All categories

3 years ago

HELP I CANT DO MATH PLEASE

Mathematics

2 answers:

zavuch27 [327]3 years ago

5 0

What he said tbh he did a paragraph

PSYCHO15rus [73]3 years ago

4 0

Answer:

A' : (-5, -8)

B' : (-5, -1)

C' : (2, -1)

Step-by-step explanation:

A is (-5, 8), B is (-5, 1), C is (2, 1), and we want to find the points after we reflect the triangle over x.

The first thing to remember is that when points are written like (x, y), the first number, x, is how many units you go right or left and the second number, y, is how many units you go up or down. The second thing to keep in mind is that the x-axis is the horizontal one that runs from left to right.

Now, just think about the information we have: if we want to flip a point like (-5, 8) over the x-axis, all we have to do is make the 8 negative. This is because the 8 is how far up the point is, so to make it flip down over the x-axis, we just change the sign. The -5 does NOT change because it's the x-value, which is how far right/left the point is. Does that make sense?

If we do this for the other two, (-5, 1) becomes (-5, -1) and (2, 1) becomes (2, -1).

(The apostrophe-like symbol the problem uses--like A', B', C'--just denotes that those are the resulting points after the reflection.)

Please mark as Brainliest! :)

You might be interested in

The market and Stock J have the following probability distributions:

denis-greek [22]

Answer:

1) $E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%$

2) $E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%$

3) $E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1$

And the variance would be given by:

$Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89$

And the deviation would be:

$Sd(M) = \sqrt{13.89}= 3.73$

4) $E(J^2) = 22^2*0.3 + 4^2*0.4 + 12^2*0.3 =194.8$

And the variance would be given by:

$Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56$

And the deviation would be:

$Sd(M) = \sqrt{55.56}= 7.45$

Step-by-step explanation:

For this case we have the following distributions given:

Probability M J

0.3 14% 22%

0.4 10% 4%

0.3 19% 12%

Part 1

The expected value is given by this formula:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%$

Part 2

$E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%$

Part 3

We can calculate the second moment first with the following formula:

$E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1$

And the variance would be given by:

$Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89$

And the deviation would be:

$Sd(M) = \sqrt{13.89}= 3.73$

Part 4

We can calculate the second moment first with the following formula:

$E(J^2) = 22^2*0.3 + 4^2*0.4 + 12^2*0.3 =194.8$

And the variance would be given by:

$Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56$

And the deviation would be:

$Sd(M) = \sqrt{55.56}= 7.45$

8 0

3 years ago

Please help!!!!!!! How many inches would a point on the outer edge of the large gear travel in a 150º rotation? (The gear has a

damaskus [11]

In a full rotation, 360°, a point on the edge of a gear would travel a distance equal to its circumference. The circumference of a circle is:

C=2πr, and since we are only traveling 150°, we need to set up an appropriate ratio for circumference to the distance the point travels:

d/C=150/360

d=5C/12 and since C=2πr

d=10πr/12

d=5πr/12, and since r=4in

d=5*4π/12 in

d=20π/12 in

d=5π/3 in

d≈5.24 in (to nearest hundredth of an inch)

4 0

2 years ago

The standard form of the equation of a circle with center (h, k) and radius r is __________________.

Naddika [18.5K]

$\bf \textit{equation of a circle}\\\\ 
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
\begin{array}{lllll}
center\ (&{{ h}},&{{ k}})\qquad 
radius=&{{ r}}
\end{array}$