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Anna71 [15]
3 years ago
12

WILL GIVE BRAINLIEST!!!!!!

Mathematics
1 answer:
Crank3 years ago
7 0

Answer:

17.738

Step-by-step explanation:

to find x we’ll use sine

sin69=x/19

(remember that sine = opposite/hypotenuse)

so x = 19sin69

when typing this in your calculator you might need to change from radian to degree mode

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How to solve. C=3p+5 solve for c
Elenna [48]

Answer:

1. Slope =0.667/2.000=0.333

2.c-intercept = 5/1=5.00000

3.p-intercept=5/-3= -1.66667

6 0
3 years ago
Cuanto le falta a 425,000 para llegar a 500,000<br><br> Ayúdenme por favor
Sauron [17]

Answer:

375,000 no se sisera esa

4 0
3 years ago
Read 2 more answers
A student downloaded a new game on her phone that was recommended by her friend. She wanted to show her friend how good
anygoal [31]

Answer:

a. 587

b. 543

Step-by-step explanation:

a. Mean = sum / number of numbers

To find the sum, we can add the numbers up.

432 + 543 + 348 + 788 + 823 = 2934

There are 5 numbers. Our mean is thus

2934/5 = 586.8, or 587 when rounding up. We round up because we have a .8 at the end, and 8 ≥ 5

b. To find the median, we first must order the numbers. The numbers ordered are this:

348 ,  432 , 543 , 788 , 823

To find the median, we can cross off one number on each side until we are left with one or two numbers.

348 ,  432 , 543 , 788 , 823

cross out one number from each endpoint

432 , 543 , 788

cross out one number from each endpoint

543

7 0
2 years ago
The sum of a number and eight is equal to 14 what is the number
musickatia [10]

Answer:

the number is 6

Step-by-step explanation:

14-8=6

good luck!

3 0
3 years ago
Expand the given power using the Binomial Theorem. (10k – m)5
agasfer [191]

Answer:

(10k - m)^{5}=100000k-50000k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}

Step-by-step explanation:

* Lets explain how to solve the problem

- The rule of expand the binomial is:

(a+b)^{n}=(a)^{n}+nC1(a)^{n-1}(b)+nC2(a)^{n-2}(b)^{2}+nC3(a)^{n-3}(b)^{3}+...............+(b)^{5}

∵ The binomial is (10k-m)^{5}

∴ a = 10k , b = -m and n = 5

∴ (10k-m)^{5}=(10k)^{5}+5C1(10k)^{4}(-m)+5C2(10k)^{3}(-m)^{2}+5C3(10k)^{2}(-m)^{3}+5C4(10k)^{1} (-m)^{4}+5C5(10k)^{0}(-m)^{5}

∵ 5C1 = 5

∵ 5C2 = 10

∵ 5C3 = 10

∵ 5C4 = 5

∵ 5C5 = 1

∴ (10k-m)^{5}=100000k^{5}+(5)(10000)k^{4}(-m)+(10)(1000)k^{3}(m^{2})+(10)(100)k^{2}(-m^{3})+5(10k)^{1} (m^{4})+(10k)^{0}(-m^{5})

∴ (10k-m)^{5}=100000k^{5}-50000)k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}

5 0
3 years ago
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