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liq [111]
3 years ago
14

Can you guys help me with this question? PLEASEEEEE

Mathematics
1 answer:
Andreyy893 years ago
6 0

Answer:

Walker County (0,4) (1, 6) (5, 14) (4,12)

Equation: y = 2x + 4

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Monica took a survey of her classmates' hair and eye color. The results are in the table below.
katrin2010 [14]

Answer: 0.5

Step-by-step explanation: it should be but  wait for other answers

3 0
3 years ago
When the polynomial P(x)=ax^3+bx^2+3x-10 is divided by x+1, the remainder is -8. P(x) has a factor of x+5. Find the values of a
Ahat [919]

Answer:

a = 1, b = 6

Step-by-step explanation:

The equation given is as follows;

P(x) = a·x³ + b·x² + 3·x - 10

The above equation has a factor of x + 5, therefore, we have;

P(-5) = 0 = a·(-5)³ + b·(-5)² + 3·(-5) - 10

-125·a + 25·b + (-15) - 10 = 0

-125·a + 25·b - 25 = 0

-125·a + 25·b  = 25...........(1)

Also, we are given that;

a·x³ + b·x² + 3·x - 10 divided by x + 1 as a remainder, R = -8, therefore;

P(-1) = -8 = a·(-1)³ + b·(-1)² + 3·(-1) - 10

-a + b - 13 = -8

-a + b = -8 + 13 = 5

-a + b = 5............................(2)

Multiply equation (2) by 25 and subtract from (1) gives

-125·a + 25·b - 25(-a + b) = 25 - 25×5

-100·a = 25 - 125 = -100

a = 1

Therefore, from equation (2) we have;

-1 + b = 5

b = 5 + 1 = 6.

7 0
3 years ago
How wide is the gate
nevsk [136]
12 cm long and 7 wide
5 0
3 years ago
An adult brain is about 140 mm wide and divided into two sections (called "hemispheres" although the brain is not truly spherica
Sedbober [7]

Answer: 3.61×10^5 A

Step-by-step explanation: Since the brain has been modeled as a current carrying loop, we use the formulae for the magnetic field on a current carrying loop to get the current on the hemisphere of the brain.

The formulae is given below as

B = u×Ia²/2(x²+a²)^3/2

Where B = strength of magnetic field on the axis of a circular loop = 4.15T

u = permeability of free space = 1.256×10^-6 mkg/s²A²

I = current on loop =?

a = radius of loop.

Radius of loop is gotten as shown... Radius = diameter /2, but diameter = 65mm hence radius = 32.5mm = 32.5×10^-3 m = 3.25×10^-2m

x = distance of the sensor away from center of loop = 2.10 cm = 0.021m

By substituting the parameters into the formulae, we have that

4.15 = 1.256×10^-6 × I × (3.25×10^-2)²/2{(0.021²) + (3.25×10^-2)²}^3/2

4.15 = 13.2665 × 10^-10 × I/ 2( 0.00149725)^3/2

4.15 = 1.32665 ×10^-9 × I / 2( 0.000058)

4.15 × 2( 0.000058) = 1.32665 ×10^-9 × I

I = 4.15 × 2( 0.000058)/ 1.32665 ×10^-9

I = 4.80×10^-4 / 1.32665 ×10^-9

I = 3.61×10^5 A

3 0
3 years ago
Which point would not be a solution to the system of linear inequalities shown below?
Anna11 [10]

Answer:

The coordinates (12,7) would NOT be a solution to the system of linear inequalities shown below.

Step-by-step explanation:

If you do the graphing and you look at all your options of coordinates, (12,7) is the only one that is not in the shaded areas or in the lines of the inequalities. Therefore, (12,7) should be the answer to that question. :)

3 0
3 years ago
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