I got 60 right out of 110 questions on a test. What was my overall grade/percentage?

Answers

Answer 1

Answer: What you do is you make it into a fraction. You got 60 out of 110, so that's 60/110. Because all fractions are practically division problems, that's 60 ÷ 110 = 0.5454545454...
Times that by 100 to get the percentage.

Answer 2

Answer: Ans-YOUR OVERALL PERCENTAGE IS 54%

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Quadrilateral A’B’C’D’ is the image of quadrilateral ABCD under a translation.A translation by ____ units to the _right/left_ and ____ units _up/down_.

Answers

A translation by 2 units to the right and 5 units down

The arrow at the maze entrance indicates that the robot will be heading east when it enters the maze. When programming the robot, let the complex number d represent the direction the robot is facing. As the robot changes direction, the value of d will also change; so, the value of d is dependent on where the robot is in the maze. At the start of the maze, what is the value of d?

Answers

Answer:

d=1

Step-by-step explanation:

If North is i and South is -i

East is 1 and West is -1

The robot is facing East as it enters the maze, it will have a starting value of 1

Answer:

The value of d at the start of the maze will be 1 because 1 represents East .

Which expression has a value of 3A. (12 + 6) ÷ (3 + 3)

B. 12 + (6 ÷ 3 + 3)

C. (12 + 6) ÷ 3 + 3

D. 12 + 6 ÷ (3 + 3)

Answers

Answer:

(12 + 6) / (3 + 3) = 18 / 6 = 3

your answer is : (12 + 6) / (3 + 3)

Answer:

Step-by-step explanation:

A) 18/6 = 3

B) 12+5 = 17

C) (18 / 3) + 3 = 9

D) 12 +(6/6) = 13

Which long division problem can be used to prove the formula for factoring the difference of two perfect cubes?

Answers

Some of the possible options of the questions are;

A) $(a - b) | \overline {a^2 + a \cdot b + b^2}$

B) $(a + b) | \overline {a^2 - a \cdot b + b^2}$

C) $(a + b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

D) $(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

The difference of two perfect cubes has a binomial factor and a trinomial factor

The option that gives the long division problem that can be used to prove the difference of two perfect cubes is option D

D) $\underline {(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}}$

Reason:

The formula for factoring the difference of twoperfect cubes is presented as follows;

a³ - b³ = (a - b)·(a² + a·b + b²)

Given that a factor of the difference of two cubes is (a - b), and that we

have; (a³ + 0·a·b² + 0·a²·b - b³) = (a³ - b³), both of which are present in

option D, by long division of option D, we have;

${} \hspace {33} a^2 + a \cdot b + b^2\n(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a^2 \cdot b - b^3}\n{} \hspace {33} \underline{a^3 - a^2 \cdot b }\n{} \hspace {55} a^2 \cdot b + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3\n {} \hspace {55} \underline{a^2 \cdot b - a \cdot b^2}\n{} \hspace {89} a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3\n{} \hspace {89} \underline{a \cdot b^2 - b^3}\n{}\hspace {89} 0$

By the above long division, we have;

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$ = a² + a·b + b²

Which gives;

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$ = (a³ + 0·a·b² + 0·a·b² - b³)/(a - b)

We get;

(a³ + 0·a·b² + 0·a·b² - b³)/(a - b) = a² + a·b + b²

(a - b)·(a² + a·b + b²) = (a³ + 0·a·b² + 0·a·b² - b³) = (a³ - b³)

(a - b)·(a² + a·b + b²) = (a³ - b³)

(a³ - b³) = (a - b)·(a² + a·b + b²)

Therefore;

The long division problem that can be used to prove the formula for

factoring the difference of two perfect cubes is

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$ , which is option D

D) $(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

Learn more here:

brainly.com/question/17022755

Answer:

The correct options, rearranged, are:

Options:

$A)(a^2+ab+b^2)/(a-b)\n\nB)(a^2-ab+b^2)/(a+b)\n\nC)(a^3+0a^2+0ab^2-b^3)/(a+b))\n\n D)(a^3+0a^2+0ab^2-b^3)/(a-b)$

And the asnwer is the last option (D).

Explanation:

You need to find which long division can be used to prove the formula for factoring the difference of two perfect cubes.

The difference of two perfect cubes may be represented by:

$a^3-b^3$

And it is, as a very well known special case:

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Then, to prove, it you must divide the left side, $a^3-b^3$ , by the first factor of the right side, $a-b$

Note that, to preserve the places of each term, you can write:

$(a^3-b^3)=(a^3+0a^2+0ab^2-b^3)$

Then, you have:

$(a^3+0a^2+0ab^2-b^3)=(a-b)(a^2+ab+b^2)$

By the division property of equality, you can divide both sides by the same factor, which in this case will be the binomial, and you get:

$(a^3+0a^2+0ab^2-b^3)/(a-b)=(a^2+ab+b^2)$

That is the last option (D).

Which transformation will always create congruent figures

Answers

Rotations, reflections, and translations are isometric. That means that these transformations do not change the size of the figure. If the size and shape of the figureis not changed, then the figures are congruent.

Two rectangles have the same width. The length of one is 1 foot longer than the width. The length of the other is 2 feet longer than the width. The larger rectangle has 6 more square feet than the smaller. What is the width of the rectangles?

Answers

$w-width\n\nl_1=w+1\nl_2=w+2\n\nwl_1=wl_2-6\n\nw(w+1)=w(w+2)-6\nw^2+w=w^2+2w-6\nw^2-w^2+w-2w=-6\n-w=-6\nw=6\leftarrow answer$

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