why was everyone laughing? And if she said “after the ban”, does that mean past tense or future tense? Please explain

Brums [2.3K]

The answer is B. Parabola is a symmetrical open plane bend framed by the crossing point of a cone with a plane parallel to its side. The way of a shot affected by gravity in a perfect world takes after a bend of this shape.

6 0

3 years ago

Read 2 more answers

Does anyone know how to solve this?

Lilit [14]

The pattern is that the numbers in the right-most and left-most squares of the diamond add to the bottom square and multiply to reach the number in the top square.

For example, in the first given example, we see that the numbers 5 and 2 add to the number 7 in the bottom square and multiply to the number 10 in the top square.

Another example is how the numbers 2 and 3 in the left-most and right-most squares add up to the number 5 in the bottom square and multiply to the number 6 in the top square.

Using this information, we can solve the five problems on the bottom of the paper.

a) We are given the numbers 3 and 4 in the left-most and right-most squares. We must figure out what they add to and what they multiply to:

3 + 4 = 7

3 x 4 = 12

Using this, we can fill in the top square with the number 12 and the bottom square with the number 7.

b) We are given the numbers -2 and -3 in the left-most and right-most squares, which again means that we must figure out what the numbers add and multiply to.

(-2) + (-3) = -5

(-2) x (-3) = 6

Using this, we can fill the top square in with the number 6 and the bottom square with the number -5.

c) This time, we are given the numbers which we typically find by adding and multiplying. We will have to use trial and error to find the numbers in the left-most and right-most squares.

We know that 12 has the positive factors of (1, 12), (2,6), and (3,4). Using trial and error we can figure out that 3 and 4 are the numbers that go in the left-most and right-most squares.

d) This time, we are given the number we find by multiplying and a number in the right-most square. First, we can find the number in the left-most square, which we will call $x$ . We know that $\frac{1}{2}x = 4$ , so we can find that $x$ , or the number in the left-most square, is 8. Now we can find the bottom square, which is the sum of the two numbers in the left-most and right-most squares. This would be $8 + \frac{1}{2} = \frac{17}{2}$ . The number in the bottom square is $\boxed{\frac{17}{2}}$ .

e) Similar to problem c, we are given the numbers in the top and bottom squares. We know that the positive factors of 8 are (1, 8) and (2, 4). However, none of these numbers add to -6, which means we must explore the negative factors of 8, which are (-1, -8), and (-2, -4). We can see that -2 and -4 add to -6. The numbers in the left-most and right-most squares are -2 and -4.

4 0

3 years ago

In a certain quiz that consists of 10 questions, each question after the first is worth 4 points more than the preceding questio

weqwewe [10]

Answer:

C. 26

Step-by-step explanation:

Let’s suppose that the first question is worth ‘’x’’ amount of points. Then the second question will be worth x+4, the third question x+8, the fourth question x+12, the fifth x+ 16, the sixth x+20, the seventh x+24, the eighth x+ 28, the ninth x+ 32, and the last one x + 36.

The sum of the points of all these questions is equal to 10x + 180. Since the quiz has a total of 360 points, then

10x + 180 = 360.

Solving for x we get:

X = 18.

This means that the first question is worth 18 points, the second one 22 points (18 +4), the third one 26 points(18+4+4), and so on.

8 0

2 years ago

Can someone help me :(

k0ka [10]

Answer:

x = 20

Step-by-step explanation:

19 < x < 25

x had to be higher than 19 but lower than 25

18 cant be the answer because it is less than 19

27 cant be the answer because it is higher than 25

that leaves 20, which is the right answer

19 < 20 < 25

4 0

2 years ago

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

<u>Step-by-step explanation:</u>

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

<u>Proof LHS → RHS</u>

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$

$\text{Identity:}\qquad \qquad \dfrac{\sin^2 A}{1-\cos A}$

$\text{LHS = RHS:}\quad \dfrac{\sin^2 A}{1-\cos A}=\dfrac{\sin^2 A}{1-\cos A}\quad \checkmark$

3 0

3 years ago