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Yuliya22 [10]
3 years ago
9

A rectangle has a length of 7.2 cm and a height of 5 cm. what is the area and perimeter of the rectangle

Mathematics
1 answer:
ratelena [41]3 years ago
8 0
Area: 36 cm squared
Perimeter: 24.4 cm
How is this high school math?
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If j(x)=3x+1, find x if j(x)=-11. I need help quick please​
Vinvika [58]

Answer:

x = -4

Step-by-step explanation:

Step 1: Define

j(x) = 3x + 1

j(x) = -11

Step 2: Substitute and solve for <em>x</em>

-11 = 3x + 1

-12 = 3x

x = -4

5 0
4 years ago
(3x^2 + 4x -7) - (x^2 - 2x + 6)
marshall27 [118]
(9x^2 +4x -7) - x^2 +2x - 6
8x^2  +6x - 13.

Get a grip on your life son. This is really simple.
8 0
3 years ago
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rjkz [21]
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7 0
3 years ago
The length of a rectangle is 7 more than the width. The area is 744 sqaure yards, find the length and width of the rectangle​
AVprozaik [17]

Answer:

  • Length of rectangle is 31 yards and Width is 24 yards.

Given:

  • The length of a rectangle is 7 more than the width.
  • The area is 744 sqaure yards

Solution:

Let's assume Width of rectangle be x and Length of rectangle be x + 7 respectively.

Using formula

\\  \:  \:  \:  \:  \pink{ \dashrightarrow \:  \:  \:  \:  \sf { \underbrace{Area_{(Rectangle)} =  Length × Width  }}} \\  \\

On Substituting the required values, we get;

\\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf (x)(x + 7) = 744 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf  {x}^{2}  + 7x = 744 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf  {x}^{2}  + 7x - 744 = 0 \\ \\ \\  \: \: \: \:  \dashrightarrow \: \: \: \: \sf  {x}^{2}  + 31x - 24x - 744 = 0 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf x(x + 31) - 24 (x + 31) = 0 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf (x + 31)(x - 24) = 0 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf x = 24 \:  or \:   - 31 \\ \\

As we know that width of the rectangle can't be negative. So x = 24

<u>Hence</u>,

  • Width of rectangle = x = 24 yards
  • Length of the rectangle = x + 7 = 31 yards

\thereforeLength of rectangle is 31 yards<u> </u>and Width is 24 yards.

8 0
2 years ago
Copy the problems onto your paper, mark the given and prove the statements asked. Given: marked Prove: ΔABC ≅ ΔDBC, ΔEHF ≅ ΔGHF
hjlf

The proof is given below. Please go through it.

Step-by-step explanation:

To solve Δ ABC ≅ Δ DBC

From Δ ABC and Δ DBC

AB = BD (given)

AC = CD (given)

BC is common side

By SSS condition Δ ABC ≅ Δ DBC ( proved)

To solve Δ EHF ≅ Δ GHF

Δ EHF and Δ GHF

EH = HG ( given)

∠ EFH = ∠ GFH ( each angle is 90°)

HF is common side

By RHS condition

Δ EHF ≅ Δ GHF

7 0
4 years ago
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