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Arlecino [84]
3 years ago
10

What is the slope of the line that contains the points (1,-1) and (-2,8)? (Also can you show me how you got it please?)​

Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
3 0
The answer is -3

You use the formula y1-y2
———
x1-x2
-1 and 8 are your y’s
1 and -2 are your x’s

Your equation should look like this
-1-8 -9
—— = —— = -3
1-(-2) 3

I plugged the y’s and x’s into the formula.
I hope this helps!!
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Can someone solve this please?
liq [111]
Multiply every number from the small octagon by 7 then add the  products to get 238.
4 0
3 years ago
Read 2 more answers
Lastima Nelson, Inc. is recruiting a new Chief Financial Officer (CFO). The recruiting budget is $44,000. Lastima has spent $10,
Musya8 [376]
The budget is $44000

Total spent so far is $10000 + $8500 = $18500

The amount left to spend = 44000 - 18500 = $25500

$25500 is the maximum 6% commission that Lastima Nelson Inc. can spend and stays within the budget

Let the maximum price of the home be x
This value of x will be the 100% before the 6% commission is calculated of it.

6% of x is 25500
1% of x is 25500 ÷ 6 = $4250
100% of x is 4250 × 100 = $425,000

So, the maximum value of the home is $425,000
5 0
3 years ago
What is the coefficient of the x^5y^5- term in the biomial expansion of (2x-3y)^10
jasenka [17]

<u>Answer:</u>

 The coefficient of x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552

<u>Solution: </u>

The given expression is (2 x-3 y)^{10}

As per binomial theorem, we know,

(x+y)^{n}=\sum n C_{k} x^{n-k} y^{k}

Now here a = 2x, b = (- 3y) and n = 10 and k = 0,1,2,….10

Now x^{5}\times y^5 will be the 6 term where k =5

Now, \mathrm{T}_{6}=10 \mathrm{C}_{5} \times(2 \mathrm{x})^{(10-5)} \times(-3 \mathrm{y})^{5}=10 \mathrm{C}_{5} 2^{5} \times \mathrm{x}^{5} \times(-3)^{5} \times \mathrm{y}^{5}

So, the coefficient of x^{5} \times y^{5} \text { is }=10\left(5 \times 2^{5} \times(-3)^{5}\right.

10 \mathrm{C}_{5}=\frac{10 !}{5 ! \times(10-5) !}=\frac{10 !}{5 !+5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=\left(\frac{30240}{120}\right)=252

The coefficient of x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552

6 0
3 years ago
Can someone please help me​
astra-53 [7]

Answer:

square 20 has 44 green squares

square 21 has 45 green squares

Step-by-step explanation:

To solve the problem, we need to observe the cases, and determine/define a rule for each case (odd number of sides, or even number of sides).

For square one, we note that the centre square is shared by two diagonals, so we saved one square from the two diagonals.

The side length is 3 for square 1, 4 for square 2, and so on.

Let

n= square number (1, 2,3...)

L = side length (3,4,5...)

G1(n) = function that gives the number of green squares for square n, n=odd

G2(n) = function that gives the number of green squares for square n, n=even

side length, L=n+2   ................(1)

G1(n) = twice the side length less one, as discussed above

G1(n) = 2L-1       now substitute L=n+2

G1(n) = 2(n+2) -1    simplify

G1(n) = 2n + 3

Check:

for n=1, square 1 has 2*1+3 = 5 green squares ... checks

for n=3, square 3 has 2*3+3 = 9... checks

for n=5, square 5 has 2*5+3 = 13 ....checks

For even squares, it is even easier, because

G2(n) = 2L = 2(n+2)

check:

for n=2, square 2 has 2(2+2) = 8 green squares........checks

for n=4, square 4 has 2(4+2) = 12 green squares........checks.

Fincally, we apply our formula to n=20 and n=21

square 20 : G2(20) = 2(n+2) = 2(20+2) = 44 green quars

square 21 : G1(21) = 2n+3 = 2(21)+3 = 45 green squares

5 0
2 years ago
Find value of x of question number F ​
a_sh-v [17]

Answer:

X=17

Step-by-step explanation:

AC^2=AB^2 +BC^2(PYTHAGOREAN THEORM)

100=64 +BC^2

36=BC^2

6CM=BC

AD^2=AB^2+BD^2(PYTHAGOREAN THEORM)

X^2 =64+225

X^2=289

X=17

6 0
2 years ago
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