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fiasKO [112]
3 years ago
6

Please help i’ll give brainliest!!! thanks

Mathematics
2 answers:
Brut [27]3 years ago
7 0
I think school is designed to help poor because it’s for free
dalvyx [7]3 years ago
3 0

Answer:

Tang instituted government - funded <u>social programs </u>, which were designed to help the poor .

hope it is helpful to you.

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Is 5/10 and 8/18 proportional
salantis [7]
5/10 = 1/2

8/18 = 4/9

1/2 > 4/9

They're not proportional.
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Which unit is the largest? centiliter liter milliter kiloliter​
V125BC [204]

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kiloliter

Step-by-step explanation:

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What fraction of 1/2 is 1/3 ? Draw a tape diagram to represent the question
sergeinik [125]

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1/1.5

Step-by-step explanation:

IM JUST SMART

7 0
3 years ago
How to write (0,8] as an inequality
larisa86 [58]
Answer: 0 \ \textless \  x \le 8

What is given to you is interval notation. We have the interval start at 0 and end at 8. The value 0 is not included in the interval as indicated by the parenthesis. So we go with a "less than" sign (instead of a "less than or equal to" sign)

The value 8 is included since a square bracket is used here. The use of "or equal to" is needed to make sure the endpoint 8 is included.

So x can be any number between 0 and 8. It can't be 0 but it can be 8. 
3 0
3 years ago
What is the result of substituting for y in the bottom equation<br> y=x+3<br> y=x^2+2x-4
8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

<em><u>Solution:</u></em>

Given that,

y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4  ----- eqn 2

<em><u>We have to substitute eqn 1 in eqn 2</u></em>

x + 3 = x^2 + 2x - 4

\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0

\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}

x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}

x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925

Also\\\\x = \frac{ -1 - 5.385 }{2}\\\\x = -3.1925

Substitute x = 2.1925 in eqn 1

y = 2.1925 + 3

y = 5.1925

Substitute x = -3.1925 in eqn 1

y = -3.1925 + 3

y = -0.1925

Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

6 0
4 years ago
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