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3 years ago

I'm not sure pls help!!!

Mathematics

2 answers:

kkurt [141]3 years ago

7 0

The answer is X=2

If you put 2 in the place of x it equals 6.

6/3 = 2

goldenfox [79]3 years ago

4 0

X= 2. If u do 3 times 2 u get the answer of 6. 3 goes into 6 twice. 2 is the answer

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Find the derivative of f(x)=4x+7 at x=5

Marizza181 [45]

The answer is 27??????

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What is the perimeter of △ABC?

timofeeve [1]

<h3>Given</h3>

ΔABC
A(-3, -1), B(0, 3), C(1, 2)

the length of the perimeter of ΔABC to the nearest tenth

<h3>Solution</h3>

The perimeter of a triangle is the sum of the lengths of its sides. The length of each side can be found using the Pythagorean theorem. Effectively, each pair of points is treated as the end-points of the hypotenuse of a right triangle with legs parallel to the x- and y-axes. The leg lengths are the differences betweeen the x- and y- coordinates of the points.

The difference of the x-coordinates of segment AB are 0-(-3) = 3. The y-coordinate difference is 3-(-1) = 4. So, the leg lengths of the right triangle whose hypotenuse is segment AB are 3 and 4. The Pythagorean theorem tells us

... AB² = 3² +4² = 9 +16 = 25

... AB = √25 = 5

You may recognize this as the 3-4-5 triangle often introduced as one of the first ones you play with when you learn the Pythagorean theorem.

LIkewise, segment AC has coordinate differences of ...

... C - A = (1, 2) -(-3, -1) = (4, 3)

These are the same leg lengths (in the other order) as for segment AB, so its length is also 5.

Segment BC has coordinate differences ...

... C - B = (1, 2) -(0, 3) = (1, -1)

The length of the line segment is figured as the root of the sum of squares, even though one of the coordinate differences is negative. The leg lengths of the right triangle used for finding the length of BC are the absolute value of these differences, or 1 and 1. Then the length BC is

... BC = √(1² +1²) = √2 ≈ 1.4

So the perimeter of the triangle ABC is

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Please be aware that the advice to "round each step" is <em>bad advice,</em> in general. For real-world math problems, you only round the final result. You always carry at least enough precision in the numbers to ensure that there will not be any error in the final rounding.

In this problem, the only number that is not an integer is √2, so it doesn't really matter.

7 0

3 years ago

the height of a cylinder is three times the diameter of the base . The surface area of thre ylinder is 126pi ft^2. What is thr r

Over [174]

The answer is 3 ft

The surface area (SA) of a cylinder with radius r and height h is:
SA = 2πr² + 2πrh = π(2r² + 2rh)

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SA = 126π ft²
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avanturin [10]

A very simple example problem to satisfy the required above is,
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3 years ago

find the parametric equations for the line of intersection of the two planes z = x + y and 5x - y + 2z = 2. Use your equations t

Kaylis [27]

Answer:

You didn't give the points in which you want the parametric equations be filled, but I have obtained the parametric equations, and they are:

x = (1/3 + t)

y = (-1/3 - 7t)

z = -6t

Step-by-step explanation:

If two planes intersect each other, the intersection will always be a line.

The vector equation for the line of intersection is given by

r = r_0 + tv

where r_0 is a point on the line and v is the vector result of the cross product of the normal vectors of the two planes.

The parametric equations for the line of intersection are given by

x = ax, y = by, and z = cz

where a, b and c are the coefficients from the vector equation

r = ai + bj + ck

To find the parametric equations for the line of intersection of the planes.

x + y - z = 0

5x - y + 2z = 2

We need to find the vector equation of the line of intersection. In order to get it, we’ll need to first find v, the cross product of the normal vectors of the given planes.

The normal vectors for the planes are:

For the plane x + y - z = 0, the normal vector is a⟨1, 1, -1⟩

For the plane 5x - y + 2z = 2, the normal vector is b⟨5, -1, 2⟩

The cross product of the normal vectors is

v = a × b =

|i j k|

|1 1 -1|

|5 -1 2|

= i(2 - 1) - j(2 + 5) + k(-1 - 5)

= i - 7j - 6k

v = ⟨1, -7, -6⟩

We also need a point on the line of intersection. To get it, we’ll use the equations of the given planes as a system of linear equations. If we set z = 0 in both equations, we get

x + y = 0

5x - y = 2

Adding these equations

5x + x + y - y = 2 + 0

6x = 2

x = 1/3

Substituting x = 1/3 back into

x + y = 0

y = -1/3

Putting these values together, the point on the line of intersection is

(1/3, -1/3, 0)

r_0= (1/3) i - (1/3) j + 0 k

r_0 = ⟨1/3, -1/3, 0⟩

Now we’ll plug v and r_0 into the vector equation.

r = r_0 + tv

r = (1/3)i - (1/3)j + 0k + t(i - 7j - 6k)

= (1/3 + t) i - (1/3 + 7t) j - 6t k

With the vector equation for the line of intersection in hand, we can find the parametric equations for the same line. Matching up r = ai + bj + ck with our vector equation,

r = (1/3 + t) i + (-1/3 - 7t) j + (-6t) k

a = (1/3 + t)

b = (-1/3 - 7t)

c = -6t

Therefore, the parametric equations for the line of intersection are

x = (1/3 + t)

y = (-1/3 - 7t)

z = -6t

3 0

3 years ago