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4 years ago

What is the solution to the equation1+1

Mathematics

1 answer:

sveta [45]4 years ago

8 0

Answer:

Step-by-step explanation:

but I would personally say FISH

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What is the next number in the series 167, 118, 82

Katyanochek1 [597]

The next number in the series are 57, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2.

<h3>What is a sequence?</h3>

It is defined as the systematic way of representing the data that follows a certain rule of arithmetic.

It is given that:

The series is:

167, 118, 82

A number is a mathematical entity that can be used to count, measure, or name things. For example, 1, 2, 56, etc. are the numbers.

As we know, the prime is divisible by 1 and itself.

The prime numbers before 82 are:

57, 41, 37, 31, 29, 23, 19.. and so on

The next number in series are 57, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2

Thus, the next number in the series are 57, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2.

Learn more about the sequence here:

brainly.com/question/21961097

#SPJ1

3 0

2 years ago

A right rectangular prism has a length of five centimeters a width of eight centimeters and a height of 4 cm what is the volume

lukranit [14]

Answer:

V=160cm^3

Step-by-step explanation:

V=lhw

V=5•4•8

V=20•8

V=160cm^3

^-^ your welcome

3 0

4 years ago

If xy=5 and x square +y square =25 ,then (x+y)square

nika2105 [10]

Answer:

(x + y)² = 35

Step-by-step explanation:

The expansion of

(x + y)² = x² + 2xy + y²

= x² + y² + 2xy ← substitute given values

= 25 + 2(5)

= 25 + 10

= 35

8 0

3 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

3 years ago

How do I do question 9 b?

Lynna [10]

You find the circumference of a full circle so 2πr =43.98
then divide by 4=10.995
so the answer is 10.995cm

3 0

3 years ago