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3 years ago

What is a counterexample for the conjecture? Any number that is divisible by four is also divisible by eight

1 answer:

3 0

The value 12 is a good counterexample. It is divisible by 4 (since 12/4 = 3 is a whole number with no remainders or decimal portions) but 12 is not divisible by 8 (note how 12/8 = 1 remainder 4 = 1.5)

To find other examples, add 8 to 12 to get 20. Then add on 8 more to get 28. This pattern continues on forever.

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Use the digits 2,3,and 5 to create a fraction and a whole number with a product greater than 2

mylen [45]

Product is multiplication.

Since the answer needs to be greater than 2 either the fraction needs to be an improper fraction ( meaning the fraction would be greater than 1) or the whole number would need to be greater than 2.

One answer would be 5 x 2/3 = 3 1/3

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3 years ago

Describe the translation of the figure ABCD. Complete the sentence to explain your answer

irina1246 [14]

No picture. Please repost

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3 years ago

The diagram shown can be used to prove the Pythagorean theorem. Which statement is TRUE about the diagram?

Margaret [11]

Answer:

Step-by-step explanation:

the answer is c

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3 years ago

How do you find the variable when it’s on both sides of the equation

ASHA 777 [7]

If, for example, you had x on one side and x on the other, then subtracting x from both sides "disappears" x. If no other power of x shows up on either side, then your equation either has no solution or is always true.

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3 years ago

Find the vector projection of B onto A if A = 5i + 11j – 2k,B = 4i + 7k

valkas [14]

Answer:

$\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\$

Step-by-step explanation:

Given A = 5i + 11j – 2k and B = 4i + 7k, the vector projection of B unto a is expressed as $proj_ab = \dfrac{b.a}{||a||^2} * a$

b.a = (5i + 11j – 2k)*( 4i + 0j + 7k)

note that i.i = j.j = k.k =1

b.a = 5(4)+11(0)-2(7)

b.a = 20-14

b.a = 6

||a|| = √5²+11²+(-2)²

||a|| = √25+121+4

||a|| = √130

square both sides

||a||² = (√130)

||a||² = 130

$proj_ab = \dfrac{6}{130} * (5i+11j-2k)\\\\proj_ab = \frac{30}{130} i+\frac{11}{130} j-\frac{12}{130} k\\\\proj_ab = \frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\\\$

<em>Hence the projection of b unto a is expressed as </em> $\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\$ <em></em>

7 0

3 years ago