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3 years ago

Evaluate the expression when x = -1 and y = 4.

Mathematics

2 answers:

Sedaia [141]3 years ago

6 0

Answer: -9/6

Step-by-step explanation:

Substitute the value of X and Y into the equation. Therefore

X-2y

When X = -1 and when Y= 4

-1-2(4)

-1-8=-9

Y-2x

When X= -1 and when Y=4

4-2(-1)

When multiplying 2 negatives it becomes a positive so...

4+2=6

-9,6 would be your answers... hope this helps you!!

Pani-rosa [81]3 years ago

5 0

Answer:

- 7/2 i'm pretty sure

Step-by-step explanation:

substitute x for - 1 and substitute y for 4.

x - 2y = - 1 - 2 (4)

y - 2x = 4 - (2 -1)

then do the math for both of them and it should either be -7/2 or 7/2

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Find x and y if angle one is 2x+10 and angle two is 4y-30

Nostrana [21]

I hope this helps you

5 0

3 years ago

a 50m long chain hangs vertically from a cunlinder attached to a winch. Assume there is no friction in the system and that the c

Finger [1]

Answer:

147000 J

Step-by-step explanation:

We are given that

Length of chain=L=50 m

Density of chain= $\rho=10kg/m^3$

We have to find the work done required to wind the chain into the cylinder if a 50 kg block is attached to the end of the chain.

Work done= $\int_{a}^{b}F(y)dy$

We have F(y)= $\rho g(50-y)dy$

a=0 and b=50

$g=9.8m/s^2$

Using the formula

Work done= $w_1=10\times 9.8\int_{0}^{50}(50-y)dy$

Where Length of chain is (50-y) has to be lifted.

Work done= $w_1=10\times 9.8[50y-\frac{y^2}{2}]^{50}_{0}$

By using the formula $\int x^ndx=\frac{x^{n+1}}{n+1}+C$

Work done= $w_1=10\times 9.8\times (50(50)-\frac{(50)^2}{2})=98\times (2500-1250)=122500 J$

When the chain is weightless then the work done required to lift the block attached to the 50 m long chain

Again using the formula

Where f(y)=mg

$w_2=\int_{0}^{50}mgdy$

We have m=50 kg

$w_2=\int_{0}^{50}50\times 9.8 dy=490[y]^{50}_{0}=490\times 50=24500 J$

The work done required to wind the chain into the cylinder if a 50 kg block is attached to the end of the chain= $w_1+w_2=122500+24500=147000 J$

4 0

4 years ago

H = 13 units and r = 4 units, then what is the approximate volume of the cone shown above?

mr Goodwill [35]

Answer:

The approximate volume of the cone = 218 units³

Step-by-step explanation:

<u>Points to remember</u>

Volume of cone = (1/3)πr²h

Where r is the radius and h the height of the cone

<u>To find the volume of given cone</u>

Here r = 4 units and h = 13 units

Volume = (1/3)πr²h

= (1/3) * 3.14 * 4² * 13

= (1/3) * 3.14 * 16 * 13

= 217.71 ≈ 218 units³

5 0

3 years ago

What is f(x)=8x^2+4x written in vertex form

nika2105 [10]

Answer:

f(x) = 8 (x + ¼)² − ½

Step-by-step explanation:

f(x) = 8x² + 4x

Divide both sides by 8.

1/8 f(x) = x² + 1/2 x

Take half of the second coefficient, square it, then add to both sides.

(½ / 2)² = (1/4)² = 1/16

1/8 f(x) + 1/16 = x² + 1/2 x + 1/6

Factor the perfect square.

1/8 f(x) + 1/16 = (x + 1/4)²

Multiply both sides by 8.

f(x) + 1/2 = 8 (x + 1/4)²

Subtract 1/2 from both sides.

f(x) = 8 (x + 1/4)² − 1/2

5 0

3 years ago

There are 28 students in Mr. Connelly's Pre-Calculus class.

enot [183]

The correct answer is 8

3 0

3 years ago