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2 years ago

What is the area of ivory’s copy

Mathematics

1 answer:

ivolga24 [154]2 years ago

3 0

Answer:

12 units

Step-by-step explanation:

formula to get the scale

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I just need help for one question

olganol [36]

(2,1) is the answer.

6 0

3 years ago

Find the slope of the line that passes through (-6,1) and (4,-3)

Aleonysh [2.5K]

$Slope\ formula=\frac{rise}{run}=\frac{y_2-y_1}{x_2-x_1}=\\\\\frac{-3-1}{4-(-6)}=\\\\\frac{-4}{10}=\\\\{\boxed{Slope=-\frac{2}{5}}$

7 0

2 years ago

PLEASE HELP!!!

vovikov84 [41]

Answer:

Both the parts of this question require the use of the "Intersecting Secant-Tangent Theorem".

Part A

The definition of the Intersecting Secant-Tangent Theorem is:

"If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment."

This, when applied to our case becomes, "The length of the secant RT, times its external segment, ST, equals the square of the tangent segment TU".

Mathematically, it can be written as:

Part B

It is given that RT = 9 in. and ST = 4 in. Thus, it is definitely possible to find the value of the length TU and it can be found using the Intersecting Secant-Tangent Theorem as:

Thus,

Thus the length of TU=6 inches

6 0

2 years ago

A one-year-old car is now worth AED 12 000.

masha68 [24]

Answer:

AED 15,000

Step-by-step explanation:

Depreciation is a reduction in the value or worth of an asset as a result of use.

Given that the car is depreciated by 20%, it means that the value of the car after the application of depreciation is the result of the original price of the car less the amount of depreciation which has been given as 20% of the original price.

Let the original price of the car (its price when it was new) be T then

T - 0.2T = 12,000

0.8T = 12,000

T = 12,000/0.8

= AED 15,000

3 0

3 years ago

I need some help with math

Irina18 [472]

Answer:

$(y-7)^2+(x-2)^2=16$

and

$(x+2)^2+(y-15)^2 = 9$

Step-by-step explanation:

The standard equation of a circle is $(x-h)^2+(y-k)^2=r^2$ where the coordinate (h,k) is the center of the circle.

Second Problem:

We can start with the second problem which uses this info very easily.
(h,k) in this problem is (-2,15) simply plug these into the equation. $(x--2)^2+(y-15)^2=r^2$ .
We can also add the radius 3 and square it so it becomes 9. The equation.
This simplifies to $(x+2)^2+(y-15)^2 = 9$ .

First Problem:

The first problem takes a different approach it is not in standard form. But we can convert it to standard form by completing the square.
$y^2-14y+x^2-4x+37=0$ first subtract 37 from both sides so the equation is now $y^2-14y+x^2-4x=-37$ .
$y^2-14y+x^2-4x+37=0$ by adding $(-\frac{b}{2a} )^2$ to both the x and y portions of this equation you can complete the squares. $(-\frac{b}{2a})^2=(-\frac{-14}{2(1)})^2$ and $(-\frac{-4}{2(1)})^2$ which equals 49 and 4.
Add 49 and 4 to both sides and the equation is now: $y^2-14y+49+x^2-4x+4=-37+49+4$ You can simplify the y and x portions of the equations into the perfect squares or factored form $(y-7)^2$ and $(x-2)^2$ .
Finally put the whole thing together. $(y-7)^2+(x-2)^2=16$ .

I hope this helps!

5 0

2 years ago