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3 years ago

I don’t know the answer

Mathematics

1 answer:

anastassius [24]3 years ago

7 0

I THINK IT IS B

B) D(X)= X-8x4

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The measure of the supplement of an angle is seven times as large as the measure of its complement find the measure of the angle

8090 [49]

Answer: original angle = 75°

Its supplement 105°

Its complement = 15°

Step-by-step explanation:

We know that,

Sum of supplementary angles is 180 degrees.
Sum of complementary angles is 90 degrees.

Let x be the original angle , then its supplement = 180° - x

its complement = 90° - x

As per given,

180° - x = 7(90° - x)

⇒ 180° - x = 630° - 7x

⇒ 7x- x = 630° -180°

⇒ 6 x = 450°

⇒ x = 75° [divide both sides by 6]

So, original angle = 75°

Its supplement = 180° - 75° = 105°

Its complement = 90° -75° = 15°

Hence, original angle = 75°

Its supplement 105°

Its complement = 15°

4 0

3 years ago

Which set of side lengths create a right triangle ?

I am Lyosha [343]

Answer:

(8, 15 and 17)

Step-by-step explanation:

$\sqrt{10^{2} +28^{2} } \neq 29$

$\sqrt{6^{2} +8^{2} } \neq 13$

$\sqrt{7^{2} +12^{2} } \neq 14$

$\sqrt{8^{2} +15^{2} } =17$

3 0

3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

Which segment is a dilation of BC using A as the center of dilation and a scale factor of 2/3

BabaBlast [244]

Answer: 45

Step-by-step explanation:

5 0

2 years ago

Please can u tell me the answer and method

stich3 [128]

160/4=40 so each side is 40. Divide the perimeter by how many sides you have.

6 0

3 years ago