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3 years ago

Volume of a right rectangular prisms the side lengths are 1ft,2ft,3ft,4ft,and 5ft.

Mathematics

2 answers:

In-s [12.5K]3 years ago

6 0

The volume would be 120sqft.

horrorfan [7]3 years ago

5 0

The Volume would be 120 Ft

First you would Multiply 1x2=2
Then, 2x3=6
Then, 6x4=24
Finally, 24x5=120

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Latoya used 4 gallons of ginger ale, 1 gallon of orange juice, and 1 gallon of pineapple juice to make punch for her brother's b

Shkiper50 [21]

Answer: 96 cups

Step-by-step explanation: A gallon holds 128 ounces there were 6 gallons used to make the punch. 128x6= 768 ounces

A cup holds 8 ounces so you would do 768/8 and that would give you 96 cups

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1 year ago

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Find the length of the unknown side. Round your answer to the nearest tenth.

Paraphin [41]

Answer:

The correct option is: 11.5

Step-by-step explanation:

Suppose, the length of the unknown side is $x$

So, here the length of hypotenuse is 14 and the lengths of other two sides are 8 and $x$ .

As the given triangle is right angle triangle, so using Pythagorean theorem, we will get.........

$8^2+x^2= 14^2\\ \\ 64+x^2=196\\ \\ x^2=196-64=132\\ \\ x=\sqrt{132}=11.489...\approx 11.5$

(Rounded to the nearest tenth)

So, the length of the unknown side is 11.5

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3 years ago

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9th grade Algebra 1 help! PLEASE HELP ME! EXPLAIN YOUR ANSWERS :)

zhannawk [14.2K]

My best bet would be d

6 0

3 years ago

How many solutions does the following equation have?

Masteriza [31]

Answer:

One solution

Step-by-step explanation:

Hi,

3(x + 5) = -4x + 8

3x + 15 = -4x + 8

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One solution :)

8 0

2 years ago

Read 2 more answers

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e^−2

Mekhanik [1.2K]

Answer:

x=1

y=s

z=1

Step-by-step explanation:

(x, y, z)=(1, 0, 1)

Substitute 0 for y

$y = e^{-2t} *sin 4t\\0 = e^{-2t} *sin 4t\\\\0 = e^{-2t} *(\frac{e^{4t}-e^{-4t}}{2j} )\\0 = e^{-2t} *(e^{4t}-e^{-4t} )\\0 = e^{-2t} *e^{4t}*(1-e^{-8t} )\\\\0 = e^{2t}*(1-e^{-8t} )\\\\\\Either \\0 = e^{2t}\\t = -inf\\or\\0 = (1-e^{-8t} )\\(e^{-8t} ) = 1\\ -8t = ln(1) =0\\t=0\\\\$

Confirming if t=0 satisfy the other equation

x = e^−2t cos 4t = e^−2(0)cos(4*0)

= e^(0)cos(0) = 1

z = e^−2t = e^−2(0) = 0

Therefore t=0 satisfies the other equation

Finding the tangent vector at t=0

$\frac{dx}{dt}=-2te^{-2t} cos4t + e^{-2t}(-4sin4t)=-2(0)e^{-2(0)} cos4(0) + e^{-2(0)}(-4sin4(0))=0\\\\ \frac{dy}{dt} =-2te^{-2t} sin4t + e^{-2t}(4cos4t)=-2(0)e^{-2(0)} sin4(0)+ e^{-2(0)(4cos4(0)) }= 1\\\\\frac{dz}{dt} = -2te^{-2t} = -2(0)e^{-2(0)} = 0$

The vector equation of the tangent line is

(1, 0, 1) +s(0,1,0)= (1, s, 1)

The parametric equations are:

x=1

y=s

z=1

6 0

2 years ago