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My name is Ann [436]
2 years ago
9

2k+5=21 answer this problem plz

Mathematics
1 answer:
Anon25 [30]2 years ago
3 0

Answer:

k=8

Step-by-step explanation:

2k+5=21  (-5 on both sides to get rid of +5)

2k=16      (divide by 2 to get rid of that 2 in front of K)

k=8

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What is the scale factor in the dilation if the coordinates of A prime are (–7, 6) and the coordinates of C prime are (–4, 3)?
olchik [2.2K]

Option A: \frac{1}{3} is the dilation factor

Explanation:

The Coordinates of the square ABCD are A(-21,18), B(-12,18), C(-12,9), D(-21,9)

Also, the coordinates of the square A'B'C'D' are A'(–7, 6), B'(-4,6), C'(-4,3), D'(-8,3)

Now, we shall determine the scale factor in the dilation if the coordinates A' and C'.

Since, the dilation is the enlargement of the image in the same shape but different size.

To determine the scale factor, let us divide A'C' by AC

Thus, we have,

\begin{aligned}A^{\prime} C^{\prime} &=\sqrt{(-4+7)^{2}+(3-6)^{2}} \\&=\sqrt{3^{2}+(-3)^{2}} \\&=\sqrt{9+9} \\&=\sqrt{18}\\&=3\sqrt{2} \end{aligned}

Also,

\begin{aligned}A C &=\sqrt{(-12+21)^{2}+(9-18)^{2}} \\&=\sqrt{9^{2}+(-9)^{2}} \\&=\sqrt{81+81} \\&=\sqrt{162}\\&=9\sqrt{2} \end{aligned}

Dividing A'C' by AC, we have,

\frac{A'C'}{AC} =\frac{3\sqrt{2} }{9\sqrt{2}} =\frac{1}{3}

Thus, \frac{1}{3} is the dilation factor

3 0
3 years ago
Read 2 more answers
Pls I need help with this
LiRa [457]

Answer:

third side = 4

Step-by-step explanation:

third side is hypoenuse as it is opposite to 90 degree.

using pythagoras theorem

(perpendicular)^2 + (base)^2 = (hypotenuse)^2

2^2 + (2\sqrt{3 )^2 = hypotenuse^2

4 + 4*3 = hypotenuse^2

16 = hypotenuse^2

\sqrt{16} = hypotenuse

4 = hypotensue

7 0
3 years ago
Help help help help what is the answer −5(4x−3)−x+5=190!!!
artcher [175]
X=8.5
190-5=185
185/-5=-37
-37-3=-34
34/4=8.5

8 0
3 years ago
8/9 + 4/7 =<br> Can you get the Anwar for this sum
attashe74 [19]

here. I don't know if its right but i hope it is.

7 0
3 years ago
Read 2 more answers
Find all values of k for which the equation 3x^2−4x+k=0 has no solutions.
dusya [7]

The given equation has no solution when K is any real number  and k>12

We have given that

3x^2−4x+k=0

△=b^2−4ac=k^2−4(3)(12)=k^2−144.

<h3>What is the condition for a solution?</h3>

If Δ=0, it has 1 real solution,

Δ<0 it has no real solution,

Δ>0 it has 2 real solutions.

We get,

Δ=k^2−144 here Δ is not zero.

It is either >0 or <0

Δ<0 it has no real solution,

Therefore the given equation has no solution when K is any real number.

To learn more about the solution visit:

brainly.com/question/1397278

5 0
2 years ago
Read 2 more answers
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