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1 year ago

Isosceles △ABC (AC=BC) is inscribed in the circle k(O). Prove that the tangent to the circle at point C is parallel to AB .

Mathematics

1 answer:

timofeeve [1]1 year ago

5 0

Explanation:

Let M be the midpoint of AB. Then CM is the perpendicular bisector of AB. As such, center O is on CM, and OC is a radius (and CM). The tangent is perpendicular to that radius (and CM), so is parallel to AB, which is also perpendicular to CM.

If you need to go any further, you can show that triangles CMA and CMB are congruent, so (linear) angles CMA and CMB are congruent, hence both 90°.

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The number of bacteria in a petri dish on the first day was 113 cells. If the number of bacteria increase at a rate of 82% per d

Tresset [83]

Answer:

4107 cells

Step-by-step explanation:

From the question, we have the following values:

Day 1 : 113 cells

Number of cells increases by day by 82%

Hence,

Day 2

113 × 82% = 92.66cells

Hence, Total number of bacteria cells for Day 2 = 113 + 92.66 = 205.66cells

Day 3

205.66 × 82% = 168.6412 cells

Hence, Total number of bacteria cells for Day 3 = 168.6412 + 205.66 = 374.3012 cells

Day 4

374.3012 × 82% = 306.926984 cells

Hence, Total number of bacteria cells for Day 4 = 306.926984 + 374.3012 = 681.228184 cells

Day 5

681.228184 × 82% = 558.60711088 cells

Hence, Total number of bacteria cells for Day 5 = 558.60711088 + 681.228184 = 1239.8352949 cells

Day 6

1239.8352949 × 82% = 1016.6649418 cells

Hence, Total number of bacteria cells for Day 5 = 1016.6649418 + 1239.8352949 = 2256.5002367 cells

Day 7

2256.5002367 × 82% = 1850.3301941 cells

Hence, Total number of bacteria cells for Day 7 = 1850.3301941 + 2256.5002367 = 4106.8304308 cells

Approximately to nearest whole number, the total number of bacteria cells that would be present after 7 days = 4107 cells

3 0

1 year ago

Consider the following expression -2m(m+n-4)+5(-2m+2n)+n(m+4n-5) which of the following is an equivalent expression

velikii [3]

The answer is:

-2m²+4n²-mn-2m+5n

aka letter c in algebra nation

3 0

2 years ago

The volumes of two similar figures are 343 mm3 and 512 mm3. If the surface area of the larger figure is 192 mm2, what is the sur

Kobotan [32]

In geometry, similar figures are those whose ratios of the corresponding sides are equal and the corresponding angles are congruent. In relation to the volume, we determine first the cube roots of the given and find the ratio as shown below.

s1 / s2 = cube root of (512/343)
= 8/7
The square of this ratio is the ratio of the areas of the figure. If we let x be the area of the smaller figure then,
(8/7)^2 = 192 mm²/ x
The value of x from the equation is 147 mm².

The area therefore of the smaller figure is 147 mm².

3 0

2 years ago

Read 2 more answers

On an algebra test, the highest grade was 42 points higher than the lowest grade. The sum of the two grades was 138. Find the lo

miv72 [106K]

you could use this equation to help you solve it;

x + (x + 42) = 138

the first step is to combine like terms;

2x = 138 -42

2x = 96

X = 96/2

X = 48

we already solved for x now substitute it in the equation I gave you.

48 + (48 + 42) = 138

48 + 90 = 138

hope it helped...if you have any concerns just let me know:)

3 0

2 years ago

Plato explains that we know geometry by _________.

sasho [114]

<span>Plato explains that we know geometry by our gain knowledge through recollection. Our soul is what recollects this place hence we came where there exist unchanging truths. Delivered the theory of Forms, according to which the world people know by means of the senses is just an imitation of the eternal, pure, eternal, and fixed world of the Forms.</span>

7 0

2 years ago