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densk [106]
3 years ago
6

What is the volume of a come that has a height of 40 millimeters and a diameter of 25 millimeters? Use 3.14 for pi.

Mathematics
1 answer:
Fantom [35]3 years ago
3 0

Answer:

6541.7 mm³ (nearest tenth)

Step-by-step explanation:

\boxed{volume \: of \: cone =  \frac{1}{3}\pi {r}^{2}h  }

Given that the height is 40mm, h= 40mm.

Diameter= 2(radius)

Radius of cone

= 25 ÷2

= 12.5mm

Volume of cone

=  \frac{1}{3} (3.14)(12.5)^{2} (40)

= 6541.7 mm³ (nearest tenth)

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A college student took 4 courses last semester. His final grades, along with the credits each class is worth, are as follow: A (
NikAS [45]

Answer:

A college student took 4 courses last semester. His final grades, along with the credits each class is worth, are as follow: A (3), B (4), C (2), and D (3). The grading system assigns quality points as follows: A: 4; B: 3; C: 2; D: 1; and F: 0. Find the student’s GPA for this semester. Round your answer to the nearest thousandth.

another way is

This is a weighted average question. You are going to "weight" each course by the number of credits it is worth and then divide by the total number of credits. In other words, you are going to multiply each grade (A=4, B=3) by the number of credits attached to that grade. This will ensure that the courses that have more credits count more in the overall average. Then you are going to divide by the total number of credits to get the overall GPA.

So,

(3*4 + 4*3 + 2 *2 + 3*1)/(3+4+2+3) = GPA

Step-by-step explanation:

bran-list please

6 0
3 years ago
According to the Holy Bible, Jesus begins to perform healings. There were many healers during Jesus’ time. What made Jesus’ heal
Gekata [30.6K]

Healing was essential to the ministry of Jesus because He envisioned healing as a physical symbol of forgiveness. He guaranteed the ultimate glory of the human body through His personal resurrection, but forecast that restoration by healing twisted, shrunken, blinded limbs and organs. The paralytic's restoration is but one of many such examples (Mark 2:1-12).

Though there were many healers, Jesus was able to even raise up the dead, and also there was one instance where the Canaanite woman struggled through His disciples' desire to dismiss her, and His own initial, courteous refusal, to get what she knew she could trust Him to grant (Matthew 16:28). The crowds "begged him to let the sick just touch the edge of his cloak," (Matthew 14:36), for "all who touched him were healed."

Healers were not able to do that, and Jesus also claimed that he was God's son, so performing miracles like this was like proving the point. Also, healers were not always able to heal the person, but Jesus was able to do so all the time (for free even!) so many people traveled to meet him so he could heal them.

The woman with a hemorrhage crept silently through the crowd to merely touch His clothes (Mark 5:28). She also claimed that she went to many healers, but she didn't get healed, in fact, she got worse! So that's an instance that proves the point.

<em>Thank you :D</em>

3 0
3 years ago
child admission is $5.90 and adult admission is $9.10 . 175 tickets were sold for a total sales of $1282.10 . How many child tic
GenaCL600 [577]

Answer:

Child tickets are 97

Step-by-step explanation:

x + y = 175 ...............(1)

x = 175 - y

5.90x + 9.10y = 1282.10. .........(2)

5.90 (175 - y) + 9.10y = 1282.10

1032.50 - 5.9y + 9.10y = 1282.10

3.2y = 249.60

y = 249.6. / 3.2

y = 78

From equation 1

x + y = 175

x + 78 = 175

x = 175 - 78

x = 97

8 0
3 years ago
Does 3.9 x 6.12 equal 238.68.
STatiana [176]
23.868 but thats close one dp. off
3 0
4 years ago
What eigen value for this matix <br> (1 -2)<br> (-2 0)
natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

  1. Compute the matrix A' = A-\lambda I, where I is the identity matrix (1s on the diagonal, 0s elsewhere)
  2. Compute the determinant of A'
  3. Set the determinant of A' equal to zero and solve for lambda.

So, in this case, we have

A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]

The determinant of this matrix is

\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4

Finally, we have

\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}

So, the two eigenvalues are

\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}

5 0
3 years ago
Read 2 more answers
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