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3 years ago

Is my answer right orrr

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

4 0

Answer:

$\huge\boxed{135\° }$

Step-by-step explanation:

m∠2 = 135° (Corresponding Angles)

m∠5 = 135° (Vertical Angles)

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AnonymousHelper1807</h3>

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Enter numbers to write 13,000 in scientific notation.<br><br><br><br><br> 13,000=1.3 × <br> =1.3×10

TiliK225 [7]

Answer:

104 = 10,000 x 1.3 = 13,000

Step-by-step explanation:

step 1

To find a, take the number and move a decimal place to the right one position.

Original Number: 13,000

New Number: 1.3000

Step 2

Now, to find b, count how many places to the right of the decimal.

New Number: 1 . 3 0 0 0

Decimal Count: 1 2 3 4

There are 4 places to the right of the decimal place.

Step 3

Building upon what we know above, we can now reconstruct the number into scientific notation.

Remember, the notation is: a x 10b

a = 1.3 (Please notice any zeroes on the end have been removed)

b = 4

Now the whole thing:

1.3 x 104

Step 4

Check your work:

104 = 10,000 x 1.3 = 13,000

4 0

3 years ago

Using principle of mathematical induction prove that 6^-1 divisble by 5 .

salantis [7]

I suppose the claim is $5 \mid 6^n - 1$ for $n\in\Bbb N$ .

When $n=1$ , we have $6^1 - 1 = 6 - 1 = 5$ , and of course 5 divides 5.

Assume the claim holds for $n=k$ , that $5 \mid 6^k - 1$ . We want to use this to show it holds for $n=k+1$ , that $5 \mid 6^{k+1} - 1$ .

We have

$6^{k+1} - 1 = \left(6^{k+1} - 6\right) + \left(6 - 1) = 6\left(6^k - 1\right) + 5$

Since $5 \mid 6^k - 1$ , we can write $6^k - 1 = 5\ell$ for some integer $\ell$ . Then

$6^{k+1} - 1 = 6\cdot5\ell + 5 = 5(6\ell + 1)$

which is clearly divisible by 5. QED

5 0

1 year ago

<img src="https://tex.z-dn.net/?f=%20%5Chuge%20%5Cboxed%20%7B%5Cmathfrak%7B%20%5Cred%7Bquestion%20%3A%20%7D%7D%7D" id="TexFormul

ioda

60 is your answer

i hope it helps

have a nice day

#Captainpower

6 0

3 years ago

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What is the value of the expression i 0 × i 1 × i 2 × i 3 × i 4?

astraxan [27]

ANSWER

The value of the expression is
$- 1$

EXPLANATION

Method 1: Rewrite as product of
${i}^{2}$

The expression given to us is,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4}$

We use the fact that
${i}^{2} = - 1$
to simplify the above expression.

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0} \times {i}^{1} \times {i}^{3} \times {i}^{2} \times {i}^{4}$

This implies,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2}$

We substitute to obtain,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = 1\times - 1 \times - 1 \times - 1\times - 1 \times - 1$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = 1\times 1 \times 1 \times - 1 = - 1$

Method 2: Use indices to solve.

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0 + 1 + 2 + 3 + 4}$

This implies that,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{10}$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = ( {{i}^{2}} )^{5}$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = ( - 1 )^{5} = - 1$

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3 years ago

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Write the expression as a single logarithm. <br> pls help me with this problem

DanielleElmas [232]

Answer:

in logarithm + works as a * for numbers with the same bases.

Step-by-step explanation:

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3 years ago

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