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Step2247 [10]
2 years ago
14

Which statement is trure?​

Mathematics
2 answers:
o-na [289]2 years ago
5 0

Answer:

Sqrt57 < sqrt50 + sqrt 7

Step-by-step explanation:

Sqrt 57 = 7.5 and sqrt 50 = 7.1 + sqrt 7 = 2.6

= 7.5 < 7.1 + 2.6

= 7.5 < 9.7

krek1111 [17]2 years ago
3 0

Answer:

I think it is A, or ,D

Step-by-step explanation:

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Name the property shown by the statement 3(42) = 3(40) + 3(2)
Hatshy [7]

Answer: i think it’s distributive don’t kill me if it’s wrong

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
SOMEONE PLEASE HELP!!!
yuradex [85]

Answer:

This is an arithmetic sequence with a common difference <em>d</em> of -5

Step-by-step explanation:

When you find out <em>d</em>, you realize the difference between the numbers to be -5. Since it is <em>d</em> and not common ratio <em>r</em>, it is an arithmetic sequence.

8 0
2 years ago
Quadrilateral HIJK has sides measuring 12 cm, 26 cm, 14 cm, and 30 cm. Which could be the side lengths of a dilation of HIJK wit
matrenka [14]

Answer:

(C) 18cm, 39cm, 21cm and 45cm.

Step-by-step explanation:

The quadrilateral HIJK has sides measuring 12 cm, 26 cm, 14 cm, and 30 cm.

When HIJK is dilated with a scale factor of 1.5, the side lengths becomes:

12 X 1.5 =18 cm

26 X 1.5 =39 cm

14 X 1.5 =21 cm

30 X 1.5 =45 cm

A dilation of HIJK with a scale factor of 1.5 will give us the side lengths:

18cm, 39cm, 21cm and 45cm.

<u>The correct option is C.</u>

6 0
3 years ago
Is the expression completely factored: 12x ^ 2 - 2x - 24
Juliette [100K]

No, It is not. Factor out 2 from it

7 0
3 years ago
Read 2 more answers
Please help me on this problem
Anna71 [15]

Answer:

The pairs of integer having two real solution forax^{2} -6x+c = 0 are

  1. a = -4, c = 5
  2. a = 1, c = 6
  3. a = 2, c = 3
  4. a = 3, c = 3

Step-by-step explanation:

Given

ax^{2} -6x+c = 0

Now we will solve the equation by putting all the 6 pairs so we get the  following

-3x^{2} -6x-5 = 0 for a = -3 , c=-5

-4x^{2} -6x+5 = 0 for a = -4 , c=5

1x^{2} -6x+6 = 0 for a = 1 , c=6

2x^{2} -6x+3 = 0 for a = 2 , c=3

3x^{2} -6x+3 = 0 for a = 3 , c=3

5x^{2} -6x+4 = 0 for a = 5 , c=4

The above  all are Quadratic equations inn general form ax^{2} +bx+c=0

where we have a,b and c constant values

So for a real Solution we must have

Disciminant , b^{2} -4\timesa\timesc \geq 0

for a = -3 , c=-5 we have

Discriminant =-24 which is less than 0 ∴ not a real solution.

for a = -4 , c=5 we have

Discriminant = 116 which is greater than 0 ∴ a real solution.

for a = 1 , c=6 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 2 , c=3 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 3 , c=3 we have

Discriminant =0 which is equal to 0 ∴ a real solution.

for a = 5 , c=4 we have

Discriminant =-44 which is less than 0 ∴ not a real solution.

7 0
3 years ago
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