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3 years ago

What's 9+10 ? twenty -one

Mathematics

2 answers:

Leya [2.2K]3 years ago

8 0

Answer:

19 or 21

Step-by-step explanation:

slamgirl [31]3 years ago

4 0

Answer: 19!!!!!!!!!!!!!!!!!!!!!

Step-by-step explanation: thanks for the points

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Help with this question, please!

uranmaximum [27]

Answer:

52 ft

Step-by-step explanation:

For each chord, the product of segment lengths is a constant. We can find that constant as the product of the segment lengths of the bisected chord:

(24 ft)·(24 ft) = 576 ft^2

Then the missing segment length (DX) of the diameter chord is ...

DX·(36 ft) = 576 ft^2

DX = (576 ft^2)/(36 ft) = 16 ft

So, the total length of the diameter chord is ...

DX +XL = 16 ft + 36 ft

DL = 52 ft

_____

We know DL is a diameter because the perpendicular bisector of any chord intersects the center of the circle.

5 0

3 years ago

8/9 + (−5/ 6) divided by 1/6

rusak2 [61]

Answer:

-4 1/9

Step-by-step explanation:

5 0

3 years ago

5 x 100 = ? for easy points

zimovet [89]

Answer:

500! Just multiply 5 x 1 then add the remaining 2 zeros :)

Step-by-step explanation:

Thank you!

8 0

3 years ago

Read 2 more answers

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,

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

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

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

8 0

4 years ago

Read 2 more answers

Find the distance between each pair of points.

dangina [55]

Okay so make a coordinate plane, insert the numbers in the right place, then count what’s in between, you got this, I’ll totally help you if you didn’t get it.

8 0

4 years ago