if u can answer all them correctly ill make a post for u and u can have all my coins just pls answer them all im begging

Nastasia [14]

Use the pythagorean theorem to solve this: a^2 + b^2 = c^2.
You already have a and b which are 3 and 4. Square them up and you will get 9 + 16 = c^2.
Add.
25 = c^2
Lastly, square root both sides.
5 = c
The length of AB is 5.

8 0

2 years ago

Can someone help me out, please?

GenaCL600 [577]

Rectangle, triangle, and I believe square

3 0

3 years ago

Through: (-3,0), perp. to y = 3x - 1

Marrrta [24]

Answer:

$y=-\dfrac{1}{3}x-1$

Step-by-step explanation:

Given:

line y=3x-1

point (-3,0)

Write: equation of the line that is perpendicular to the given and passes through the point (-3,0)

Solution:

The slope of the given line is $m=3$

If $m_1$ is the slope of perpendicular line, then

$m\cdot m_1=-1\\ \\3m_1=-1\\ \\m_1=-\dfrac{1}{3}$

So, the equation of the needed line is $y=-\dfrac{1}{3}x+b.$

Find b. This line passes through the point (-3,0), so its coordinates satisfy the equation:

$0=-\dfrac{1}{3}\cdot (-3)+b\\ \\0=1+b\\ \\b=-1\\ \\y=-\dfrac{1}{3}x-1$

7 0

3 years ago

Geometry problem help! Please refer to the image below...

Vinil7 [7]

Answer:

A. 1/3

B. √10

C. -1, 1

D. √8, 6

E. congruent and opposite pairs parallel

F. perpendicular, not congruent

G. rhombus, explanation below

Step-by-step explanation:

Hey there! I'm happy to help!

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A.

Slope is the rise over the run. Let's look at F to G.

We are going from -1 to 2 on our x-axis (run), so our run is 3 units.

Our rise is 1 unit as we go from 2 to 3 on the y-axis.

$slope=\frac{rise}{run} =\frac{1}{3}$

This slope is the same for all of the sides.

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B.

We will use the distance formula (which is basically just the Pythagorean Theorem) to calculate the length of each side. Let's go between F and G again, but this distance is the same for all the sides.

$\sqrt{(x_{2}-x_1)^2+(y_{2}-y_1)^2 } \\\\(x_1,y_1)=(-1,2)\\\\(x_2,y_2)=(2,3)\\\\\\\sqrt{(2+1)^2+(3-2)^2 } \\\\\sqrt{(3)^2+(1)^2 }\\\\\sqrt{9+1 }\\\\\sqrt{10}$

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C.

The diagonals are the lines that connect the non-adjacent vertices.

Our two diagonals are FH and GE.

-----------------------------

FH

We go from x-value -1 to 1 from F to H, so our run is 2.

We go from y-value 2 to 0. so our rise is -2.

$slope=\frac{rise}{run} =-\frac{2}{2} =-1$

-----------------------------

GE

We go from x-value -2 to 2 from E to G, so our run is 4.

We go from y-value -1 to 3. so our rise is 4.

$slope=\frac{rise}{run} =\frac{4}{4} =1$

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D.

Let's use the distance formula on each of our diagonals.

-----------------------------

FH

$\sqrt{(x_{2}-x_1)^2+(y_{2}-y_1)^2 } \\\\(x_1,y_1)=(-1,2)\\\\(x_2,y_2)=(1,0)\\\\\\\sqrt{(1+1)^2+(0-2)^2 } \\\\\sqrt{(2)^2+(-2)^2 }\\\\\sqrt{4+4 }\\\\\sqrt{8}$

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GE

$\sqrt{(x_{2}-x_1)^2+(y_{2}-y_1)^2 } \\\\(x_1,y_1)=(-2,-1)\\\\(x_2,y_2)=(2,3)\\\\\\\sqrt{(2+2)^2+(3+1)^2 } \\\\\sqrt{(4)^2+(4)^2 }\\\\\sqrt{16+16 }\\\\\sqrt{36}\\\\6$

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E.

They are congruent as they all have the same length (√10) and the opposite sides are parallel as they have the same slope (1/3)

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F.

They are perpendicular diagonals as their slopes are negative reciprocals (1 and -1), and they are not congruent as they have different lengths (√8 and 6).

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G.

Parallelogram- quadrilateral with opposite pairs of parallel sides.

Rhombus- a parallelogram with four equal sides

Square- a rhombus with four right angles

We can see that this is a parallelogram as we saw that the opposite sides are parallel due to having the same slope, and the perpendicular diagonals show that as well. This is also a rhombus because if we use that distance formula on all the sides, it will be the same. It is not a square though because it does not have four right angles, so this is a rhombus.

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Have a wonderful day and keep on learning!

8 0

3 years ago

Show your work with x box method

pashok25 [27]

Yessssssssssssssssss

7 0

3 years ago