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Elan Coil [88]
3 years ago
10

Select the three ratios that are equivalent to 2 adults/5 children

Mathematics
2 answers:
IceJOKER [234]3 years ago
3 0
2:5
4:10
8:20
16:40
Hope that helps:)
morpeh [17]3 years ago
3 0

This a screen shot of the answer.

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Nth term fo 3,11,25,45
alexandr1967 [171]
I think the answer may be 291
7 0
3 years ago
Read 2 more answers
Add using a number​ line: -9 + (-3) = ___
Tomtit [17]
-12

Mark brainliest please

Hope this helps
3 0
3 years ago
Pls help I don’t understand
dmitriy555 [2]

The answer is c. \frac{46}{9}

a. \frac{11}{5}=2.2 (2.2≠5.11) (2.2 is not equal to 5.11)

b. 5\frac{1}{8}=\frac{5}{8} or 0.625 (0.625≠5.11) (0.625 is not equal to 5.11)

c. \frac{46}{9} =5.11 (5.11=5.11) (5.11 is equal to 5.11)

Therefore c is the answer

4 0
3 years ago
On a coordinate plane, a parabola opens up. Solid circles appear on the parabola at (negative 4, 14), (negative 3, 9. 5), (negat
umka2103 [35]

The rate of change for the interval between 2 and 6 on the x-axis is 3.

Given

On a coordinate plane, a parabola opens up.

Solid circles appear on the parabola at (negative 4, 14), (negative 3, 9. 5), (negative 2, 6), (0, 2), (1, 1. 5), (2, 2), (4, 6), (5, 9. 5), (6, 14).

<h3>What is the rate of change?</h3>

A rate of change defines how one quantity changes in relation to another quantity.

The rate of change for the interval between 2 and 6 on the x-axis is;

\rm Rate \ of \ change =\dfrac{2-14}{2-6}\\\\Rate \ of \ change =\dfrac{-12}{-4}\\\\Rate \ of \ change =3

Hence, the rate of change for the interval between 2 and 6 on the x-axis is 3.

To know more about Parabola click the link given below.

brainly.com/question/16549411

8 0
2 years ago
Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k​
masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

2

−2k−4k+12

D=k^2-6k+13D=k

2

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0
3 years ago
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