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Serga [27]
3 years ago
14

Find the area of the shape shown below pls

Mathematics
1 answer:
Afina-wow [57]3 years ago
6 0

Step-by-step explanation:

area of missed triangle =1/2×2×4=4 cm2

area of whole rectangle =4×8=32 cm2

area of shape =32_4 =28cm2

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Can anyone help me and explain this to me
boyakko [2]

Answer:

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Step-by-step explanation:

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5 0
3 years ago
Read 2 more answers
PLZ HELP ME!<br> Directions up above the picture<br> **MATH (Algebra)
Vikentia [17]

Answer:

The answer to your question is t = 1.3 s

Step-by-step explanation:

Data

Equation    h(t) = -4.9t² + v₀t + h₀

v₀ = 0 m/s

h₀ = 8 m

t = ?

h = 0 m

Process

1.- Substitute the values in the formula

                  0 = -4.9t² + 0t + 8

2.- Simplification

                  0 = -4.9t² + 8

3.- Solve for t

                 4.9t² = 8

                      t² = 8/4.9

                      t² = 1.63

4.- Result

                      t = 1.27 ≈ 1.3 s

6 0
3 years ago
The base of a cylindrical candle has an area of
Fudgin [204]

The volume of the cylindrical candle that has a base area of 12.56 inches is 12.56h inches³.

<h3>Volume of of cylinder</h3>

volume of a cylinder = πr²h

where

  • r = radius
  • h = height

Therefore,

volume of a cylinder = πr²h

Area of the base = πr²

Therefore,

Area of the base = 12.56 inches²

volume of the cylindrical candle = 12.56h inches³

learn more on volume here: brainly.com/question/11459769

#SPJ1

7 0
2 years ago
What percent of 88 is 33
Damm [24]
The percent that is 33 is (33/88)*100 = 37.5%
4 0
3 years ago
Find the solution set of the following quadratic equations using the quadratic formula.
oksian1 [2.3K]

Answer:

<h3>                1)  x_1=\dfrac{7+\sqrt{13}}2\,,\quad x_2=\dfrac{7-\sqrt{13}}2</h3><h3>                2)   x_1=-\dfrac13\,,\quad x_2=-3    </h3>

Step-by-step explanation:

<h3>1)</h3>

x^2 - 7x + 9 = 0\quad\implies\quad a=1\,,\ b = -7\,,\ c=9\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-7)\pm\sqrt{(-7)^2-4\cdot1\cdot9}}{2\cdot1}=\dfrac{7\pm\sqrt{49-36}}2\\\\x_1=\dfrac{7+\sqrt{13}}2\,,\quad x_2=\dfrac{7-\sqrt{13}}2

<h3>2)</h3><h3>3x^2 + 10x=-3\\\\3x^2+10x+3=0\quad\implies\quad a=3\,,\ b =10\,,\ c=3\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-10\pm\sqrt{10^2-4\cdot3\cdot3}}{2\cdot3}= \dfrac{-10\pm\sqrt{100-36}}6\\\\x_1=\dfrac{-10+\sqrt{64}}6=\dfrac{-10+8}6=-\dfrac13\,,\qquad x_2=\dfrac{-10-8}6=-3</h3>
7 0
3 years ago
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