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2 years ago

PLEASE HELP ME WITH THIS QUESTION 6

Mathematics

1 answer:

nata0808 [166]2 years ago

5 0

Answer:

Step-by-step explanation:

Assuming this is a trapezoid,

$\frac{29+WZ}{2}=37 \\ \\ 29+WZ=74 \\ \\ WZ=45$

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PLEASE HELP!! I WILL NAME THE BRAINLEST!!!!

Brrunno [24]

Where is the table at

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3 years ago

What is the value of f(3)?

Anni [7]

Answer:

-12

Step-by-step explanation:

Note that when x < 6, you will use the expression -4x

f(x) = -4x

Plug in 3 for x

f(3) = -4(3)

Simplify. Multiply

f(3) = -4(3)

f(3) = -12

-12 is your answer

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3 years ago

Solve for x.<br> 144°<br> 18 degrees<br> 30 degrees<br> 48 degrees<br> 120 degrees

disa [49]

Answer:

Step-by-step explanation:

you have a rotation angle of 360 °, take out the known values (360 ° -90 ° -144 ° = 126), 144 ° is part of a flat angle of 180 °, from 180 ° remove 144 and you have 36 °, they are opposite angles therefore the same. therefore 360 - 144 - 90 - 36 - 36 = 54.

54 is 3x, so x = 18 ° (54 : 3 = 18)

7 0

2 years ago

Help please! this is my third time posting this!! T.T

ArbitrLikvidat [17]

Answer:

see below

Step-by-step explanation:

24 - 3 x = -27

Subtract 24 from each side

24-24 - 3 x = -27-24

-3x =-51

Divide each side by -3

-3x/-3 = -51/-3

x = 17

Check

24 -3(17) = -27

24 -51 = -27

-27 = -27

8 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