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Karolina [17]
4 years ago
14

A rectangle and a square have equal areas. The rectangle has length of 9 cm and width 4 cm. What is the perimeter of the square

in centimeters?
Mathematics
1 answer:
pshichka [43]4 years ago
3 0

Answer:

p = 24 cm

Step-by-step explanation:

9 * 4 = 36

a square with an area of 36 must be 6 inches on each side so P = 6 * 4= 24 cm

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(PLEASE HELP!!)
Flauer [41]

Answer:

b

Step-by-step explanation:

5 0
3 years ago
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Please helppp will mark brainlyist with right answer
Len [333]
An angle bisector is a line or ray that divides an angle into two congruent angles
so
in this problem
m∠RST=m∠RSV+m∠VST
m∠RSV=m∠VST
hence
m∠RST=2*m∠RSV --------> equation 1
we have
m∠RST=(6x-24)
m∠RSV=(2x+8)
Substitute in the equation 1
6x-24=2*(2x+8)
6x-24=4x+16
6x-4x=16+24
2x=40
x=40/2
x=20 degrees
therefore
the value of x is 20 degrees
Tell me if I’m wrong
7 0
3 years ago
I need help please (18 points)
inysia [295]

Answer:

128

Step-by-step explanation:

8 0
3 years ago
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On a coordinate plane, a curved line with a minimum value of (5.1, negative 7) and a maximum value of (0, 25), crosses the x-axi
liubo4ka [24]

Answer:

<u>The true statement is D</u>

Step-by-step explanation:

The rest of the question is the attached figure and the statement options.

  • A. Over the interval [–4, –2], the local minimum is 0.
  • B. Over the interval [–2, –1], the local minimum is 25.
  • C. Over the interval [–1, 4], the local minimum is 0.
  • D. Over the interval [4, 7], the local minimum is -7.

============================================================

According to the graph, we will check the options:

A. Over the interval [–4, –2], the local minimum is 0.  (<u>Wrong</u>)

Because the minimum is -12

B. Over the interval [–2, –1], the local minimum is 25.   (<u>Wrong</u>)

Because the minimum is 18

C. Over the interval [–1, 4], the local minimum is 0.  (<u>Wrong</u>)

Because the minimum is at x = 4 less than zero

D. Over the interval [4, 7], the local minimum is -7.    (<u>True</u>)

5 0
3 years ago
Read 2 more answers
Check answer please
Cerrena [4.2K]
The fourth or the D) Option is correct.

To find the new induced matrix via a scalar quantified multiplication we have to multiply the scalar quantity with each element surrounded and provided in a composed (In this case) 3×3 or three times three matrix comprising 3 columns and 3 rows for each element which is having a valued numerical in each and every position.

Multiply the scalar quantity with each element with respect to its row and column positioning that is,

Row × Column. So;

(1 × 1) × 7, (2 × 1) × 7, (3 × 1) × 7, (1 × 2) × 7, (2 × 2) × 7, (3 × 2) × 7, (1 × 3) × 7, (2 × 3) × 7 and (3 × 3) × 7. This will provide the final answer, that is, the D) Option.

To interpret and make it more interesting in LaTeX form. Here is the solution with LaTeX induced matrix.

\mathcal{A = \begin{bmatrix}1 & 0 & 3 \\ 2 & -1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}

\mathbf{\therefore \quad 7A = 7 \times \begin{bmatrix}1 & 0 & 3 \\ 2 & - 1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}

\mathbf{\therefore \quad \begin{bmatrix}7 \times 1 & 7 \times 0 & 7 \times 3 \\ 7 \times 2 & 7 \times -1 & 7 \times 2 \\ 7 \times 0 & 7 \times 2 & 7 \times 1 \\ \end{bmatrix}}

\therefore \quad \begin{\bmatrix}7 & 14 & 0 \\ 0 & -7 & 14 \\ 21 & 14 & 7 \end{bmatrix}

Hope it helps.
5 0
3 years ago
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