6) A student scored an 88, S2, and 76 on three math tests. What does the student need to get on the fourth test to have an average of 85 for all four tests? James wants to use algebra to solve this problem. Which equation should he use? 256 4x 85 256 x 85 85 x 256 256 85x

Answers

Answer 1

Answer: Equation B.
The average of the sum of the existing 3 scores plus the score of the new test should equal 85.

To solve this, multiply both sides by 4 and subtract 256. The student needs an 84 in order to have an average of 85.

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A coordinate grid with 2 lines. The first line labeled f(x) passes through (negative 3, 3), (0, 2), and (3, 1). The second line labeled g(x) passes through the points at (negative 3, 0) and (0, 2) What is the solution to the system of linear equations? (–3, 0) (–3, 3) (0, 2) (3, 1)

Read the picture above to answer. Thanks

Answers

The answer and explanation is down below. HTH

Evaluate.
5 P 3 x 6 C 4

a. 150
b. 300
c. 900

Answers

Answer:

Option (c) is correct.

$^5P_3*^6C_4$ is 900

Step-by-step explanation:

Given :Expression $^5P_3*^6C_4$

We have to find the value of given expression $^5P_3*^6C_4$

Consider the given expression $^5P_3*^6C_4$

The possibility of choosing an ordered set of r object from n object is given by

$nPr=(n!)/(\left(n-r\right)!)$

and The number of subset of r elements from n elements $nCr=(n!)/(r!\left(n-r\right)!)$

Thus, $^5P_3=(5!)/(\left(5-3\right)!)==(5!)/(2!)=5\cdot \:4\cdot \:3=60$

and $^6C_4=(6!)/(4!\left(6-4\right)!)==(6!)/(4!\cdot \:2!)==(6\cdot \:5)/(2!)=15$

Thus, $^5P_3*^6C_4=60* 15=900$

Thus, $^5P_3*^6C_4$ is 900

The answer is 300 because if u x the 5x3=15 and divide 6 and 4 is 1.5 then multipy that 22 then you put the letters in the equations

Sparkle is 464 mm long and twilight is 46 cm long which of Amelia's cats is longer

Answers

If Sparkle is 464 mm long and twilight is 46 cm long then Sparkle is longer than Twilight by 4 mm.

What is Unit of Measurement?

A unit of measurement is a definite magnitude of a quantity, defined and adopted by convention or by law, that is used as a standard for measurement of the same kind of quantity.

To compare the length of Sparkle and Twilight, we need to convert their lengths to the same unit of measurement.

Sparkle's length is given in millimeters (mm), while Twilight's length is given in centimeters (cm).

To convert Twilight's length to millimeters, we can multiply by 10

1 centimeter = 10 milli meters

Twilight's length in millimeters

= 46x 10

= 460 mm

Now we can compare the lengths:

Sparkle: 464 mm

Twilight: 460 mm

Therefore, Sparkle is longer than Twilight by 4 mm.

To learn more on Unit of Measurement click:

brainly.com/question/15402847

#SPJ2

464 mm is longer because it is 46.4 cm and twilight is 46 cm. Change mm to cm by moving the decimal point of the mm back one point.

the sum of three numbers is 16. the largest number is equal to the sum of the other two and 3 times the smallest number is 1 more than the largest

Answers

Answer:

8, 5, 3 are the numbers.

Step-by-step explanation:

Let the three numbers are x, y , and z.

out of which x is the largest and z is the smallest number.

Sum of three numbers is 16.

x + y + z = 16------(1)

The largest number is equal to the sum of the other two.

x = y + z --------(2)

3 times the smallest number is 1 more than the largest.

3z = x + 1 ---------(3)

Now we put the value of x from equation (2) to equation (1)

x + x = 16

2x = 16

x = 8

By putting x = 8 in equation (3)

3z = 8 + 1

3z = 9

z = 3

Now we put the values of x and z in equation (2)

8 = y + 3

y = 8 - 3

y = 5

Therefore, the numbers are 8, 5, 3.

x = 8

Z = 7/3

y = 17/3

A field goal kicker lines up to kick a 44 yard (40m) field goal. He kicks it with an initial velocity of 22m/s at an angle of 55∘. The field goal posts are 3 meters high.Does he make the field goal?What is the ball's velocity and direction of motion just as it reaches the field goal post

Answers

Answer:

The ball makes the field goal.

The magnitude of the velocity of the ball is approximately 18.166 meters per second.

The direction of motion is -45.999º or 314.001º.

Step-by-step explanation:

According to the statement of the problem, we notice that ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical uniform accelerated motion, whose equations of motion are described below:

$x = x_(o)+v_(o)\cdot t\cdot \cos \theta$ (Eq. 1)

$y = y_(o) + v_(o)\cdot t \cdot \sin \theta +(1)/(2)\cdot g\cdot t^(2)$ (Eq. 2)

Where:

$x_(o)$ , $y_(o)$ - Coordinates of the initial position of the ball, measured in meters.

$x$ , $y$ - Coordinates of the final position of the ball, measured in meters.

$\theta$ - Angle of elevation, measured in sexagesimal degrees.

$v_(o)$ - Initial speed of the ball, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $x_(o) = 0\,m$ , $y_(o) = 0\,m$ , $v_(o) = 22\,(m)/(s)$ , $\theta = 55^(\circ)$ , $g = -9.807\,(m)/(s)$ and $x = 40\,m$ , the following system of equations is constructed:

$40 = 12.618\cdot t$ (Eq. 1b)

$y = 18.021\cdot t -4.904\cdot t^(2)$ (Eq. 2b)

From (Eq. 1b):

$t = 3.170\,s$

And from (Eq. 2b):

$y = 7.847\,m$

Therefore, the ball makes the field goal.

In addition, we can calculate the components of the velocity of the ball when it reaches the field goal post by means of these kinematic equations:

$v_(x) = v_(o)\cdot \cos \theta$ (Eq. 3)

$v_(y) = v_(o)\cdot \cos \theta + g\cdot t$ (Eq. 4)

Where:

$v_(x)$ - Final horizontal velocity, measured in meters per second.

$v_(y)$ - Final vertical velocity, measured in meters per second.

If we know that $v_(o) = 22\,(m)/(s)$ , $\theta = 55^(\circ)$ , $g = -9.807\,(m)/(s)$ and $t = 3.170\,s$ , then the values of the velocity components are:

$v_(x) = \left(22\,(m)/(s) \right)\cdot \cos 55^(\circ)$

$v_(x) = 12.619\,(m)/(s)$

$v_(y) = \left(22\,(m)/(s) \right)\cdot \sin 55^(\circ) +\left(-9.807\,(m)/(s^(2)) \right)\cdot (3.170\,s)$

$v_(y) = -13.067\,(m)/(s)$

The magnitude of the final velocity of the ball is determined by Pythagorean Theorem:

$v =\sqrt{v_(x)^(2)+v_(y)^(2)}$ (Eq. 5)

Where $v$ is the magnitude of the final velocity of the ball.

If we know that $v_(x) = 12.619\,(m)/(s)$ and $v_(y) = -13.067\,(m)/(s)$ , then:

$v = \sqrt{\left(12.619\,(m)/(s) \right)^(2)+\left(-13.067\,(m)/(s)\right)^(2) }$

$v \approx 18.166\,(m)/(s)$

The magnitude of the velocity of the ball is approximately 18.166 meters per second.

The direction of the final velocity is given by this trigonometrical relation:

$\theta = \tan^(-1)\left((v_(y))/(v_(x)) \right)$ (Eq. 6)

Where $\theta$ is the angle of the final velocity, measured in sexagesimal degrees.

If we know that $v_(x) = 12.619\,(m)/(s)$ and $v_(y) = -13.067\,(m)/(s)$ , the direction of the ball is:

$\theta = \tan^(-1)\left((-13.067\,(m)/(s) )/(12.619\,(m)/(s) ) \right)$

$\theta = -45.999^(\circ) = 314.001^(\circ)$

The direction of motion is -45.999º or 314.001º.

The ball makes the field goal.

The magnitude of the velocity of the ball is approximately 18.166 meters per second.

The direction of motion is -45.999º or 314.001º.

According to the statement of the problem, we notice that ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical uniform accelerated motion, whose equations of motion are described below:

X=Xo+Vo*t*cosФ (Eq. 1)

Y=Yo+Vo*t*sinФ +(1/2)*g*t²(Eq. 2)

Where:

Xo,Yo - Coordinates of the initial position of the ball, measured in meters.

X,Y - Coordinates of the final position of the ball, measured in meters.

Ф- Angle of elevation, measured in sexagesimal degrees.

Vo - Initial speed of the ball, measured in meters per square second.

t - Time, measured in seconds.

If we know that Xo = 0m, Yo = 0m, Vo = 22m/s, Ф = 55°,g = -9.807m/s and X = 40m, the following system of equations is constructed:

40 = 12.618*t (Eq. 1b)

Y = 18.021*t-4.904*t² (Eq. 2b)

From (Eq. 1b):

t = 3.170s

And from (Eq. 2b):

Y = 7.847m

Therefore, the ball makes the field goal.

In addition, we can calculate the components of the velocity of the ball when it reaches the field goal post by means of these kinematic equations:

Vx = Vo*cosФ (Eq. 3)

Vy = Vo*cosФ+g*t (Eq. 4)

Where:

Vx - Final horizontal velocity, measured in meters per second.

Vy- Final vertical velocity, measured in meters per second.

If we know that Vo = 22m/s, Ф= 55°, g = -9.807m/s and t = 3.170s, then the values of the velocity components are:

Vx = (22m/s)*cos55°

Vx = 12.619m/s

Vy = (22m/s)*sin55°+(-9.807m/s²)*3.170s

Vy = -13.067m/s

The magnitude of the final velocity of the ball is determined by Pythagorean Theorem:

V = √(Vx²+Vy²) (Eq. 5)