Jim has gotten scores of 99 and 93 on his first two tests. what score must he get on his third test to keep an average of 90 or greater

Answers

Answer 1

Answer:

Jim must get ≥ 78 score to maintain his average ≥ 90 .

What is an average ?

Every thing has a central tendency , Average is one of the measure of Central Tendency.

It is the mean of all the given numbers and is determined by summing all the numbers or data and then dividing by the total number of data.

It is given that

Jim has gotten scores of 99 and 93 on his first two tests.

He has to maintain an average of 90 or greater

The third test score = ?

Average = (99+93+x)/3

90≤ (99+93+x)/3

270 ≤ (99+93+x)

270 ≤ 192 + x

Therefore x ≥ 78

Jim must get ≥ 78 score to maintain his average ≥ 90 .

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Answer 2

Answer: For the average to be 90,

(99 + 93 + x) / 3 = 90

or 192 + x = 270
or x = 270 - 192 = 78

Jim must score at least 78 in the third test.

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If 3x^2-2x+7=0,then (x-1/3)^2= please help with detailed steps because i dont really understand it. I know the answer is -20/9 but please explain
How to solve this system of equation 3x + y= 17 and 4x + 2 y = 20

A is the point (2, 3) and B is the point (7,-5).Find the equation of the linethrough A that is perpendicular to AB. Give your answer in the form y =
mx+C.

Answers

The equation of the line that is perpendicular to line AB will be y = (5/8) x + c.

What is the equation of a perpendicular line?

Let the equation of the line be ax + by + c = 0. Then the equation of the perpendicular line that is perpendicular to the line ax + by + c = 0 is given as bx - ay + d = 0. If the slope of the line is m, then the slope of the perpendicular line will be negative 1/m.

A is the point (2, 3) and B is the point (7,-5). Then the slope of the line AB is given as,

m = (3 + 5) / (2 - 7)

m = - 8 / 5

Then the slope of the perpendicular line is given as,

⇒ -1/m

⇒ - 1/(- 8/5)

⇒ 5/8

Then the equation of the perpendicular line is given as,

y = (5/8) x + c

The equation of the line that is perpendicular to line AB will be y = (5/8) x + c.

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Answer:

y=0.625x+1.75

Step-by-step explanation:

The equation of the line that goes through (2, 3) and (7, -5) needs to be found before finding the equation perpendicular to it. The formula y=mx+c will be used for finding this equation.

Calculating the slope between the data points:

$m1=(3-(-5))/(2-7)=(8)/(-5)=-1.6$

Finding the first y-intercept:

$y=mx+c$

$c=y-mx$

$c=3-(-1.6*2)=3+3.2=6.2$

The equation for the line that goes through the points is:

$y=-1.6x+6.2$

The perpendicular slope will be the opposite reciprocal of the original slope.

The second slope is found by first multiplying the first slope (m1) by -1:

$m2=m1*-1=-1.6*-1=1.6$

Then by taking the reciprocal:

$m2=(1)/(1.6) =0.625$

The second equation must have an intercept that goes through point A:

$c=y-mx$

$c=3-(0.625*2)=3-1.25=1.75$

The perpendicular equation is:

$y=0.625x+1.75$

The Halloween candy bowl of chocolates had 80 pieces of candy. On average 3 pieces were eaten every hour.The bowl of sour candy had 120 pieces and 5 pieces were eaten each hour.

Write and solve the system. Write each equation in the order that it appears in the problem.

At what hour will the bowls have the same amount? How many pieces will each bowl have at this hour?

Answers

Answer:

The Halloween candy bowl of chocolates had 80 pieces of candy. 3 pieces were eaten every hour.

80 divide 3 =26.6666667

idk how to round this so you need to do that

Step-by-step explanation:

The bowl of sour candy had 120 pieces and 5 pieces were eaten each hour.

120 divide 5 = 24

hope this helps

Find the values of x and y if

Answers

Answer:

see explanation

Step-by-step explanation:

Since the triangles are congruent then corresponding sides are congruent, then

CA = NE , that is

4x + 3 = 11 ( subtract 3 from both sides )

4x = 8 ( divide both sides by 4 )

x = 2

----------------------------------------

AR = EW, that is

4y - 12 = 10 ( add 12 to both sides )

4y = 22 ( divide both sides by 4 )

y = 5.5

-------------------------------------------

NW = x + y = 2 + 5.5 = 7.5

CR = NW = 7.5

Find the slope of the line through the given points.

(1,2), (-3,2)

Answers

the slope can be found by using the equation y2-y1 over x2-x1
2-2 over -3-1
= 0 over -4
your slope is 0

The slope=
$y= (y2-y1)/(x2-x1) (2-2)/(-3-1) (0)/(-4) }$

Arrange the circles (represented by their equations in general form) in ascending order of their radius lengths.x2 + y2 − 2x + 2y − 1 = 0
x2 + y2 − 4x + 4y − 10 = 0
x2 + y2 − 8x − 6y − 20 = 0
4x2 + 4y2 + 16x + 24y − 40 = 0
5x2 + 5y2 − 20x + 30y + 40 = 0
2x2 + 2y2 − 28x − 32y − 8 = 0
x2 + y2 + 12x − 2y − 9 = 0

Answers

The correct answer is:

x²+y²-2x+2y-1 = 0;
x²+y²-4x+4y-10 = 0;
5x²+5y²-20x+30y+40 = 0;
x²+y²-8x-6y-20 = 0;
x²+y²+12x-2y-9 = 0;
4x²+4y²+16x+24y-40 = 0; and
2x²+2y²-28x-32y-8 = 0

Explanation:

For each of these, we want to write the equation in the form
(x+h)²+(y+k)² = r².

To do this, we evaluate the terms 2hx and 2ky in each equation. We will take half of this; this will tell us what h and k are for each equation.

For the first equation:
2hx = -2x and 2ky = 2y.

Half of -2x = -1x and half of 2y = 1y; this means h = -1 and k = 1:
(x-1)² + (y+1)² + ___ - 1 = 0

When we multiply (x-1)², we get
x²-2x+1.
When we multiply (y+1)², we get
y²+2y+1.

This gives us 1+1 = 2 for the constant. We know we must add something to 2 to get -1; 2 + ___ = -1; the missing term is -3. Add that to each side (to have r² on the right side of the equals) and we have
(x-1)² + (y+1)² = 3
This means that r² = 3, and r = √3 = 1.732.

For the second equation, 2hx = -4x and 2ky = 4y; this means h = -4/2 = -2 and k = 4/2 = 2. This gives us
(x-2)² + (y+2)² -10 + ___ = 0.

Multiplying (x-2)² gives us
x²-4x+4.
Multiplying (y+2)² gives us
y²+4x+4.
This gives us 4+4= 8 for our constant so far.

We know 8 + ___ = -10; this means the missing term is -18. Add this to each side of the equation to have
(x-2)²+(y+2)² = 18; r² = 18; r = √18 = 3√2 = 4.243.

For the third equation, 2hx = -8x and 2ky = -6y. This means h = -8/2 = -4 and k = -6/2 = -3. This gives us:
(x-4)²+(y-3)²-20 = 0

Multiplying (x-4)² gives us
x²-8x+16.
Multiplying (y-3)² gives us
y²-6y+9.

This gives us 16+9 = 25 for the constant. We know that 25+___ = -20; the missing term is -45. Add this to each side for r², and we have that
r²=45; r = √45 = 3√5 = 6.708.

For the next equation, we factor 4 out of the entire equation:
4(x²+y²+4x+6y-10)=0.
This means 2hx = 4x and 2ky = 6y; this gives us h = 4/2 = 2 and k = 6/2 = 3. This gives us
4((x+2)²+(y+3)² - 10) = 0.

Multiplying (x+2)² gives us
x²+4x+4.
Multiplying (y+3)² gives us
y²+6y+9.

This gives us a constant of 4+9 = 13. We know 13+__ = -10; this missing value is -23. Since we had factored out a 4, that means we have 4(-23) = -92. Adding this to each side for r², we have
r²=92; r = √92 = 2√23 = 9.59.

For the next equation, we factor out a 5 first:
5(x²+y²-4x+6y+8) = 0. This means that 2hx = -4x and 2ky = 6y; this gives us h = -4/2 = -2 and k = 6/2 = 3:

5((x-2)²+(y+3)²+8) = 0.

Multiplying (x-2)² gives us
x²-4x+4.
Multiplying (y+3)² gives us
y²+6y+9.

This gives us a constant of 4+9 = 13. We know that 13+__ = 8; the missing value is -5. Since we factored a 5 out, we have 5(-5) = -25. Adding this to each side for r² gives us
r²=25; r = √25 = 5.

For the next equation, we first factor a 2 out:
2(x²+y²-14x-16y-4) = 0. This means 2hx = -14x and 2ky = -16y; this gives us h = -14/2 = -7 and k = -16/2 = -8:

2((x-7)²+(y-8)²-4) = 0.

Multiplying (x-7)² gives us
x²-14x+49.
Multiplying (y-8)² gives us
y²-16x+64.

This gives us a constant of 49+64=113. We know that 113+__ = -4; the missing value is -117. Since we first factored out a 2, this gives us 2(-117) = -234. Adding this to each side for r² gives us
r²=234; r = √234 = 3√26 = 15.297.

For the last equation, 2hx = 12x and 2ky = -2; this means h = 12/2 = 6 and k = -2/2 = -1:
(x+6)²+(y-1)²-9 = 0

Multiplying (x+6)² gives us
x²+12x+36.
Multiplying (y-1)² gives us
y²-2y+1.

This gives us a constant of 36+1 = 37. We know that 37+__ = -9; the missing value is -46. Adding this to each side for r² gives us
r² = 46; r=√46 = 6.78.

Find the radius of each equation:

1.
$x^2 + y^2-2x+2y-1 = 0, \n x^2-2x+1-1 + y^2+2y+1-1-1 = 0, \n (x-1)^2+(y+1)^2=3$ , then $r_1= √(3)$ .

2.
$x^2 + y^2-4x + 4y- 10 = 0, \n x^2 -4x+4-4+ y^2 + 4y+4-4- 10 = 0, \n (x-2)^2+(y+2)^2=18$ , then $r_2= √(18)=3 √(2)$ .

3.
$x^2 + y^2-8x- 6y- 20 = 0, \n x^2-8x+16-16+ y^2- 6y+9-9- 20 = 0, \n (x-4)^2+(y-3)^2=45$ , then $r_3= √(45) =3 √(5)$ .

4.
$4x^2 + 4y^2+16x+24y- 40 = 0, \n 4x^2+16x+16-16+ 4y^2+24y+36-36- 40 = 0, \n 4(x+2)^2+4(y+3)^2=92,\n (x+2)^2+(y+3)^2=23$ , then $r_4= √(23)$ .

5.
$5x^2 + 5y^2-20x+30y+ 40 = 0, \n 5x^2-20x+20-20+ 5y^2+30y+45-45- 40 = 0, \n 5(x-2)^2+5(y+3)^2=105,\n (x-2)^2+(y+3)^2=21$ , then $r_5= √(21)$ .

6.
$2x^2 + 2y^2-28x-32y- 8= 0, \n 2x^2-28x+98-98+ 2y^2-32y+128-128- 8= 0, \n 2(x-7)^2+2(y-8)^2=234,\n (x+2)^2+(y+3)^2=117$ , then $r_6= √(117)=3√(13)$ .

7.
$x^2 + y^2+12x-2y-9 = 0, \n x^2+12x+36-36+ y^2-2y+1-1- 9 = 0, \n (x+6)^2+(y-1)^2=46$ , then $r_7= √(46)$ .

Hence
$r_1= √(3)$ , $r_2=3 √(2)$ , $r_3=3 √(5)$ , $r_4= √(23)$ , $r_5= √(21)$ , $r_6= 3√(13)$ , $r_7= √(46)$ and $r_1\ \textless \ r_2\ \textless \ r_5\ \textless \ r_4\ \textless \ r_3\ \textless \ r_7\ \textless \ r_6$ .

Use the intercepts to graph the equation. -x+2y=4