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bulgar [2K]
3 years ago
10

The average of five numbers is 27. Four of the five numbers are 48, 8, 43, and 10. What is the other number?

Mathematics
1 answer:
jarptica [38.1K]3 years ago
4 0

<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>

<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em>

<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>

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1/2ac2 where a =5 and b=2​
Evgen [1.6K]
I’ll assume the c is b. (1/2) x (5) x (2) x (2) is 10
8 0
3 years ago
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Is the following statement true or false?<br> A line is zero-dimensional.<br> true<br> false
garik1379 [7]
Answer:
False

Explanation:
Zero-dimensional would be a point.
One-Dimensional would be a line.
Two-Dimensional would be a plane.
Three-Dimensional would be a shape.
7 0
3 years ago
Read 2 more answers
A ball is thrown into the air from a height of 4 feet at time t = 0. The function that models this situation is h(t) = -16t2 + 6
katrin2010 [14]

Answer:

Part a) The height of the ball after 3 seconds is 49\ ft

Part b) The maximum height is 66 ft

Part c) The ball hit the ground for t=4 sec

Part d) The domain of the function that makes sense is the interval

[0,4]

Step-by-step explanation:

we have

h(t)=-16t^{2} +63t+4

Part a) What is the height of the ball after 3 seconds?

For t=3 sec

Substitute in the function and solve for h

h(3)=-16(3)^{2} +63(3)+4=49\ ft

Part b) What is the maximum height of the ball? Round to the nearest foot.

we know that

The maximum height of the ball is the vertex of the quadratic equation

so

Convert the function into a vertex form

h(t)=-16t^{2} +63t+4

Group terms that contain the same variable, and move the constant to the opposite side of the equation

h(t)-4=-16t^{2} +63t

Factor the leading coefficient

h(t)-4=-16(t^{2} -(63/16)t)

Complete the square. Remember to balance the equation by adding the same constants to each side

h(t)-4-16(63/32)^{2}=-16(t^{2} -(63/16)t+(63/32)^{2})

h(t)-(67,600/1,024)=-16(t^{2} -(63/16)t+(63/32)^{2})

Rewrite as perfect squares

h(t)-(67,600/1,024)=-16(t-(63/32))^{2}

h(t)=-16(t-(63/32))^{2}+(67,600/1,024)

the vertex is the point (1.97,66.02)

therefore

The maximum height is 66 ft

Part c) When will the ball hit the ground?

we know that

The ball hit the ground when h(t)=0 (the x-intercepts of the function)

so

h(t)=-16t^{2} +63t+4

For h(t)=0

0=-16t^{2} +63t+4

using a graphing tool

The solution is t=4 sec

see the attached figure

Part d) What domain makes sense for the function?

The domain of the function that makes sense is the interval

[0,4]

All real numbers greater than or equal to 0 seconds and less than or equal to 4 seconds

Remember that the time can not be a negative number

6 0
3 years ago
A scientist has two solutions, which she has labeled Solution A and Solution Each contains salt. She knows that Solution A is 40
Alexxandr [17]

Answer:

Step-by-step explanation:

Let the number of ounces of

A = x

B = y

A scientist has two solutions, which she has labeled Solution A and Solution Each contains salt. She knows that Solution A is 40% salt and Solution B is 55% salt. She wants to obtain 90 ounces of a mixture that is 45% salt. How many ounces of each solution should she use?

Our system of equations is given as

x + y = 90.... Equation 1

x = 90 - y

40% × x + 55% × y = 45% × 90 ounces

0.4x + 0.55y = 40.5....Equation 2

We substitute 90 - y for x in Equation 2

6 0
3 years ago
Sketch the equilibrium solutions for the following DE and use them to determine the behavior of the solutions.
GREYUIT [131]

Answer:

y=\dfrac{1}{1-Ke^{-t}}

Step-by-step explanation:

Given

The given equation is a differential equation

\dfrac{dy}{dt}=y-y^2

\dfrac{dy}{dt}=-(y^2-y)

By separating variable

⇒\dfrac{dy}{(y^2-y)}=-t

\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-dt

Now by taking integration both side

\int\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-\int dt

⇒\ln (y-1)-\ln y=-t+C

Where C is the constant

\ln \dfrac{y-1}{y}=-t+C

\dfrac{y-1}{y}=e^{-t+c}

\dfrac{y-1}{y}=Ke^{-t}

y=\dfrac{1}{1-Ke^{-t}}

from above equation we can say that

When t  will increases in positive direction then e^{-t} will decreases it means that {1-Ke^{-t}} will increases, so y will decreases. Similarly in the case of negative t.

4 0
3 years ago
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