PLS help 20 min
Which of the following correctly uses absolute value to show the distance between -80 and 15? (5 points) a |-80 - 15| = |-95| = -95 units b |-80 + 15| = |-65| = 65 units c |-80 - 15| = |-95| = 95 units d |-80 + 15| = |-65| = -65 units

Answer:

x=−y−14

Step-by-step explanation:

Answer:

x = none y = -27

Step-by-step explanation:

2x + 2y = -28 + 2x - 26

-2x                  -2x

__________________

2y = -28 - 26

2y = -54

y = -27

Answer:

y=1/2x-2

Step-by-step explanation:

perpendicular lines have a neg. reciprocal for the slope.

for example:

in this equation y=-2x+1

-2 is the slope and the negative reciprocal would be 1/2

hence, the slop of our new line is 1/2

y=1/2x+b

we can substitute the coordinates given for x and y:

2=1/2(8)+b

2=4+b

-2=b

hence the new equation is y=1/2x-2

Answer:

C.16, 32, 38, 39, 41, 42, 58

and

E.18, 30, 34, 38, 43, 45, 68

its C. & E.

they both have outliers lol

Answer:

y=-6

Step-by-step explanation:

the line needs to be vertical to be parallel to y=2, and pass through (-1,-6)

Answer:

A

Step-by-step explanation:

the equivalent pair of expression are 2(x+5)and 2x+10

Answer:

$1608.48

Step-by-step explanation:

A = P(1 + r/n)^nt

P: 1100   3: 3.8% / year     n: 365/year   t: 10 years

A = 1100 (1 + 0.038/365)³⁶⁵*¹⁰ = 1608.48

Answer:

$100

Step-by-step explanation:

since 1 hockey mask costs $10 and you need 10 masks you can multiply 10x10 and you will get 100

Answer:

-10

Step-by-step explanation:

5(-6 + 4)

=-30+20

=-10

Answer:

-10 or just 10 in general

Step-by-step explanation:

uh- this is the correct way to answer the problem....

−4x+12=−6x

Step 1: Add 6x to both sides.

−4x+12+6x=−6x+6x

2x+12=0

Step 2: Subtract 12 from both sides.

2x+12−12=0−12

2x=−12

Step 3: Divide both sides by 2.

[tex]\frac{2x}{2}[/tex] = [tex]\frac{-12}{2}[/tex]

x=−6

Answer:

[tex]T(ln2)[/tex][tex]= <\sqrt{145} , \frac{6\sqrt{2} }{\sqrt{145} } , \frac{6\sqrt{2} }{\sqrt{145} } >[/tex]

Step-by-step explanation:

Given : [tex]r(t)=<7-\sqrt{e^t}, 3e^t,3e^t > .........(i)[/tex]

We first have to differentiate of equation [tex](i)[/tex]

[tex]r'(t) = < \frac{\sqrt{e^t} }{2} , 3e^t,3e^t> ..........(ii)[/tex]

Now to get a unit tangent vector at the given value of [tex]t[/tex], we put [tex]t = ln2[/tex] in equation [tex](ii)[/tex]

[tex]r'(ln2) = < \frac{\sqrt{e^{ln2}} }{2} , 3e^{ln2},3e^{ln2}>[/tex]

            [tex]= <\frac{1}{\sqrt{2} } ,6,6>[/tex]        [∵[tex]e^{lna}=a[/tex]]

Now to get a unit tangent vector , we will divide our vector [tex]r'(ln2)[/tex] by its magnitude. So let's first find the magnitude.

[tex]|r'(ln2)|=\sqrt{(\frac{1}{\sqrt{2} } )^2 +6^2+6^2}[/tex]

             [tex]= \sqrt{\frac{1}{2}+36+36 }[/tex]

             [tex]= \sqrt{\frac{145}{2} }[/tex]

Now we can find the our unit tangents vector.

      [tex]T(ln2)=\frac{r'(ln2)}{|r'(ln2)|}[/tex]

                  [tex]= \frac{<\frac{1}{\sqrt{2}} ,6,6> }{\sqrt{\frac{145}{2}} }[/tex]

                  [tex]= <\frac{\frac{1}{\sqrt{2} } }{\sqrt{\frac{145}{2}} } ,\frac{6}{{\sqrt{\frac{145}{2}} }}, \frac{6}{{\sqrt{\frac{145}{2}} }} >[/tex]

                   [tex]= <\sqrt{145} , \frac{6\sqrt{2} }{\sqrt{145} } , \frac{6\sqrt{2} }{\sqrt{145} } >[/tex]

Hence, [tex]T(ln2)[/tex] [tex]= <\sqrt{145} , \frac{6\sqrt{2} }{\sqrt{145} } , \frac{6\sqrt{2} }{\sqrt{145} } >[/tex]

Answer: 52

Step-by-step explanation: Friend me and yes that is right!

Answer: 25/6

Step-by-step explanation:

Answer:

? = 4.1667

Step-by-step explanation:

300/72 = 4.1667

Hope this Helps! :)

Have any questions? Ask below in the comments and I will try my best to answer.

-SGO

Answer:

Zero is nor a positive or negative number, so im guessing the answer is 0.

Step-by-step explanation:

Answer:  4.5 units

Step-by-step explanation:

OD and OC are both radii of the circle O.  All radii of the same circle are equal.  Thus,

[tex]6x+2=10x-16[/tex]

18=4x

x=4.5

Hope it helps :)

Answer:

positive

Step-by-step explanation:

even negatives = positive

whatever positives = positive

positive*positive = positive

Answer:

They are equal

Step-by-step explanation:

Given

[tex]\frac{3}{27}\ and\ \frac{1}{9}[/tex]

Required

Determine if they are equal

We start by equating both:

[tex]\frac{3}{27} = \frac{1}{9}[/tex]

Cross Multiply

[tex]3 * 9 = 1 * 27[/tex]

[tex]27 = 27[/tex]

Because both sides of the equation are equal, then the ratios are equal

Answer:

A because you have to mutiply 3 x 2 and it gives you 6

Answer:

required answer is 9 of your question

Answer:

11(4-1)

Step-by-step explanation:

11 is the greatist common factor of 44 and 11 so you factor an 11 out of both terms

44/11=4

11/11=4

Therefore the answer is 11(4-1)

Answer:

1/2

Step-by-step explanation:

Answer:

1st graph on line 2

Step-by-step explanation:

Answer:

Yes, Yes, No, Yes

Step-by-step explanation:

Answer:

b

Step-by-step explanation:

i just took the quiz plz mark brainliest

Answer:

The correct answer is: Option 1: x≤-3

Step-by-step explanation:

Given inequality is:

[tex]-6x-4\geq -2(x-4)[/tex]

First of all, we will distribute -2 into the bracket

[tex]-6x-4\geq -2x+8[/tex]

Then we will add 4 on both sides

[tex]-6x-4+4\geq -2x+8+4\\-6x \geq -2x+12[/tex]

Adding 2x on both sides

[tex]-6x+2x\geq -2x+12+2x\\-4x\geq 12[/tex]

Dividing both sides by 4

[tex]\frac{-4x}{4} \geq \frac{12}{4}\\-x \geq 3[/tex]

As there should be only x on the left side of the inequality we will multiply the inequality with -1.

Multiplying the inequality with -1 reverses the sign of inequality

So,

[tex]x \leq -3[/tex]

Hence,

The correct answer is: Option 1: x≤-3

Step-by-step explanation:

Since area of Square A = 7 ft²,

Side length of Square A = √7 ft.

Length of Rectangle B = √7 ft

Width of Rectangle B = 4/√7 ft

Hence Area of Square C

= (4)² / (√7)² ft² = 16/7 ft².

The area of Square C is the square of the height of the Rectangle B.

Correct responses:

The given area of square A = 7 ft.²

The given area of rectangle, B = 4 ft.²

Let s represent the side length of square C and let S represent the side length of square A, we have;

Area of rectangle B = S × s = S·s

Area of square A = S × S = S²

Area of square C = s × s = s²

[tex]\mathbf{\dfrac{(Area \ of \ rectangle \ B)^2}{Area \ of \ square \ A } } = \dfrac{\left(S \cdot s \right)^2}{S^2} = \dfrac{S^2 \cdot s^2}{S^2} = s^2 = Area \ of \ square \ C[/tex]

Therefore, the expression for the area of Square C in terms of the area

of Square A and Rectangle B is presented as follows;

Which gives;

By simplification, we have;

[tex]\mathbf{\dfrac{\left(4 \, ft.^2 \right)^2}{7 \, ft.^2}} = \dfrac{16}{4} \, ft.^2[/tex]

Learn more about squares and rectangles here:

https://brainly.com/question/2288475

https://brainly.com/question/14044544

https://brainly.com/question/8428727

Answer:

sorry im doing a challenge but i will have someone answer it    

Step-by-step explanation: