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4 years ago

In circle D, which is a secant? A. EF B. DC C. AB D. GH

Mathematics

1 answer:

Drupady [299]4 years ago

7 0

I think it’s D) GH, I’m sorry if it’s wrong

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Whoever answers will get 60 points. Pls answer based off of the answer choices.

AnnyKZ [126]

Hello from MrBillDoesMath!

Answer:

Discussion:

$350 at 6% = => as "percent" means per 100

350 * (6/100) = => as 350 * 6 = 2100

2100/100 =

Thank you,

MrB

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3 years ago

Describe two methods for finding the sale price of an items that is discounted 30%

madam [21]

you could write an equation or make a table.

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4 years ago

Emma says that Function A has a greater initial value. Is Emma correct? Justify your response. Function A Function B Alternative

netineya [11]

Answer: B

Step-by-step explanation: Function A starts at 0,2 and function B starts at 2,2

6 0

2 years ago

Read 2 more answers

Find the measure of arc DF (with an explanation of how to solve please!)

allochka39001 [22]

Answer:

mDF = 90°

(Assuming arc mCE is 5x + 10)

Step-by-step explanation:

When we have two chords crossing in a circle, there is a property where the angle in the cross point is equal half the sum of the corresponding arcs.

So in this case, we have:

70° = (mCE + mDF)/2

Assuming mCE is 5x + 10, we have:

70 = (5x + 10 + 11x + 2)/2

70 = (16x + 12)/2

70 = 8x + 6

8x = 64

x = 8

So the arc mDF is:

mDF = 11x + 2 = 11 * 8 + 2 = 90°

4 0

3 years ago

Use induction to show that 12 + 22 + 32 + ... + n2 = n(n+1)(2n+1)/6, for all n > 1.

dlinn [17]

Answer with Step-by-step explanation:

Let $P(n)=1^2+2^2+3^2+.....+n^2=\frac{n(n+1)(2n+1)}{6}$

Substitute n=2

Then $P(2)=1+2^2=5$

$P(2)=\frac{2(2+1)(4+1)}{6}=5$

Hence, P(n) is true for n=2

Suppose that P(n) is true for n=k >1

$P(k)=1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$

Now, we shall prove that p(n) is true for n=k+1

$P(k+1)=1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS

$P(k+1)=1^2+2^2+3^2+.....+k^2+(k+1)^2$

Substitute the value of P(k)

$P(k+)=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$

$P(k+1)=(k+1)(\frac{k(2k+1}{6})+k+1)$

$P(k+1)=(k+1)(\frac{2k^2+k+6k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+7k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+4k+3k+6}{6})$

$P(k+1)=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS=RHS

Hence, P(n) is true for all n >1.

Hence, proved

4 0

3 years ago