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Mamont248 [21]
2 years ago
10

Akira recives a prize at a science fair for having the most information project her trophy is in the shape of a square pyramid a

nd is covered in shinny gold foil how much gold foil did it take to cover the trophy including the bottom
Mathematics
1 answer:
liberstina [14]2 years ago
4 0

Answer:

12

Step-by-step explanation:

Because I think it is

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Solve for bbb.<br> -3b+2.5 = 4−3b+2.5=4
Anna11 [10]
The answer is not possible because both most common factors cancel each other out
7 0
3 years ago
Read 2 more answers
Points E, F, and D are on circle C, and angle G measures 60°. The measure of arc EF equals the measure of arc FD.
galben [10]

Answer:

∠EFD ≅ ∠EGD ⇒ A

Arc ED ≅ arc FD ⇒ C

m arc FD = 120° ⇒ E

Step-by-step explanation:

Let us revise some facts

  1. Equal chords subtended equal arcs
  2. The measure of an inscribed angle is one-half the measure of the central angle which subtended by the same arc
  3. The measure of a central angle is equal to the measure of its subtended arc
  4. If one angle of an isosceles triangle measure 60° then the triangle is equilateral
  5. The sum of the measures of the interior angles of any quadrilateral is 360°

In the quadrilateral  CDGE

∵ m∠G = 60°

∵ m∠GDC = m∠GEC = 90°

- By using the 5th rule above

∴ m∠G + m∠GDC + m∠DCE + m∠GEC = 360°

∴ 60 + 90 + m∠DCE + 90 = 360

∴ 240 + m∠DCE = 360

- Subtract 240 from both sides

∵ m∠DCE = 120°

In circle C

∵ ∠DCE is a central angle subtended by arc DE

∵ ∠DFE is an inscribed angle subtended by arc DE

- By using the 2nd rule above

∴ m∠DFE = \frac{1}{2} m∠∠DCE

∵ m∠DCE = 120°

∴ m∠DFE = \frac{1}{2} (120)

∴ m∠DFE = 60°

- That means ∠EFD ≅ ∠EGD because their measure is 60°

∴ ∠EFD ≅ ∠EGD

In Δ EFD

∵ EF = FD

∵ m∠DFE = 60°

- By using the 4th rule above

∴ Δ EFD is an equilateral triangle

∴ ED = FD = FE

In circle C

∵ Side ED subtended by arc ED

∵ Side FD subtended by FD

∵ Side ED ≅ side FD ⇒ proved

- By using the 1st rule above

∴ Arc ED ≅ arc FD

∵ m∠ECD = 120°

∵ ∠ECD is a central angle subtended by arc ED

- By using the 3rd rule above

∴ m∠ECD = m arc ED

∴ m of arc ED = 120°

∵ Arc ED ≅ arc FD

∴ m arc ED = m arc FD

∴ m arc FD = 120°

5 0
3 years ago
Read 2 more answers
Multiply 3x (2x - 1)​
Volgvan

Answer:

6x^2 - 3x

6x^2-3x

Step-by-step explanation:

3x (2x-1)

multiply 3x by 2x -> 6x^2

multiply 3x by -1 -> -3x

6 0
2 years ago
Read 2 more answers
What type of polynomial is: 3x+x^2+4 <br>A.quadratic<br> B. quartic<br> C. linear <br>D. cubic
Svetradugi [14.3K]

Answer:

A.quadratic

Step-by-step explanation:

3x+x^2+4

Put in order from largest to smallest power of x

x^2 +3x+4

The highest power is 2

A.quadratic   highest power 2

B. quartic   highest power 4

C. linear  highest power 1

D. cubic highest power 3

3 0
3 years ago
Integrate this two questions ​
tankabanditka [31]

Simplify the integrands by polynomial division.

\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)

\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)

Now computing the integrals is trivial.

5.

\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}

where we use the power rule,

\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)

and a substitution to integrate the last term,

\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)

8.

\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}

using the same approach as above.

5 0
2 years ago
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