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3 years ago

1x-4=2xanwser need the written equation

1 answer:

4 0

1x - 4 = 2x
(minus the 1x from the left to cancel it out and minus it to other side making the 2x a 1x or just x)⤵️

-4 = x

and that is the answer

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Pablo Picasso’s Head of a Women, is an example of a(n) ______________.

lina2011 [118]

Answer:

Sculpture in the round.

Step-by-step explanation:

Option A, sculpture in the round, is the right answer.

This is Picasso’s initial Cubist sculpture illustrations the master made of his girlfriend Fernande Olivier in the year 1909. After the return of the couple from a tour to Spain Picasso created the bust in Paris. As in his old Cubist pictures, the frame of her sculpted head is faceted into miniature pieces. Meant to be viewed in the round, the production switches form when seen from various corners. Therefore, it is clear that Pablo Picasso’s "Head of a Women" was a sculpture in the round.

8 0

3 years ago

Read 2 more answers

A bricklayer is able to set 2.5 bricks in one minute. How many bricks can he set in 8 hours?

pochemuha

Given:

A bricklayer is able to set 2.5 bricks in one minute.

Required:

To find the number of bricks can he set in 8 hours.

Explanation:

8 hours =480 minutes.

2.5 bricks in one minute.

So for 480 minutes,

$\begin{gathered} =2.5\times480 \\ \\ =1200 \end{gathered}$

Final Answer:

1,200 bricks can he set in 8 hours.

5 0

1 year ago

Can someone please help!!

anastassius [24]

X=f
x×1+1-2+3=3234557901028292

8 0

2 years ago

.. Which of the following are the coordinates of the vertices of the following square with sides of length a?

atroni [7]

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Step-by-step explanation:

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

To find the sides of a square, let us use the distance formula,

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } S T=\sqrt{(a-0)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } T W=\sqrt{(a-a)^{2}+(0-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,a), T(a,a), W(a,0)$ has sides of length a.

Option B: $O(0,0), S(0,a), T(2a,2a), W(a,0)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text {Length } S T=\sqrt{(2 a-0)^{2}+(2 a-a)^{2}}=\sqrt{5 a^{2}}=a \sqrt{5}\\&\text {Length } T W=\sqrt{(a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{2 a^{2}}=a \sqrt{2}\\&\text {Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

This is not a square because the lengths are not equal.

Option C: $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length OS }=\sqrt{(0-0)^{2}+(2 a-0)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } S T=\sqrt{(2 a-0)^{2}+(2 a-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } T W=\sqrt{(2 a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } O W=\sqrt{(2 a-0)^{2}+(0-0)^{2}}=\sqrt{4 a^{2}}=2 a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$ has sides of length 2a.

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length OS }=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } S T=\sqrt{(a-a)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } T W=\sqrt{(0-a)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } O W=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

Thus, the square with vertices $O(0,0), S(a,0), T(a,a), W(0,a)$ has sides of length a.

Thus, the correct answers are option a and option d.

8 0

2 years ago

There are 4 green marbles and 2 red marbles in the jar. You just randomly draw one by one without replacement and stop when you

zalisa [80]

Answer:

Then the probability distribution is:

P(0) = 1/3

P(1) = 4/15

P(2) = 1/5

P(3) = 2/15

P(4) = 1/15

The expected value for X is:

EV = 1.33...

Step-by-step explanation:

We have a total of 6 marbles in the jar.

The probability of getting a red marble in the first try (X = 0) is equal to the quotient between the number of red marbles and the total number of marbles, this is:

P(0) = 2/6 = 1/3

The probability of drawing one green marble (X = 1)

is:

First, you draw a green marble with a probability of 4/6

Then you draw the red one, but now there are 5 marbles in the jar (2 red ones and 3 green ones), then the probability is 2/5

The joint probability is:

P(1) = (4/6)*(2/5) = (2/3)*(2/5) = 4/15

The probability of drawing two green marbles (X = 2)

Again, first we draw a green marble with a probability of 4/6

Now we draw again a green marble, now there are 3 green marbles and 5 total marbles in the jar, so this time the probability is 3/5

Now we draw the red marble (there are 2 red marbles and 4 total marbles in the jar), with a probability of 2/4

The joint probability is:

P(2) = (4/6)*(3/5)*(2/4) = (2/6)*(3/5) = 1/5

The probability of drawing 3 green marbles (X = 3)

At this point you may already understand the pattern:

First, we draw a green marble with a probability 4/6

second, we draw a green marble with a probability 3/5

third, we draw a green marble with a probability 2/4

finally, we draw a red marble with a probability 2/3

The joint probability is:

P(3) = (4/6)*(3/5)*(2/4)*(2/3) = (2/6)*(3/5)*(2/3) = (1/5)*(2/3) = (2/15)

Finally, the probability of drawing four green marbles (X = 4) is given by:

First, we draw a green marble with a probability 4/6

second, we draw a green marble with a probability 3/5

third, we draw a green marble with a probability 2/4

fourth, we draw a green marble with a probability 1/3

Finally, we draw a red marble with a probability 2/2 = 1

The joint probability is:

P(4) = (4/6)*(3/5)*(2/4)*(1/3)*1 = (1/5)*(1/3) = 1/15

Then the probability distribution is:

P(0) = 1/3

P(1) = 4/15

P(2) = 1/5

P(3) = 2/15

P(4) = 1/15

The expected value will be:

EV = 0*P(0) + 1*P(1) + 2*P(2) + 3*P(3) + 4*P(4)

EV = 1*(4/15) + 2*( 1/5) + 3*( 2/15) + 4*(1/15 ) = 1.33

So we can expect to draw 1.33 green marbles in this experiment.

5 0

3 years ago