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2 years ago

2. 3(x + 2) = 3x + 6

Mathematics

1 answer:

Lorico [155]2 years ago

3 0

Answer: The answer to this question is zero.

Step-by-step explanation: This is the linear equation, on the left hand side we have to multiply by 3 in order to remove the brackets.

3 (x+2)= 3x+6

3x+6=3x+6

Take 6 to the other side and 3x to the other side also

3x-3x=6-6

0=0

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sleet_krkn [62]

Answer:

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Step-by-step explanation:

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2 years ago

What is the fourth term of an arithmetic sequence whose first term is 23 and whose seventh term is 5?

mixas84 [53]

We have that the fourth term of an arithmetic sequence is

a_4=14

Option C

From the question we are told

What is the fourth term of an arithmetic sequence whose first term is 23 and whose seventh term is 5?

A) 78

B) 32

C) 14

(Explain your work)

Generally the equation for the arithmetic sequence is mathematically given as

$a_4=a_1+(n-1)d$

Therefore

For seventh term

$5=23+(6)d\\\\d=5-23/(6)\\\\d=-3$

Therefore

For Fourth term

$a_4=23+(3*(-3))$

a_4=14

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2 years ago

Find the area of a circle with a radius of 0.4 inches

AleksandrR [38]

Answer:

0.16 π

Step-by-step explanation:

Using the area formula pi*r^2 we can find the area of the circle, plugging in the numbers gets us pi*(0.4)^2 = 0.16 π

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2 years ago

Question 3: Which of the following constructions would help to construct a line passing through point C that is perpendicular to

lakkis [162]

Answer:

Construction of a perpendicular bisector through C

Step-by-step explanation:

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2 years ago

Consider the functions below. f(x, y, z) = x i − z j + y k r(t) = 4t i + 6t j − t2 k (a) evaluate the line integral c f · dr, wh

fredd [130]

With

$\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k$

we have

$\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

The vector field evaluated over this parameterization is

$\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k$

so the line integral is

$\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

$=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4$

6 0

2 years ago