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3 years ago

Which of the following conjectures is false?Please give a counterexample!Thank youuu

Mathematics

1 answer:

MakcuM [25]3 years ago

6 0

Which of the following conjectures is false?

Solution: The false conjecture is:

J. The sum of two odd numbers is odd

Explanation:

F. The product of two even numbers is even.

For example: Let us take two even numbers 4 and 6.

The product of two even numbers is:

$4 \times 6 = 24$

Therefore, the product is also even number. Hence the conjecture is true.

G. The sum of two even numbers is even.

For example: Let us take two even numbers 4 and 6.

The sum of two even numbers is:

$4+6 = 10$

Therefore, the sum is also even number. Hence the conjecture is true.

H. The product of two odd numbers is odd.

For example: Let us take two odd numbers 3 and 5.

The product of two odd numbers is:

$3\times5 = 15$

Therefore, the product is also odd number. Hence the conjecture is true.

J. The sum of two odd numbers is odd.

For example: Let us take two odd numbers 3 and 5.

The sum of two odd numbers is:

$3+5= 8$

Therefore, the sum is not an odd number. Hence the conjecture is false.

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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

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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 $\large \boxed{\textbf{507 in}^{\mathbf{2}}}$}$

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neonofarm [45]

Answer:

Please check the explanation and attached graph.

Step-by-step explanation:

Given the parent function

y = |x|

In order to translate the absolute function y = |x| vertically, we can use the function

g(x) = f(x) + h

when h > 0, the graph of g(x) translated h units up.

Given that the image function

y=|x|+4

It is clear that h = 4. Since 4 > 0, thus the graph y=|x|+4 translated '4' units up.

The graph of both parent and translated function is attache below.

In the graph,

The blue line represents the parent function y=|x|.

The red line represents the image function y=|x| + 4.

It is clear from the graph that the y=|x| + 4 translated '4' units up.

Please check the attached graph.

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