Can someone please quickly help me out.

The side of a square are 3 cm long. One vertex of the square is at (2,0) on a square coordinate grid marked in centimeters units

yuradex [85]

(2,0) (5.0) (5,3) (2,3)
(2,0) (-1,0) (2,3) (-1,3)

3 0

3 years ago

I need help! I'm *so* confused! I will mark Brainliest

Naddik [55]

Answer:

(-2,21)

Step-by-step explanation:

2 days ago, Paulo's commute time was halfway between his commute times 8 and 9 days ago.

8 days ago means that x-coordinate -8 represents this day. Count 8 units to the left and find that when x = -8, y = 24 minutes.

9 days ago means that x-coordinate -9 represents this day. Count 9 units to the left and find that when x = -9, y = 18 minutes.

Find halfway commute time between points (-8,24) and (-9,18):

$y=\dfrac{24+18}{2}=\dfrac{42}{2}=21,$

then coordinates that Paulo should graph are

(-2, 21)

(-2 means 2 days ago, 21 is the halfway)

5 0

3 years ago

Pls help solve!!!!!!!!

Svetradugi [14.3K]

Its also 40°30' because the median is also a bisector in isosceles triangle

5 0

3 years ago

A furniture store received an order for 8,367 tables they can fit 7 tables in the lower shipping box how many shipping boxes whe

-BARSIC- [3]

1195 boxes are needed to ship all the tables

<em><u>Solution:</u></em>

Given that, furniture store received an order for 8,367 tables

They can fit 7 tables in the lower shipping box

To find: Number of shipping boxes needed to ship all the tables

From given,

Total number of tables = 8367

Number of tables fit in one shipping box = 7

Thus, we can find the number of shipping boxes needed to ship all the tables by dividing total number of tables by number of tables fit in one shipping box

$\text{Number of shipping boxes needed } = \frac{\text{Total number of tables}}{\text{Number of table fit in one box}}$

Substituting the values we get,

$\text{Number of shipping boxes needed } = \frac{8367}{7}=1195.28 \approx 1195$

Thus 1195 boxes are needed to ship all the tables

8 0

3 years ago

Use the Binomial Theorem/Pascal's Triangle to expand (2a + 2b)^5

pochemuha

Answer:

32a^5+160a^4b+320a^3b^2+320a^2b^3+160ab^4+32b^5

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Explanation:

Let's use Pascal's Triangle

In the row that starts with 1,5,... we have the values 1,5,10,10,5,1

These will be the coefficients of the terms.

Let x = 2a and y = 2b

We want to expand out (x+y)^5

Using pascals triangle, we get the following expansion

(x+y)^5 = 1x^5y^0+5x^4y^1+10x^3y^2+10x^2y^3+5x^1y^4+1x^0y^5

The numbers in bold are the coefficients 1,5,10,10,5,1 found earlier.

Note how the exponents for x start at 5 and count down to 0; while the y exponents start at 0 and count up to 5. For any term, the x and y exponents always add to 5 for the expansion of (x+y)^5. In general, the exponents of any term will add to n for (x+y)^n.

At this point, we plug in x = 2a and y = 2b

Since this will clutter things a bit, I'll do it term by term

1x^5y^0 = 1(2a)^5(2b)^0 = 1(32a^5)(1) = 32a^5
5x^4y^1 = 5(2a)^4(2b)^1 = 5(16a^4)(2b) = 160a^4b
10x^3y^2 = 10(2a)^3(2b)^2 = 10(8a^3)(4b^2) = 320a^3b^2
10x^2y^3 = 10(2a)^2(2b)^3 = 10(4a^2)(8b^3) = 320a^2b^3
5x^1y^4 = 5(2a)^1(2b)^4 = 5(2a)(16b^4) = 160ab^4
1x^0y^5 = 1(2a)^0(2b)^5 = 1(1)(32b^5) = 32b^5

So in the end, the expression (2a+2b)^5 expands out to

32a^5+160a^4b+320a^3b^2+320a^2b^3+160ab^4+32b^5

The binomial theorem uses the same basic idea, but instead of using Pascal's Triangle to get the coefficients, you'll use the nCr combination formula.

8 0

2 years ago