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Kamila [148]
3 years ago
15

The sum of the three consecutive numbers is33 more than the smallest of the number. Find the numbers

Mathematics
1 answer:
DiKsa [7]3 years ago
5 0

Answer:

15 udhffjfjgjgjjgvjjgjgjgvjjgjc

You might be interested in
19. Given that angle BFC = 58°. Angle BXG = 48° and angle CBF 22 Find () 2BGX (iv) BCG (i) 2BGF (iv) ZBFG (ii) BCF H EE 58 48° D
RoseWind [281]

The unknown angles in the cyclic quadrilateral is as follows:

∠BGX = 74° (sum of angles in a triangle)

∠BGF = 180° (opposite angles of cyclic quadrilateral are supplementary)

∠BCF = 100°(sum of angles in a triangle)

∠BCG  = 26°

∠BFG = 22°

<h3>Cyclic Quadrilateral</h3>

A cyclic quadrilateral has all its angles equal to 360 degrees. The sum of angles in a cyclic quadrilateral is equals to 360 degrees.

Let's find the missing angles as follows:

∠BGX = 180 - 48 - 58 = 74° (sum of angles in a triangle)

∠BGF = 180 - 100 = 80° (opposite angles of cyclic quadrilateral are supplementary)

∠BCF = 180 - 22 - 58 = 100°(sum of angles in a triangle)

∠BCG = 100 - 74 = 26°

∠BFG ≅ CBF = 22°(alternate angles)

learn more on angles here: brainly.com/question/19430381

6 0
2 years ago
I am confused please help
klemol [59]

Answer:

volume = lenght x width x height

v= 12x20x14

v=3,360 in³

3 0
2 years ago
Will mark brainliest!!<br><br> picture included ^^^^<br><br> please and THANK YOU<br><br><br> ^^!!
Alisiya [41]

Answer:

shift the graph up 1 unit

Step-by-step explanation:

d is the vertical shift

+1 means we shift the graph up 1 unit

5 0
3 years ago
What has the sum of 20 but the difference is 8?
arlik [135]
X + y = 20---- 1
x - y = 8----- 2

From 1+2
2x = 28
x = 14
Substitute x in eq.2
14 -y = 8
y = 6

Ans : x= 14 , y= 6
6 0
3 years ago
Find the error with this proof and explain how it mat be corrected in order to clearly prove the equation.
enyata [817]

Answer:

Step-by-step explanation:

It is true that for any given odd integer, square of that integer will also be odd.

i.e if m is and odd integer then m^{2} is also odd.

In the given proof the expansion for (2k + 1)^{2} is incorrect.

By definition we know,

(a+b)^{2} = a^{2} + b^{2} + 2ab

∴ (2k + 1)^{2} = (2k)^{2} + 1^{2} + 2(2k)(1)\\(2k + 1)^{2} = 4k^{2} + 1 + 4k

Now, we know 4k^{2} and 4k will be even values

∴4k^{2} + 1 + 4k will be odd

hence (2k + 1)^{2} will be odd, which means m^{2} will be odd.

7 0
3 years ago
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