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Zarrin [17]
3 years ago
14

Can someone help me with this ASAP please I’m being timed

Mathematics
1 answer:
SOVA2 [1]3 years ago
3 0
Top right goes down five then up 5
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Explain why it works to break apart a number by place values to multiply
IrinaK [193]
Idononoooont know boi
3 0
3 years ago
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What is the measure of < DBC
notsponge [240]

Answer:

∠D B C = 41°

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

Given ∠ABC = 90°

 In diagram ∠D B C + ∠A B D = 90°

                       6 x+5 + 8 x +1 = 90

                       14 x + 6 = 90

                      14 x = 90 -6

                      14 x = 84

                    <em>     x = 6</em>

<u><em>Step(ii):-</em></u>

∠D B C = 6 x + 5 = 6 (6) +5 = 36 +5 = 41

<em>∠D B C = 41°</em>

<em>  ∠A B D = 8(6) +1  = 49°</em>

<em>∠D B C + ∠A B D =  41° + 49° = 90°</em>

3 0
2 years ago
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Convert 149g/cm to kilograms per meter
marishachu [46]
Set it up kind of like a picket fence. Put 149 g over one cm and multiply that by 1kg over 1000g then multiply that by 1 cm over .01 m. The answer is 14.9 kg/m
5 0
2 years ago
What is the result of adding -3x2 - 5x + 1 and 8x2 - 2x - 9
Fudgin [204]

(-3x^2 - 5x + 1) + (8x^2 - 2x - 9)

1st remove the parenthesis

-3x^2 - 5x + 1 + 8x^2 - 2x - 9

then combine like terms

5x^2 - 7x - 8 is the final answer

6 0
2 years ago
*Find the formula of the series 1+2²+3²+4²+....+n²*<br>​
Sophie [7]

The formula is \frac{n(n+1)(2n+1)}{6}

What are series?

In mathematics, we can describe a series as adding infinitely many numbers or quantities to a given starting number or amount.

We will find the formula as shown as below:

Let S=1+2^2+3^2+4^2+................+n^2

We know (n+1)^3=n^3+3n^2+3n+1

(1+1)^3=1^3+3(1)^2+3(1)+1

(2+1)^3=2^3+3(2)^2+3(2)+1

(3+1)^3=3^3+3(3)^2+3(3)+1

.

.

(n+1)^3=n^3+3(n)^2+3(n)+1

On adding

2^3+3^3+4^3......(n+1)^3=(1^3+2^3+3^3+.....+n^3)+3(1^2+2^2+.....+n^2)+3(1+2+3....n)+(1+1+1+....+1)

2^3+3^3+4^3......(n+1)^3-(1^3+2^3+3^3+.....+n^3)=3S+\frac{3n(n+1)}{2}+(1+1+1+....+1)

(n+1)^3-1^3=3S+\frac{3n(n+1)}{2} +n

n^3+3(n)^2+3(n)+1-1=3S+\frac{3n(n+1)}{2} +n

2n^3+6n^2+6n=6S+3n^2+3n+2n

6S=2n^3+3n^2+n

6S=2n^2(n+1)+n(n+1)

6S=(n+1)(2n^2+n)

6S=n(n+1)(2n+1)

S=\frac{n(n+1)(2n+1)}{6}

Hence, the formula is \frac{n(n+1)(2n+1)}{6}

Learn more about Series here:

brainly.com/question/24643676

#SPJ1

3 0
2 years ago
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