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Olenka [21]
3 years ago
8

Simplify the ratio , pls I need help ASAP!!

Mathematics
2 answers:
anastassius [24]3 years ago
7 0

Answer:

10:11

Step-by-step explanation:

IRISSAK [1]3 years ago
6 0

Answer:

its option b

9:10

..........................

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wite the answer to the problem below, using the correct number of significant figures 18.432+ 5.55 + 19.3
Marysya12 [62]
18.432 + 5.55 + 19.3 
We can notice that the last number in the process of addition is 19.3 | with 1 decimal place value / 3 significant figures |
It is better to change all values to 1 decimal place value / 3 significant figures
18.432 → 18.4
5.55 → 5.6 ( 5 rounds up )
19.3 → 19.3
Let us add:
18.4 + 5.6 + 19.3  = 43.3
If you add the original numbers, you will get
18.432 + 5.55 + 19.3 = 43.282
43.282 ≈ 43.3
8 0
3 years ago
Find the 12th term of geometric 5, 15, 45
lesantik [10]

Answer:

An=a1*r^n-1  n = 12  Ratio=3  A1=5    Answer =885735

Step-by-step explanation:

hope this helps  :)

7 0
3 years ago
Use elimination to solve the system 3x+5y=16 and 8x-y=28 for x?
Mekhanik [1.2K]

Answer:

x = 156/43

Step-by-step explanation:

3x + 5y = 16

8x - y = 28

3x + 5y = 16

5(8x - y = 28)

3x + 5y = 16

40x - 5y = 140

43x = 156

x = 156/43

*sigh, an ugly answer*

5 0
3 years ago
Help me please boyos
luda_lava [24]

Answer:

para wala nang makasagot

3 0
3 years ago
Read 2 more answers
Graph the solution to the system<br> -x+y&gt;2<br> X+2y&gt;2
mariarad [96]

Answer:

\begin{bmatrix}y>2+x\\ y>\frac{2-x}{2}\end{bmatrix}

Step-by-step explanation:

<u>Step 1: Solve the system of equations</u>

<u />\begin{bmatrix}-x+y>2\\ x+2y>2\end{bmatrix}<u />

Isolate y for -x+y>2

\mathrm{Add\:}x\mathrm{\:to\:both\:sides}

-x+y+x>2+x

y>2+x

Isolate y for x+2y>2

\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}

2y>2-x

\mathrm{Divide\:both\:sides\:by\:}2

\frac{2y}{2}>\frac{2}{2}-\frac{x}{2}

y>\frac{2-x}{2}

\bold{=\begin{bmatrix}y>2+x\\ y>\frac{2-x}{2}\end{bmatrix}}

<u>Step 2: Graph</u>

Instructions for graphing:

\mathrm{1.\:Graph\:each\:inequality\:separately.}

\mathrm{2.\:Choose\:a\:test\:point\:to\:determine\:which\:side\:of\:the\:line\:needs\:to\:be\:shaded.}

3. \mathrm{The\:solution\:to\:the\:system\:will\:be\:the\:area\:where\:the\:shadings\:}\\\mathrm{from\:each\:inequality\:overlap\:one\:another.}

When done, it should look like this:

6 0
3 years ago
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