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tatiyna
3 years ago
5

Quadrilateral ABCD is translated to get quadrilateral A′B′C′D′. Vertex A is at (-5, 2), and vertex A′ is at (2, -2).

Mathematics
1 answer:
fgiga [73]3 years ago
4 0
Then vertex B’ is at (1,1)
proof: the translation from A to A’ is (7,-4) therefore applying the same conditions to B would give you the outcome of (1,1)
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I need help!!!! <br> Which represents the inverse of the function f(x)=4x?
Mrac [35]

Answer:

h(x)=\frac{1}{4}x.

Step-by-step explanation:

The inverse is basically you just swapping x and y.  People want to solve it for y to make it a function of x.

y=4x

Swap x and y:

x=4y

Now solve for y by dividing y on both sides:

\frac{x}{4}=y

\frac{1}{4}x=y

y=\frac{1}{4}x.

4 0
3 years ago
Find the area of the triangle with the given measurements. Round the solution to the nearest hundredth if necessary. B = 107°, a
Archy [21]

Answer:    105.19 \text{ square cm}      

Step-by-step explanation:

Since, by the SAS formula of finding the area of triangle,

The area of triangle, A = \frac{1}{2}\times s_1\times s_2\times sin\theta

Where, s_1 and  s_2 are adjacent sides and \theta  is the included angle of these sides

Here, s_1=22 s_2=10

And, \theta=107^{\circ}

Thus, the area of the triangle ABC,

A = \frac{1}{2}\times 22\times 10\times sin (107)^{\circ}

A = 110\times 0.95630475596

A = 105.193523156\approx 105.19\text { square cm}

Therefore, Third Option is correct.

3 0
4 years ago
Solving a Two-Step Matrix Equation<br> Solve the equation:
Cloud [144]

Answer:

\boxed {x_{1} = 3}

\boxed {x_{2} = -4}

Step-by-step explanation:

Solve the following equation:

\left[\begin{array}{ccc}3&2\\5&5\\\end{array}\right] \left[\begin{array}{ccc}x_{1}\\x_{2}\\\end{array}\right] + \left[\begin{array}{ccc}1\\2\\\end{array}\right] = \left[\begin{array}{ccc}2\\-3\\\end{array}\right]

-In order to solve a pair of equations by using substitution, you first need to solve one of the equations for one of variables and then you would substitute the result for that variable in the other equation:

-First equation:

3x_{1} + 2x_{2} + 1 = 2

-Second equation:

5x_{1} + 5x_{2} + 2 = -3

-Choose one of the two following equations, which I choose the first one, then you solve for x_{1} by isolating

3x_{1} + 2x_{2} + 1 = 2

-Subtract 1 to both sides:

3x_{1} + 2x_{2} + 1 - 1 = 2 - 1

3x_{1} + 2x_{2} = 1

-Subtract 2x_{2} to both sides:

3x_{1} + 2x_{2} - 2x_{2} = -2x_{2} + 1

3x_{1} = -2x_{2} + 1

-Divide both sides by 3:

3x_{1} = -2x_{2} + 1

x_{1} = \frac{1}{3} (-2x_{2} + 1)

-Multiply -2x_{2} + 1 by \frac{1}{3}:

x_{1} = \frac{1}{3} (-2x_{2} + 1)

x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}

-Substitute -\frac{2x_{2} + 1}{3} for x_{1} in the second equation, which is 5x_{1} + 5x_{2} + 2 = -3:

5x_{1} + 5x_{2} + 2 = -3

5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3

Multiply -\frac{2x_{2} + 1}{3} by 5:

5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3

-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3

-Combine like terms:

-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3

\frac{5}{3}x_{2} + \frac{11}{3} = -3

-Subtract \frac{11}{3} to both sides:

\frac{5}{3}x_{2} + \frac{11}{3} - \frac{11}{3} = -3 - \frac{11}{3}

\frac{5}{3}x_{2} = -\frac{20}{3}

-Multiply both sides by \frac{5}{3}:

\frac{\frac{5}{3}x_{2}}{\frac{5}{3}} = \frac{-\frac{20}{3}}{\frac{5}{3}}

\boxed {x_{2} = -4}

-After you have the value of x_2, substitute for x_{2} onto this equation, which is x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}:

x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}

x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}

-Multiply -\frac{2}{3} and -4:

x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}

x_{1} = \frac{8 + 1}{3}

-Since both \frac{1}{3} and \frac{8}{3} have the same denominator, then add the numerators together. Also, after you have added both numerators together, reduce the fraction to the lowest term:

x_{1} = \frac{8 + 1}{3}

x_{1} = \frac{9}{3}

\boxed {x_{1} = 3}

5 0
3 years ago
If 3t=5t-8 evaluate -4t+5
Sergio [31]

3t = 5t - 8

-2t = -8

t = 4

Now put that into -4t + 5:

-4t + 5

-4(4) + 5

-16 + 5

-11

⭐ Please consider brainliest! ⭐

✉️ If any further questions, inbox me! ✉️

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