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Inessa [10]
3 years ago
5

Which numbers are solutions of the inequality? X < –8; –10, –5, 0 –10 –10, –5, and 0 0 –5 and

Mathematics
1 answer:
lozanna [386]3 years ago
5 0

Answer:

-10

Step-by-step explanation:

Here the given inequality is

x

This means that the numbers that have values less than -8 will satisfy the inequality. This means it will be towards the left of -8

The numbers in the options are -10,-5,0

The only number less than -8 is -10. The others are greater than -8.

So, the number that is the solution of inequality is -10.

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nlexa [21]

Answer:

864 cards

Step-by-step explanation:

72 cards per set times 15 sets = 1080 cards - 216 cards he already has =864 cards

5 0
3 years ago
MULTIPLE CHOICE A thermometer is
pav-90 [236]

The absolute value equation which could be used to obtain the greatest and least possible temperature values when the temperature reading is 17°F is (17 ± 2)°F

  • Thermometer accuracy = ±2°F
  • Thermometer reading = 17°F

<u>The least possible value of </u><u>temperature</u><u> </u><u>can</u><u> </u><u>be</u><u> </u><u>calculated</u><u> </u><u>thus</u> :

  • Thermometer reading - thermometer accuracy

  • 17 - 2 = 15°F

<u>The highest possible </u><u>temperature</u><u> </u><u>can</u><u> </u><u>be</u><u> </u><u>calculated</u> thus :

  • Themometer reading - thermometer accuracy

  • 17 + 2 = 19°F

Hence, the absolute value equation which represents the scenario is (17 ± 2)°F

Learn more :brainly.com/question/15748955

4 0
2 years ago
What mixed number between is 1 and 1 1/2?
EastWind [94]
1 1/4

Pretty easy. For me at least.

6 0
3 years ago
Read 2 more answers
If a shape is a rhombus then the diagonal are perpendicular
Delvig [45]

Question

If a shape is a rhombus then the diagonal are perpendicular?

Answer:

The Correct Is True Or Yes.

the diagonals of a rhombus are perpendicular, and that they intersect at the midpoints of both.

if the diagonals of a parallelogram are perpendicular, then it's a rhombus

{Hope this helps!!}

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5 0
2 years ago
8.
Ahat [919]

Answer:    y= 4x² +2x -4

C. y = 4x2 + 2x − 4

Step-by-step explanation:  f(x)=4x^2+2x-4

Given three points

P1(-2,8)

P2(0,-4)

P3(4,68)

We need the quadratic equation that passes through all three points.

Solution:

We first assume the final equation to be

f(x)=ax^2+bx+c .............................(0)

Observations:

1. Points are not symmetric, so cannot find vertex visually.

2. Using the point (0,-4) we substitute x=0 into f(x) to get

f(0)=0+0+c=-4, hence c=-4.

3. We will use the two other points (P1 & P3) to set up a system of two equations to find a and b.

f(-2)=a(-2)^2+b(-2)-4=8 => 4a-2b-4=8.................(1)

f(4)=a(4^2)+b(4)-4=68 =>  16a+4b-4=68.............(2)

4. Solve system

2(1)+(2) => 24a+0b-12=84 => 24a=96 => a=96/24 => a=4 ......(3)

substitute (3) in (2) => 16(4)+4b-4=68 => b=8/4 => b=2  ..........(4)

5. Put values c=-4, a=4, b=2 into equation (0) to get

f(x)=4x^2+2x-4

Check:

f(-2)=4((-2)^2)+2(-2)-4=16-4-4=8

f(0)=0+0-4 = -4

f(4)=4(4^2)+2(4)-4=64+8-4=68

So all consistent, => solution ok.

...............................................................................................................................................

Answer:

y = 4x² + 2x − 4

Step-by-step explanation:

The y-values go down and up again, so the parabola opens upward. That means the x² coefficient must be positive. All answer choices are eliminated except one.

y = 4x² +2x -4

 There are other ways to get the answer. One of them is to use the quadratic regression function of a spreadsheet or graphing calculator. The one attached shows the points are matched by the equation ...

y = 4x² +2x -4 . . . . . as above

...............................................................................................................................................

8 0
2 years ago
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