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Mashcka [7]
2 years ago
14

Show that a% of b = b% of a

Mathematics
1 answer:
Dafna1 [17]2 years ago
4 0

Answer:

See below

Step-by-step explanation:

a% of b

= b * a/100 = ab/100

b% of a

= a * b/100 = ab/100

So they are the same.

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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
Use the identity (x+y)(x^2−xy+y^2)=x^3+y^3 to find the sum of two numbers if the product of the numbers is 28, the sum of the sq
lara31 [8.8K]

Step-by-step explanation:

We have xy = 28, x² + y² = 65 and x³ + y³ = 407.

Since (x + y)(x² - xy + y²) = x³ + y³,

x + y = (x³ + y³)/(x² + y² - xy)

= (407) / [(65) - (28)]

= 407 / 37

= 11.

Hence the sum of the numbers is 11.

4 0
3 years ago
Read 2 more answers
Which point is located at (-3,6) <br><br> A) A<br><br> B) B<br><br> C) C<br><br> D) D
Andrej [43]

Answer is A

Step-by-step explanation:

Go over to the left 3 times then up 6

3 0
2 years ago
Hcf of 14 154 and 28​
anyanavicka [17]

Answer:

14

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i
Anit [1.1K]

Answer:

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The n^{th} term of geometric sequence is given by:

a_n = ar^{n-1}

a_4 = 16 = ar^3\\a_6 = 4 = ar^5

Dividing the two equations, we get,

\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}

the first term can be calculated as:

16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128

Thus, the required geometric sequence is

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

4 0
3 years ago
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