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SashulF [63]
2 years ago
13

May anyone help me with this?

Mathematics
1 answer:
Verdich [7]2 years ago
5 0

Answer:

31

Step-by-step explanation:

You might be interested in
What is the sum of all positive integers less than 300 which are divisible by 3
drek231 [11]

Answer:

14,850

Step-by-step explanation:

You need the sum of

3 + 6 + 9 + 12 + ... + 294 + 297

Factor out a 3 from the sum

3 + 6 + 9 + 12 + ... + 294 + 297 = 3(1 + 2 + 3 + 4 + ... + 98 + 99)

You need to add all integers from 1 to 99 and multiply by 3.

The sum of all consecutive integers from 1 to n is:

[n(n + 1)]/2

The sum of all consecutive integers from 1 to 99 is

[99(99 + 1)]/2

The sum you need is 3 * [99(99 + 1)]/2

3 + 6 + 9 + 12 + ... + 294 + 297 =

= 3 * [99(99 + 1)]/2

= 3 * [99(100)]/2

= 3 * 9900/2

= 14,850

4 0
2 years ago
PLEASE HELP!!<br> just look at the picture
arlik [135]

Answer:

its the  math that solve

Step-by-step explanation:

4 0
2 years ago
Can someone help me with this problem please?
MrRissso [65]
If you're going to ask for help, you may as well ask for help from an app designed for the purpose. This app is available for both Android and iOS devices. My TI-83/84 calculator also has a triangle solver app.

(A) There is only one possible solution

b = 16.3
C = 26°
B = 87°

_____
If you want to do this yourself, you can use the Law of Sines to find angle C.
.. C = arcsin(7.1/15*sin(67°)) ≈ 25.8302° . . . . don't round intermediate values
Then angle B is 180° -67° -angle C = 87.1698°
and side b is found from the Law of Sines:
.. b = 15*sin(C)/sin(A) ≈ 16.2755

6 0
3 years ago
Could anyone help answer the question in the photo?
Kipish [7]

Answer:

n = -3/10

Step-by-step explanation:

Given the expression \frac{\sqrt[5]{b} }{\sqrt[]{b} }

\frac{\sqrt[5]{b} }{\sqrt[]{b} } \\= \frac{b^{1/5}}{b^{1/2}} \\= b^{1/5-1/2}\\= b ^{2-5/10}\\= b^{-3/10}\\Compare \ b^n \ with \  b^{-3/10}\\\\n = -3/10

7 0
2 years ago
What is the missing factor of 10 Times what =200
Assoli18 [71]
The missing factor is 20. 10 x 20 = 200
8 0
2 years ago
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