What are the equations for these graphs

The point (-7,4) is reflected over the line x=-3. Then the resulting point is reflected over the line y=x. Where is the point lo

Ne4ueva [31]

Answer:

(4, 1)

Step-by-step explanation:

Since the first reflection is over the vertical line x=-3, the y-coordinate remains the same. The x-coordinate of A' will make the point (-3, 4) on the line of reflection be the midpoint between A and A':

(-3, 4) = (A +A')/2

2(-3, 4) -A = A' = (-6-(-7), 8 -4) = (1, 4)

The reflection over the line y=x simply interchanges the two coordinate values:

A'' = (4, 1)

3 0

2 years ago

Pls work out and give detail. Use the slope formula to determine the slope of each line.

kolbaska11 [484]

Answer:

$m=0$

Step-by-step explanation:

Given

See attachment

Required

Determine the slope of the line

From the attachment, we have the following points

$(x_1,y_1) = (-4,1)$

and

$(x_2,y_2) = (3,1)$

The slope m is then calculated using:

$m = \frac{y_2- y_1}{x_2 - x_1}$

Substitute values for the x's and y's

$m = \frac{1-1}{3 - (-4)}$

$m = \frac{1-1}{3 +4}$

$m = \frac{0}{7}$

$m=0$

Hence, the slope is 0

7 0

2 years ago

What is the height ? Please help, I need it! Urgent !

Vika [28.1K]

Answer:

207.24 inch^2

Step-by-step explanation:

A=2*3.14*9+2.3.14*3*8

=56.52+150.72

=207.24 square inches

4 0

2 years ago

Please help!! MathsWatch, Surd Expressions, Clip 207c

dexar [7]

Answer:

$A=(72+36\sqrt{3})\ mm^2$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

Let

a ----> the height of rectangle in mm

b ---> the base of rectangle in mm

step 1

Find the base of rectangle

$b=AB+BC+CD+DE$ ----> by segment addition postulate

substitute

$b=3+3+3+3=12\ mm$

step 2

Find the height of rectangle

$a=FB+CG+GH$ ---> by segment addition postulate

substitute the given values

$a=3+CG+3\\a=6+CG$

Find the length sides CG

Applying the Pythagorean Theorem

$BG^2=BC^2+CG^2$

substitute the given values

$6^2=3^2+CG^2$

$CG^2=6^2-3^2$

$CG=\sqrt{27}\ mm$

simplify

$CG=3\sqrt{3}\ mm$

therefore

$a=(6+3\sqrt{3})\ mm$

step 3

Find the area of rectangle

$A=ab$

we have

$a=(6+3\sqrt{3})\ mm$

$b=12\ mm$

substitute

$A=(6+3\sqrt{3})(12)=(72+36\sqrt{3})\ mm^2$

8 0

3 years ago

Solve the inequality and show your solution on a number line: −4x≥−1

kiruha [24]

*To solve an inequality, it's the same as solving an equation; isolate the variable to one side.

For this, all you need to do is divide both sides by -4. And since we are dividing by a negative, flip the inequality sign and your inequality is $x\leq\frac{1}{4}$

Now to graph this. Since x can be "equal to" 1/4 ( ≤ = less than or equal to), you will have a closed circle on 1/4. And since x can also be "less than" 1/4, the arrow will be going to the left of 1/4.

6 0

2 years ago

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