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3 years ago

Part c

Mathematics

1 answer:

Sergeu [11.5K]3 years ago

6 0

circurference of gear = 2πr

= 2π(10)

= 62.83185307 inches

= 62.83 inches (4sf)

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hello my frands teheehee free answers maybe even giving out brainliest I dont know how to spell don't attack me

Aleks04 [339]

What’s the question

7 0

3 years ago

Read 2 more answers

In the accompanying diagram, ABCD is a

Romashka [77]

Given:

m∠DAE = 30°

m∠CBE = 20°

To find:

The value of x.

Solution:

In rectangle, all angles are right angle.

m∠A = m∠B = m∠C = m∠D = 90°

m∠EDA + m∠EAB = 90°

30° + m∠EAB = 90°

m∠EAB = 90° - 30°

m∠EAB = 60°

Similarly, m∠CBE + m∠EBA = 90°

20° + m∠EBA = 90°

m∠EBA = 90° - 20°

m∠EBA = 70°

In triangle AEB,

Sum of all angles in a triangle = 180°

m∠EAB + m∠AEB +m∠EBA = 180°

60° + x° + 70° = 180°

130° + x° = 180°

Subtract 130° from both sides, we get

x° = 50°

x = 50

The value of x is 50.

7 0

3 years ago

What is the product of 2x+ 3 and 4x2 - 5x+ 6?

elena-s [515]

Answer:

8x³ - 25x² - 3x + 18

Step-by-step explanation:

(2x + 3)(4x²- 5x + 6)

multiply 2x by each of the three terms in the second expression to get:

8x³ - 10x² + 12x

now multiply 3 by each of the three terms in the second expression to get:

12x² - 15x + 18

Combine 'like terms': -10x² + (-15x²) = -25x²

Combine 'like terms': 12x + (-15x) = -3x

Put all terms in decreasing exponent order:

8x³ - 25x² - 3x + 18

7 0

3 years ago

What is the sum of the infinite geometric series 9−6+4−83+...9-6+4-83+...??

vivado [14]

Successive terms have a common factor of $-\dfrac23$ , which means the $n$ th partial sum of the series can be written as

$S_n=9-6+4-\dfrac83+\cdots+9\left(-\dfrac23\right)^{n-1}$
$S_n=9\left(1+\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\cdots+\left(-\dfrac23\right)^{n-1}$

$\implies-\dfrac23S_n=9\left(\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\left(-\dfrac23\right)^4+\cdots+\left(-\dfrac23\right)^n\right)$

$\implies S_n-\left(-\dfrac23\right)S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$\dfrac53S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$S_n=\dfrac{27}5\left(1-\left(-\dfrac23\right)^n\right)$

As $n\to\infty$ , the geometric term vanishes, leaving you with

$S=\displaystyle\lim_{n\to\infty}S_n=\dfrac{27}5$

7 0

3 years ago

Quadrilateral ABCD is similar to Quadrilateral EFGH. Diagonal AC has length 7 and diagonal EG has length 13. What is the scale f

olasank [31]

Answer:

$\large \boxed{\text{A. }\dfrac{13}{7}\text{; B. }\dfrac{17}{14} \text{; C. 507 in}^{2}}$

Step-by-step explanation:

A. Scale factor

When you dilate an object by a scale factor, you multiply its line lengths by the same number.

If EF/AB = 13/7, the scale factor is 13/7.

B. Length of EF

$\begin{array}{rcl}\dfrac{EF}{AB} & = & \dfrac{13}{7}\\\\\dfrac{EF}{\frac{17}{26}} & = & \dfrac{13}{7}\\\\EF & = & \dfrac{13}{7}\times\dfrac{17}{26}\\\\ & = &\dfrac{1}{7}\times\dfrac{17}{2}\\\\ & = & \mathbf{\dfrac{17}{14}}\\\end{array}\\\text{The length of EF is $\large \boxed{\mathbf{ \dfrac{17}{14}}}$}$

C. Area of EFGH

If the lengths in a shape are all multiplied by a scale factor, then the areas will be multiplied by the scale factor squared.

ABCD is dilated by a scale factor of 13/7, so its area is dilated by a scale factor of

$\left(\dfrac{13}{7} \right)^{2} = \dfrac{169}{49}$

The area of its dilated image EFGH is

$\text{Area of EFGH} = \text{147 in}^{2} \times \dfrac{\text{169}}{\text{49}} = 3 \times 169\text{ in}^{2} = 507 \text{ in}^{2}\\\\\text{The area of EFGH is $\large \boxed{\textbf{507 in}^{\mathbf{2}}}$}$

8 0

3 years ago