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leonid [27]
3 years ago
15

in a xy-coordinate system line 1 has a slope of -3. If the points (2,16 and (5,t) are on line 1, what is the value of t?

Mathematics
1 answer:
Dimas [21]3 years ago
8 0
We can use the formula y2-y1/x2-x1 to find our answer. So, we have : t-16/5-2 = -3. We simplify that to t-16/3 = -3. We can then multiply both sides by 3 to get t-16 = -9. Then we can add 16 to both sides to get t = 7.
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Write the coordinate of the point in the diagram.
mash [69]

Answer:

(2, 2, 1)

Step-by-step explanation:

You can see that the point is moved 2 units (x axis), 2 units (y axis) and 1 unit (z axis). Because you write coordiantes as (x,y,z), you write (2,2,1)

8 0
3 years ago
Which describes how square S could be transformed to square S prime in two steps? Assume that the center of dilation is the orig
gayaneshka [121]

Answer:

The correct option is;

A dilation by a scale factor of Two-fifths and then a translation of 3 units up

Step-by-step explanation:

Given that the coordinates of the vertices of square S are

(0, 0), (5, 0), (5, -5), (0, -5)

Given that the coordinates of the vertices of square S' are

(0, 1), (0, 3), (2, 3), (2, 1)

We have;

Length of side, s, for square S is s = √((y₂ - y₁)² + (x₂ - x₁)²)

Where;

(x₁, y₁) and (x₂, y₂) are the coordinates of two consecutive vertices

When (x₁, y₁) = (0, 0) and (x₂, y₂) = (5, 0), we have;

s = √((y₂ - y₁)² + (x₂ - x₁)²) = s₁ = √((0 - 0)² + (5 - 0)²) = √(5)² = 5

For square S', where (x₁, y₁) = (0, 1) and (x₂, y₂) = (0, 3)

Length of side, s₂, for square S' is s₂ = √((3 - 1)² + (0 - 0)²) = √(2)² = 2

Therefore;

The transformation of square S to S' involves a dilation of s₂/s₁ = 2/5

The after the dilation (about the origin),  the coordinates of S becomes;

(0, 0) transformed to (remains at) (0, 0) ....center of dilation

(5, 0) transformed to (5×2/5, 0) = (2, 0)

(5, -5) transformed to (2, -2)

(0, -5) transformed to (0, -2)

Comparison of (0, 0), (2, 0), (2, -2), (0, -2) and (0, 1), (0, 3), (2, 3), (2, 1) shows that the orientation is the same;

For (0, 0), (2, 0), (2, -2), (0, -2) we have;

(0, 0), (2, 0) the same y-values, (∴parallel to the x-axis)

(2, -2), (0, -2) the same y-values, (∴parallel to the x-axis)

For (0, 1), (0, 3), (2, 3), (2, 1) we have;

(0, 3), (2, 3) the same y-values, (∴parallel to the x-axis)

(0, 1), (2, 1) the same y-values, (∴parallel to the x-axis)

Therefore, the lowermost point closest to the y-axis in (0, 0), (2, 0), (2, -2), (0, -2) which is (0, -2) is translated to the lowermost point closest to the y-axis in (0, 1), (0, 3), (2, 3), (2, 1) which is (0, 1)

That is (0, -2) is translated to (0, 1) which shows that the translation is T((0 - 0), (1 - (-2)) = T(0, 3) or 3 units up

The correct option is therefore a dilation by a scale factor of Two-fifths and then a translation of 3 units up.

7 1
3 years ago
Find the area of the composite figure below.
pashok25 [27]

Answer: D is the correct answer.

8 0
2 years ago
Let the equation C = 2.32 N + 34,180 represent the cost of raising a child, C, on an income, N.
ivanzaharov [21]

Answer:

C = 2.32 N + 34,180

C = 2.32 × 60000 + 34,180

C = 139200 + 34180

C = $173,380

4 0
3 years ago
Evaluate the difference quotient for the given function. Simplify your answer.
nasty-shy [4]

I suppose you mean

f(x) = \dfrac{x+5}{x+1}

Then

f(3) = \dfrac{3+5}{3+1} = \dfrac84 = 2

and the difference quotient is

\dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5}{x+1}-2}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5-2(x+1)}{x+1}}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{-x+3}{(x+1)(x-3)} \\\\ \dfrac{f(x)-f(3)}{x-3} = \boxed{-\dfrac{x-3}{(x+1)(x-3)}}

If it's the case that <em>x</em> ≠ 3, then (<em>x</em> - 3)/(<em>x</em> - 3) reduces to 1, and you would be left with

\dfrac{f(x)-f(3)}{x-3}\bigg|_{x\neq3} = -\dfrac1{x+1}

4 0
2 years ago
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