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2 years ago

HELP RIGHTTT AWAYYYYY PLSSSSSS

Mathematics

1 answer:

poizon [28]2 years ago

3 0

Answer:

64 pretzels = 16 oz

16 pretzels = 4 oz

1 pretzel = .25 oz

4 pretzels = 1 oz

Step-by-step explanation:

If you know that 64 equals 16 ounces than you can divide 64 by 16 to get 4. This means that one ounce equals four pretzels. Then you can multiply 1 oz by 4 and 4 pretzels by 4 to get 16 pretzels that equal 4 oz. If 4 pretzels equals 1 oz then 1 pretzel equals 1 oz divided by 4.

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The graph of the function f(x) = –(x + 3)(x – 1) is shown below.

pav-90 [236]

Answer:

The function is negative for all real values of x where

x < –3 and where x > 1.

Step-by-step explanation:

This quadratic function is downward parabola.

It's root are -3 and 1.

Quadratic is positive between -3 and 1.

And negative on R-(-3,1).

4 0

2 years ago

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Let θ

konstantin123 [22]

Answer:

θ is decreasing at the rate of $\frac{21}{16}$ units/sec

or $\frac{d}{dt}$ (θ) = $\frac{-21}{16}$

Step-by-step explanation:

Given :

Length of side opposite to angle θ is y

Length of side adjacent to angle θ is x

θ is part of a right angle triangle

At this instant,

x = 8 , $\frac{dx}{dt}$ = 7

( $\frac{dx}{dt}$ denotes the rate of change of x with respect to time)

y = 8 , $\frac{dy}{dt}$ = -14

( The negative sign denotes the decreasing rate of change )

Here because it is a right angle triangle,

tanθ = $\frac{y}{x}$ -------------------------------------------------------------------1

At this instant,

tanθ = $\frac{8}{8}$ = 1

Therefore θ = π/4

We differentiate equation (1) with respect to time in order to obtain the rate of change of θ or $\frac{d}{dt}$ (θ)

$\frac{d}{dt}$ (tanθ) = $\frac{d}{dt}$ (y/x)

( Applying chain rule of differentiation for R.H.S as y*1/x)

$sec^{2}$ θ $\frac{d}{dt}$ (θ) = $\frac{1}{x}$ $\frac{dy}{dt}$ - $\frac{y}{x*x}$ $\frac{dx}{dt}$ -----------------------2

Substituting the values of x , y , $\frac{dx}{dt}$ , $\frac{dy}{dt}$ , θ at that instant in equation (2)

2 $\frac{d}{dt}$ (θ) = $\frac{1}{8}$ *(-14)- $\frac{8}{8*8}$ *7

$\frac{d}{dt}$ (θ) = $\frac{-21}{16}$

Therefore θ is decreasing at the rate of $\frac{21}{16}$ units/sec

or $\frac{d}{dt}$ (θ) = $\frac{-21}{16}$

3 0

2 years ago

A rectangular box is to have a square base and a volume of 20 ft3. If the material for the base costs $0.35 per square foot, the

denpristay [2]

Answer:

Length = 2 ft

Width = 2 ft

Height = 5 ft

Step-by-step explanation:

Let the square base of the box has one side = x ft

Therefore, area of the base = x² ft²

Cost of the material to prepare the base = $0.35 per square feet

Cost to prepare the base = $0.35x²

Let the height of the box = y ft

Then the volume of the box = x²y ft³ = 20

$y=\frac{20}{x^{3} }$ -----(1)

Cost of the material for the sides = $0.10 per square feet

Area of the sides = 4xy

Cost to prepare the sides of the box = $0.10 × 4xy

= $0.40xy

Cost of the material to prepare the top = $0.15 per square feet

Cost to prepare the top = $0.15x²

Total cost of the box = 0.35x² + 0.40xy + 0.15x²

From equation (1),

Total cost $C=0.35x^{2}+(0.40x)\times \frac{20}{x^{2} }+0.15x^{2}$

$C=0.35x^{2}+\frac{8}{x}+0.15x^{2}$

$C=0.5x^{2}+\frac{8}{x}$

Now we take the derivative of C with respect to x and equate it to zero,

$C'=0.5(2x)-\frac{8}{x^{2}}$ = 0

$x-\frac{8}{x^{2}}=0$

$x=\frac{8}{x^{2} }$

$x^{3}=8$

x = 2 ft.

From equation (1),

$(2)^{2}y=20$

4y = 20

y = 5 ft

Therefore, Length and width of the box should be 2 ft and height of the box should be 5 ft for the minimum cost to construct the rectangular box.

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2 years ago

What is 1/3 written as a decimal?

Aleks04 [339]

Answer:

0.3

Step-by-step explanation:

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2 years ago

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Suppose Q is the midpoint of line segment PR, PQ = x + 10, and QR = 4x - 2

xz_007 [3.2K]

answer and step by step in the attachment

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2 years ago

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