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alekssr [168]
4 years ago
8

Can someone please hep me with this question Jordan is proofreading his personal narrative.

Mathematics
1 answer:
charle [14.2K]4 years ago
7 0

Answer:

When proofreading, you should make sure all details are accurate and writing mechanics are correctly used.

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What is the surface area of the square pyramid represented by the net? Enter your answer in the box. m²
Alborosie

Answer:

1.75 meters² or 175 centimeters²

Step-by-step explanation:

Step 1: Find Area of the Base.

  • Base is a square with dimensions of 7cm by 7cm
  • 7*7
  • 49 cm²

Step 2: Find the Area of the Triangles.

  • Each triangle has a base of 7cm and a height of 9cm.
  • 7 x 9 = 63, and 63/2 = 31.5
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Step 3: Add the areas together.

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Thus, the surface area is 1.75 m^2 or 175 cm^2.

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(X-4) (x+2)=16 <br> What’s the standard form
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Step-by-step explanation:

(x - 4)(x + 2) = 16 \\  {x}^{2}  + (- 4 + 2)x + ( - 4)  \times 2= 16 \\  {x}^{2}  - 2x - 8  - 16 = 0 \\  {x}^{2}  - 2x - 24 = 0 \\ is \: the \: standard \: form.

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3 years ago
Read 2 more answers
Determine what shape is formed for the given coordinates for ABCD, and then find the perimeter and area as an exact value and ro
Helga [31]

Answer:

Part 1) The shape is a trapezoid

Part 2) The perimeter is 25(4+\sqrt{2})\ units   or approximately  135.4\ units

Part 3) The area is 937.5\ units^2

Step-by-step explanation:

step 1

Plot the figure to better understand the problem

we have

A(-28,2),B(-21,-22),C(27,-8),D(-4,9)

using a graphing tool

The shape is a trapezoid

see the attached figure

step 2

Find the perimeter

we know that

The perimeter of the trapezoid is equal to

P=AB+BC+CD+AD

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Find the distance AB

we have

A(-28,2),B(-21,-22)

substitute in the formula

d=\sqrt{(-22-2)^{2}+(-21+28)^{2}}

d=\sqrt{(-24)^{2}+(7)^{2}}

d=\sqrt{625}

d_A_B=25\ units

Find the distance BC

we have

B(-21,-22),C(27,-8)

substitute in the formula

d=\sqrt{(-8+22)^{2}+(27+21)^{2}}

d=\sqrt{(14)^{2}+(48)^{2}}

d=\sqrt{2,500}

d_B_C=50\ units

Find the distance CD

we have

C(27,-8),D(-4,9)

substitute in the formula

d=\sqrt{(9+8)^{2}+(-4-27)^{2}}

d=\sqrt{(17)^{2}+(-31)^{2}}

d=\sqrt{1,250}

d_C_D=25\sqrt{2}\ units

Find the distance AD

we have

A(-28,2),D(-4,9)

substitute in the formula

d=\sqrt{(9-2)^{2}+(-4+28)^{2}}

d=\sqrt{(7)^{2}+(24)^{2}}

d=\sqrt{625}

d_A_D=25\ units

Find the perimeter

P=25+50+25\sqrt{2}+25

P=(100+25\sqrt{2})\ units

simplify

P=25(4+\sqrt{2})\ units ----> exact value

P=135.4\ units

therefore

The perimeter is 25(4+\sqrt{2})\ units   or approximately  135.4\ units

step 3

Find the area

The area of trapezoid is equal to

A=\frac{1}{2}[BC+AD]AB

substitute the given values

A=\frac{1}{2}[50+25]25=937.5\ units^2

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