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PSYCHO15rus [73]
3 years ago
8

10 points please help

Mathematics
1 answer:
jolli1 [7]3 years ago
4 0
6* 8.4
7* 3.5
8* -6
9* 10.899
10* 11.1
11* 2.4
12* 0.06
13* 2.2
14* 15.2
15* 11.7
16* -3.4

explanation: Sorry i couldn’t finish and i hope these are correct! I’m not that good at math but good luck
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2 years ago
What is the least possible value of 8x-13 when 9 is less than or equal to 3/4 + x?
Kruka [31]

The least possible value of 8x - 13 when 9 is less than or equal to 3/4 + x is;

53.

According to the question;

  • 9<=3/4 + x.

Therefore,

  • 9 -3/4 <= x

  • 8 1/4 <= x

Therefore, x >= 33/4.

The least possible value of x is therefore, 33/4

The expression 8x -3 becomes;

  • 8(33/4) -13

  • =66 - 13

  • = 53.

Therefore, the least possible value of 8x - 13 is;

53.

Read more;

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6 0
3 years ago
Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real
garik1379 [7]

The complete question is

"Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real number solutions? (The related quadratic function will have two x-intercepts.) Check all that apply.

0 = 2x^2 – 7x – 9

0 = 4x^ 2 – 3x – 1

The quadratic equations that have real number solutions are; 4x^2 – 3x – 1, and 2x^2 – 7x – 9.

<h3>What is the formula for Discriminant?</h3>

The formula for finding the discriminant is

b^2 - 4ac

The solution contains the term \sqrt{b^2 - 4ac} which will be:

Real and distinct if the discriminant is positive

Real and equal if the discriminant is 0

Non-real and distinct roots if the discriminant is negative

For the quadratic equation 2x^2 - 7x - 9

b^2 - 4ac

= (-7) ^2 - 4( 2) ( -9)\\\\= 49 + 72 = 121

This equation has two real number solutions.

For the quadratic equation 4x^ 2 - 3x- 1

b^2 - 4ac

= (-3) ^2 - 4( 4) ( -1)\\\\= 9 + 16 = 25

This equation will have two real number solutions.

Learn more on discriminant here:

brainly.com/question/1537997

#SPJ1

5 0
2 years ago
Rewrite the set M by listing its elements. Make sure to use the appropriate set notation
Anika [276]
We are given the rule for the set,
M= {x | x is an integer and -2 < x < 1}In words, this set translates to numbers that are integers and greater than -2 but less than 1.
The actual elements of the set are
M = { -1, 0}
There are only two elements.
8 0
3 years ago
Find the equation of locus of a point which moves such that its distance from (0,2) is one third distance from (-2,3). ( I WILL
scoray [572]

Answer:

8(x^2+y^2)-4x-30y+23=0

Step-by-step explanation:

<u />

<u>Distance formula</u>

\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Let P(x, y) = any point on the locus

Let A = (0, 2)    

Let B = (-2, 3)

If a point moves such that its distance from (0, 2) is one third distance from (-2, 3):

PA=\dfrac{1}{3}PB

Therefore, using the distance formula:

\implies \sqrt{(x-0)^2+(y-2)^2}=\dfrac{1}{3}\sqrt{(x-(-2))^2+(y-3)^2}

Square both sides:

\implies x^2+(y-2)^2=\dfrac{1}{9}[(x+2)^2+(y-3)^2]

\implies x^2+y^2-4y+4=\dfrac{1}{9}(x^2+4x+4+y^2-6y+9)

Multiply both sides by 9:

\implies 9x^2+9y^2-36y+36=x^2+4x+4+y^2-6y+9

\implies 8x^2+8y^2-4x-30y+23=0

\implies 8(x^2+y^2)-4x-30y+23=0

3 0
2 years ago
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