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Citrus2011 [14]
3 years ago
13

Solve for x can't remember how to do this for a circle and triangle through it.

Mathematics
2 answers:
Gala2k [10]3 years ago
6 0

Answer:

x = 2

Step-by-step explanation:

For secants that intersect at an external point, the product of the external length and total length is a constant.

... 3·(3 +5) = 4·(4 +x) . . . . . write the two products, and set them equal

... 24 = 16 +4x . . . . . . . . . . eliminate parentheses

... 8 = 4x . . . . . . . . . . . . . . .subtract 16

... 2 = x . . . . . . . . . . . . . . . . divide by 4

vivado [14]3 years ago
5 0

Answer:

x=2

Step-by-step explanation:

(Refer the attachment to find ABCD and O points)

When two secant lines AD and CB intersects at a point outside the circle "O"

then,

OA*OD=OB*OC

Now we can substitute the values and find x.

3*8=4*(4+x)

24=16+4x

8=4x

x=2



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The length of a rectangular garden is 7 meters longer than its width. if its perimeter is ( 12 x + 14) meters, find its length i
cluponka [151]
L=7+w
P=(12x+14)=2(6x+7)
l+w= P/2 = 2(6x+7)/2= 6x+7
l+w=6x+7
2l = 6x+7
l = (6x+7)/2
l=7+w
-> w= l-7
= [(6x+7)/2] -7
= (6x+7-14)/2
= (6x-7)/2

7 0
3 years ago
Someone please help!!
Nastasia [14]
1.46 is what google is saying just convert that to a fraction
4 0
3 years ago
Convert 7 kilometers into miles
ladessa [460]
4.3496 miles.

HTH (Hope This Helps)
6 0
3 years ago
Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10
zvonat [6]

The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0
2 years ago
HELPPP!!
katrin [286]

Answer:

It is (-3-(-6)+5)^4=4096

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
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