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3 years ago

I need help can you help me

Mathematics

2 answers:

patriot [66]3 years ago

8 0

Answer:

2 : 3

Step-by-step explanation:

BRAINLIEST PLEASE

steposvetlana [31]3 years ago

3 0

The answer is: " 2 :5 " ; or, write as: " 2/5 " .

________________________________________________________

The ratio of 'girls' to 'all students' is: "2: 5 " ; or, write as: " 2/5 ".

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Explanation:

________________________________________________________

Given: The ratio of boys to girls is: " 3:2 " .

Problem: Find the ratio of "girls" to "all students:

________________________________________________________

Note: This ratio of "boys to girls", which is " 3 : 2 " ;

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→ can be expressed as " 3x: 2x" ;

in which the total number of students is: " 3x + 2x " = 5x " .

→ The total number of students is represented as: " 5x " .

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→ The ratio of "girls to boys" is: "2x : 3x" .

→ {that is; the "inverse" of the ratio of "boys to girls"} ;

→ {that is; the "inverse" of " 3x: 2x" } ; → which is: " 2x : 3x " .

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The ratio of "girls" to "all students" is: "2x : 5x " ; or " 2x/5x " ;

→ Both "x" values cancel ; {since: " x/x = 1 "} ;

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→ and we have the answer: " 2 :5 " ; or, write as: " 2/5 " .

_________________________________________________________

The ratio of 'girls' to 'all students' is: " 2 :5 " ; or, write as: " 2/5 ".

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3 years ago

Compare 2.3 and 2.03

Andru [333]

2.3= 2.03 so there for it is equal

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3 years ago

Simplify. -10m+2m^4-13m-20m^4

Ulleksa [173]

First one:

you can add -10m and -13m but you can't add -10m and 2m^4 becuase the powers aren't the same so

when adding the like terms
look at the:
powers, (x^3 adds with x^3)
placehloder letter (x adds with x and y adds with y and so on)

-10m+2m^4-13m-20m^4
powers: m^1 and M^4
placeholders: all m

add
-10m-13m+2m^4-20m^4
-23m-18m^4

second one:
when multiplying exponents, you add with like
so if you multipliy
x^2yz^3 times x^4y^2z^2 thne you would get x^6y^3z^5
when multiply with coeficients
2x^2yz^3 times 4x^4y^2z^2=8x^6y^3z^5
so using associative property a(bc)=(ab)c
2/3 times p^4 times y^3 times y^4 times s^5 times 6 times p^2 times s^3
group like terms

(2/3 times 6) times (p^4 times p^2) times (y^3 times y^4) times (s^5 times s^3)
(4) times (p^6) times (y^7) times (s^8)
4p^6y^7s^8

8 0

3 years ago

Solve the following quadratics. State the FACTORS AND SOLUTIONS. 1. 2x^2 - 7x + 3 2. 3x^2 + 7x +2

tekilochka [14]

Answer:

1. x = 3, 1/2 (solutions); (x - 3)(2x - 1) (factors)

2. x = -1/3, -2 (solutions); (3x + 1)(x + 2) (factors)

Step-by-step explanation:

1. 2x^2 - 7x + 3

To solve problem 1, you will need to identify your a, b, and c values in this quadratic function.

Since this problem is in standard form, it will be easy to identify these values. The standard form of a quadratic function is ax^2 + bx + c.

The a value is 2, the b value is -7, and the c value is 3 if we use our standard form and see which numbers are plugged into it.

Since we know that

a = 2
b = -7
c = 3

we can use the quadratic formula: $x = \frac{-b~\pm~\sqrt{b^2~-~4ac} }{2a}$

Substitute the a, b, and c values into the quadratic formula: $x=\frac{-(-7)\pm\sqrt{(-7)^2-4(2)(3)} }{2(2)}$

Now simplify using the laws of pemdas: $x=\frac{7\pm\sqrt{(49)-(24)} }{4}$

Simplify even further: $x=\frac{7\pm\sqrt{(25)} }{4} \rightarrow x=\frac{7\pm (5) }{4}$

Now split this equation into two equations to solve for x: $x=\frac{12 }{4} ~~and~~ x=\frac{2 }{4}$

12/4 can be simplified to 3, and 2/4 can be simplified to 1/2.

This means your solutions to problem 1 is 3, 1/2.

$\boxed {x=3,\frac{1}{2} }$

There is also another way to solve for the quadratic functions, and this was by factoring.

If you factor 2x^2 - 7x + 3 using the bottoms-up method, you will get (x - 3)(2x - 1).

After factoring, solving for the solutions is simple because all you have to do is set each factor to 0.

x - 3 = 0
2x - 1 = 0

After solving for x by adding 3 to both sides, or by adding 1 to both sides then dividing by 2, you will end up with the same solutions: x = 3 and x = 1/2.

2. 3x^2 + 7x + 2

To save time I'll be using the bottoms-up factoring method, but remember to refer back to problem 1 (quadratic formula) if you prefer that method.

Factor this quadratic function using the bottoms-up method. After factoring you will have (3x + 1)(x + 2). These are your factors.

Now to solve for x and find the solutions of the quadratic function, you will set both factors equal to 0.

3x + 1 = 0
x + 2 = 0

Solve.

First factor: 3x + 1 = 0

Subtract 1 from both sides.

3x = -1

Divide both sides by 3.

x = -1/3

Second factor: x + 2 = 0

Subtract 2 from both sides.

x = -2

Your solutions are x = -1/3 and x = -2.

$\boxed {x = -\frac{1}{3} , -2}$

7 0

3 years ago

It rained 18 days during the month of April what percentage of day did it not rain during the month of April

kotykmax [81]

There are 30 days in April.

If 18 days had rain, it means 12 didn't.

18/30=x/100

Make sure both fractions can be made into fractions with a denominator of 100:

180/300=x/100

Divide 180/300 by 3:

60/100=x/100

This shows that the 18 days that did rain take up 60% of the whole month.

So then the remaining 12 days that didn't have rain must take up 40% of the month.

Solution:
40% of the days did not have rain during the month of April

6 0

3 years ago