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2 years ago

1 Negative numbers do not need thenegative sign (-) in front of them.O TrueO False

Mathematics

2 answers:

NeTakaya2 years ago

7 0

Answer:

FALSE ALL NEGATIVE NUMBERS NEED A NEGATIVE SIGHT(-)

Step-by-step explanation:

-1,-2,-3, PLEASE MARK BRAINLEST

katrin [286]2 years ago

6 0

Answer:

The answer is False

Step-by-step explanation:

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Your bank account has a starting value of $105. You deposit $45 and get $20 cash back. How much do you have in the bank account

Olenka [21]

Answer:

$80?

Step-by-step explanation:

I hope its right

3 0

2 years ago

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

2 years ago

Solve the inequality and graph the solution. −0.8b + 2.3 ≥ 8.7

zlopas [31]

The solution of the inequality is -8 ≥ b, and the correct graph is the one in option D.

"number line with a closed circle plotted at negative eight and arrow pointing left."

<h3>How to solve the inequality?</h3>

Here we have the inequality:

-0.8*b + 2.3 ≥ 8.7

And we want to solve this, to do so, we need to isolate the variable b in one of the sides of the inequality.

-0.8*b + 2.3 ≥ 8.7

2.3 - 8.7 ≥ 0.8*b

-6.4 ≥ 0.8*b

-6.4/0.8 ≥ b

-8 ≥ b

So the solution is the set of all numbers equal to or smaller than -8, then the correct graph will be the one described by D.

Learn more about inequalities:

brainly.com/question/24372553

#SPJ1

5 0

10 months ago

3/1 + 1/3<br><br> What is the answer to this?

Vika [28.1K]

Answer:

10/3, or 3 1/3.

Step-by-step explanation:

3/1 + 1/3

3 + 1/3

9+1/3

10/3, or 3 1/3.

6 0

1 year ago

Solve for x using the quadratic formula x^2-6x +9=0

Rufina [12.5K]

Answer:

3 both ways x = 3

EXPLANATION:

plug in inputs and do a little simplifying we get

x = 6 ± √(36-36) all of that over 2

So we simplify and we get 6±sqrt(0) / 2

Then we get 6/2 = 3

x-0 = x+0

thats why there is only one answer

4 0

2 years ago

Read 2 more answers

1 Negative numbers do not need thenegative sign (-) in front of them.O TrueO False​

1 Negative numbers do not need thenegative sign (-) in front of them.O TrueO False