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3 years ago

Geometry math question no Guessing and Please show work

Mathematics

2 answers:

Annette [7]3 years ago

7 0

If the quadrants are disposed like this pattern:

I. II.

III. IV.

And the y- achsis is in middle of I./II. and III./IV.

then parallel of IV. is III

sergiy2304 [10]3 years ago

6 0

the answer is D.

I dont know how to add work for a question like this, however, I know how I'd show work....

Draw a graph and stick a point in the IV quadrant and label it "original". Then next to it, draw another graph and stick a point in the III quadrant and label it "reflection"

TIP: if you do that, make sure the points are in the same location in each quadrant

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A used car has a value of 15,250 when it’s purchased in 2012. The value of the car decreases at a rate of 7.5% per year

kotegsom [21]

p=15250

r=7.5/100=0.075

y = 15250*(1 - 0.075)^x

after 8 years it would be

x = 8

y = 15250*(1 - 0.075)^8

y = $8,173.42

after 8 years the value of the car is going to be $8,173.42

8 0

3 years ago

Use the geometric mean to find the 7th term in a geometric sequence if the 6th term is 12 and the 8th term is 27. 20 19.5 18 17

Lesechka [4]

A₆ =12

a₇ = ?? = x

a₈ 27

In this case x is the geometric mean & as per the formula
geometric mean of 2 numbers =√(a₆ * a₈) = √(12*27)
& x=18.
Proof:
Find the common ratio between a₇ & a₆ ==> = ( x) / 12
now find the common ration between a₈ & a₇ ===> =27 / (x)
x/12 = 27/x ===> x² = 12*27 ==> x² = 324 & x =√324 = 18

7 0

4 years ago

The graph of a bicycle ramp contains the points (-6,0) and (3,3). What is the slope of the ramp?

olasank [31]

The slope of the ramp would be 5/9

7 0

3 years ago

Can someone help me with 10-15 plz

polet [3.4K]

Answer:

it is in the picture

Step-by-step explanation:

you just find the coordinates

4 0

3 years ago

Given a rectangle with length of (2x+9)cm and width of (3x+1)cm.Two squares, each with sides x cm is removed from the rectangle.

il63 [147K]

Answer: The length is 13cm and the width is 7cm

Step-by-step explanation:

For a rectangle of length L and width W, the area is:

A = W*L

In this case we have:

L = (2*x + 9) cm

W=(3*x + 1) cm

Then the area of the rectangle is:

A = (2*x + 9)*(3*x + 1) cm^2

A = (6*x^2 + 2*x + 27*x + 9) cm^2

A = (6*x^2 + 29*x + 9) cm^2

now we remove two squares with sides of x cm

The area of each one of these squares is (x cm)*(x cm) = x^2 cm^2

Then the area of the figure will be:

area = (6*x^2 + 29*x + 9) cm^2 - (2*x^2 ) cm^2

area = (4*x^2 + 29*x + 9) cm^2

Now we know that the area of this shape is 83 cm^2, then we need to solve:

83 cm^2 = (4*x^2 + 29*x + 9) cm^2

0 = (4*x^2 + 29*x + 9) cm^2 - 83 cm^2

0 = (4*x^2 + 29*x - 74) cm^2

Then we need to solve:

0 = 4*x^2 + 29*x - 74

Here we can use Bhaskara's equation, the solutions of this equation are given by:

$x = \frac{-29 \pm \sqrt{29^2 - 4*4*(-74)} }{2*4} = \frac{-29 \pm 45}{8}$

Then the two solutions are:

x = (-29 - 45)/8 = -9.25 (for how the length and width are defined, we can not have x as a negative number, then this solution can be discarded).

The other solution is:

x = (-29 + 45)/8 = 2

x = 2

Then the length and width of the rectangle are:

Length = (2*2 + 9)cm = 13 cm

Width = (3*2 + 1)cm = 7cm

4 0

3 years ago