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Katen [24]
2 years ago
11

11 3/8 + 2 1/2 pleaseee help its missing

Mathematics
1 answer:
Rasek [7]2 years ago
7 0

Answer:

The answer is 13 7/8

Step-by-step explanation:

First, you must add 2 to 11 3/8, which gives you 13 3/8. Then you change 1/2 to 4/8. Add 4/8 to 13 3/8 equals 13 7/8.

You might be interested in
The first term of an arithmetic sequence is -5, and the tenth term is 13. Find the common difference.
zzz [600]
Hello,

u_{1} =-5\\
 u_{2} =-5+d\\
 u_{3} =-5+2d\\
 u_{4} =-5+3d\\
...\\
 u_{10} =-5+9d=13\\
==\ \textgreater \  9d=18 \\
==\ \textgreater \  d=2


6 0
3 years ago
How many 4 digit numbers are there with the sum of the four digits equal to 34
Masteriza [31]

Answer:

Step-by-step explanation:

THE LARGEST 4 DIGIT NUMBER IS 9999 WHICH HAS SUM OF DIGITS 9+9+9+9 =36 SO TO HAVE A SUM OF DIGITS 34 ,WE HAVE TO TAKE 2 AWAY THERE ARE TWO WAYS TO DO THAT IS

1) TAKE 2 AWAY FROM 1 OF THE 9"S AND THE SET OF THE DIGITS (7,9,9,9)

2)TAKE 1 AWAY FROM 2 OF THE 9"S AND THE SET OF DIGITS (8,8,9,9)

THERE ARE 4 ARRANGEMENTS OF THE DIGITS (7,9,9,9) BECAUSE THERE ARE FOUR POSITIONS FOR THE DIGITS THOUSANDS,HUNDREDES,TENS AND UNITS SO WE CAN PICK A P[OSITION FOR THE 7 IN (4,1) = FOUR WAYS ( THERE ARE 7,9,9,9, 9,7,9,9, 9,9,7,9, 9,9,9,7

THERE ARE SIX ARRANGEMENTS OF THE DIGITS (8,8,9,9, BECAUSE THERE ARE FOUR POSITIONF FOR THE DIGITS THOUSANDS , HUNDRED , TENS AND UNITS SO WE CAN PICK A POSITIONS FOR THE 28"S IN (4,2) = 6 WAYS THEY ARE 8,8,9,9, 8,9,8,9, 8,9,9,8, SO THAT 4+6 OR 10 WAYS

3 0
3 years ago
T-5 &gt;3(-2+t) <br> please help me
Vladimir79 [104]

Answer:

t < 1/2

Step-by-step explanation:

Isolate the <u>v</u><u>a</u><u>r</u><u>i</u><u>a</u><u>b</u><u>l</u><u>e</u> by dividing each side by <u>f</u><u>a</u><u>c</u><u>t</u><u>o</u><u>r</u><u>s</u> that don't contain the <u>v</u><u>a</u><u>r</u><u>i</u><u>a</u><u>b</u><u>l</u><u>e</u>.

Inequality Form: t < 1/2

Interval Notation: ( -infinity, 1/2)

3 0
2 years ago
Find x so that the points (x,x+1), (x+2,x+3) and (x+3,2x+4) form a right-angled triangle.
azamat

Let <em>a</em>, <em>b</em>, and <em>c</em> be vectors each starting at the origin and terminating at the points (<em>x</em>, <em>x</em> + 1), (<em>x</em> + 2, <em>x</em> + 3), and (<em>x</em> + 3, 2<em>x</em> + 4), respectively.

Then the vectors <em>a</em> - <em>b</em>, <em>a</em> - <em>c</em>, and <em>b</em> - <em>c</em> are vectors that point in directions parallel to each of the legs formed by the triangle with these points as its vertices.

If this triangle is to contain a right angle, then exactly one of these pairs of vectors must be orthogonal. In other words, one of the following must be true:

(<em>a</em> - <em>b</em>) • (<em>a</em> - <em>c</em>) = 0

<em>or</em>

(<em>a</em> - <em>b</em>) • (<em>b</em> - <em>c</em>) = 0

<em>or</em>

(<em>a</em> - <em>c</em>) • (<em>b</em> - <em>c</em>) = 0

We have

<em>a</em> - <em>b</em> = (<em>x</em>, <em>x</em> + 1) - (<em>x</em> + 2, <em>x</em> + 3) = (-2, -2)

<em>a</em> - <em>c</em> = (<em>x</em>, <em>x</em> + 1) - (<em>x</em> + 3, 2<em>x</em> + 4) = (-3, -<em>x</em> - 3)

<em>b</em> - <em>c</em> = (<em>x</em> + 2, <em>x</em> + 3) - (<em>x</em> + 3, 2<em>x</em> + 4) = (-1, -<em>x</em> - 1)

Case 1: If (<em>a</em> - <em>b</em>) • (<em>a</em> - <em>c</em>) = 0, then

(-2, -2) • (-3, -<em>x</em> - 3) = (-2)×(-3) + (-2)×(-<em>x</em> - 3) = 2<em>x</em> + 12 = 0   ==>   <em>x</em> = -6

which would make <em>a</em> - <em>c</em> = (-3, 3) and <em>b</em> - <em>c</em> = (-1, 5), and their dot product is not zero. Then the triangles vertices are at the points (-6, -5), (-4, -3), and (-3, -8).

Case 2: If (<em>a</em> - <em>b</em>) • (<em>b</em> - <em>c</em>) = 0, then

(-2, -2) • (-1, -<em>x</em> - 1) = (-2)×(-1) + (-2)×(-<em>x</em> - 1) = 2<em>x</em> + 4 = 0   ==>   <em>x</em> = -2

which would make <em>a</em> - <em>c</em> = (-3, -1) and <em>b</em> = (-1, 1), and their dot product is also not zero. The vertices are the points (-2, -1), (0, 1), and (1, 0).

Case 3: If (<em>a</em> - <em>c</em>) • (<em>b</em> - <em>c</em>) = 0, then

(-3, -<em>x</em> - 3) • (-1, -<em>x</em> - 1) = (-3)×(-1) + (-<em>x</em> - 3)×(-<em>x</em> - 1) = <em>x</em> ² + 4<em>x</em> + 6 = 0

but the solutions to <em>x</em> here are non-real, so we throw out this case.

So there are two possible values of <em>x</em> that make a right triangle, <em>x</em> = -6 and <em>x</em> = -2.

3 0
3 years ago
What is the LEAST possible integer value and the greatest
vagabundo [1.1K]

Answer:

hey mate I guess ur question is incomplete......after greatest what is it? plz check

8 0
3 years ago
Read 2 more answers
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