}{0-3}$
Subtract: $m=\frac{1}{-3}$
Rewrite: $m=\frac{-1}{3}$

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3 years ago

Write an Equation and Solve. Put your answer, not the equation. in the blank.

nekit [7.7K]

Answer:

1.-6

2.135

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Step-by-step explanation:

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3 years ago

A box with a square base and an open top is being constructed out of A cm2 of material. If the volume of the box is to be maximi

viktelen [127]

Answer:

Side length = $\sqrt{\frac{A}{3} }$ cm , Height = $\frac{1}{2} \sqrt{\frac{A}{3} }$ cm , Volume = $\frac{A\sqrt{A}}{6\sqrt{3} }$ cm³

Step-by-step explanation:

Assume

Side length of base = x

Height of box = y

total material required to construct box = A ( given in question)

So it can be written as

A = x² + 4xy

4xy = A - x²

$y = \frac{A - x^{2} }{4x}$

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V = x² ₓ y

V = x² ₓ ( $\frac{A - x^{2} }{4x}$ )

V = $\frac{Ax - x^{3} }{4}$

To find max volume put V' = 0

So taking derivative equation becomes

$\frac{A - 3 x^{2} }{4} = 0$

A = 3 $x^{2}$

$x^{2}$ = $\frac{A}{3}$

x = $\sqrt{\frac{A}{3\\} }$

put value of x in equation 1

y = $\frac{A - \frac{A}{3} }{4\sqrt{\frac{A}{3} } }$

y = $\frac{2 \sqrt{\frac{A}{3} } }{4 \sqrt{\frac{A}{3} } }$

y = $\frac{1}{2} \sqrt{\frac{A}{3} }$

So the volume will be

V = $x^{2}$ × y

Put values of x and y from equation 2 & 3

V = $\frac{A}{3} (\frac{1}{2} \sqrt{\frac{A}{3} } )$

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4 years ago

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damaskus [11]

Answer:

Step-by-step explanation:

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