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Lunna [17]
3 years ago
5

HI. I really Need help. I will Mark Brainliest!

Mathematics
2 answers:
Reptile [31]3 years ago
4 0

Answer:

the first 1 and the last 1 i think

Step-by-step explanation:

lorasvet [3.4K]3 years ago
3 0

Answer:

I think is the first one 1/1÷ m/6  and the last 1/6 m

and if I'm correct the second one too. (m/6) (1)

Explanation:

Hope it helps!~

Good luck on your assignment!!

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M=-1/2 because you do y1-y2 over x1-x2 and end up with 2/-4 but simplified is -1/2
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At pizza hut, 9% of the pizzas made last night had extra cheese. if 27 pizzas has extra cheese, how many pizzas in all were made
True [87]
Answer: 300

You need to change 9% into a decimal so it would be 0.09

Then divide 27 ÷ 0.09 = 300

So 9% of 300 is 27
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3 years ago
(x+2/x-7) - (x^2+4x+13/x^2-4x-21)
olya-2409 [2.1K]

Answer:

x = -2.98079 or x = -1.15272 or x = 0.892002 or x = 4.24151

Step-by-step explanation:

Solve for x:

-x^2 + x + 14 + 2/x - 13/x^2 = 0

Bring -x^2 + x + 14 + 2/x - 13/x^2 together using the common denominator x^2:

(-x^4 + x^3 + 14 x^2 + 2 x - 13)/x^2 = 0

Multiply both sides by x^2:

-x^4 + x^3 + 14 x^2 + 2 x - 13 = 0

Multiply both sides by -1:

x^4 - x^3 - 14 x^2 - 2 x + 13 = 0

Eliminate the cubic term by substituting y = x - 1/4:

13 - 2 (y + 1/4) - 14 (y + 1/4)^2 - (y + 1/4)^3 + (y + 1/4)^4 = 0

Expand out terms of the left hand side:

y^4 - (115 y^2)/8 - (73 y)/8 + 2973/256 = 0

Add (sqrt(2973) y^2)/8 + (115 y^2)/8 + (73 y)/8 to both sides:

y^4 + (sqrt(2973) y^2)/8 + 2973/256 = (sqrt(2973) y^2)/8 + (115 y^2)/8 + (73 y)/8

y^4 + (sqrt(2973) y^2)/8 + 2973/256 = (y^2 + sqrt(2973)/16)^2:

(y^2 + sqrt(2973)/16)^2 = (sqrt(2973) y^2)/8 + (115 y^2)/8 + (73 y)/8

Add 2 (y^2 + sqrt(2973)/16) λ + λ^2 to both sides:

(y^2 + sqrt(2973)/16)^2 + 2 λ (y^2 + sqrt(2973)/16) + λ^2 = (73 y)/8 + (sqrt(2973) y^2)/8 + (115 y^2)/8 + 2 λ (y^2 + sqrt(2973)/16) + λ^2

(y^2 + sqrt(2973)/16)^2 + 2 λ (y^2 + sqrt(2973)/16) + λ^2 = (y^2 + sqrt(2973)/16 + λ)^2:

(y^2 + sqrt(2973)/16 + λ)^2 = (73 y)/8 + (sqrt(2973) y^2)/8 + (115 y^2)/8 + 2 λ (y^2 + sqrt(2973)/16) + λ^2

(73 y)/8 + (sqrt(2973) y^2)/8 + (115 y^2)/8 + 2 λ (y^2 + sqrt(2973)/16) + λ^2 = (2 λ + 115/8 + sqrt(2973)/8) y^2 + (73 y)/8 + (sqrt(2973) λ)/8 + λ^2:

(y^2 + sqrt(2973)/16 + λ)^2 = y^2 (2 λ + 115/8 + sqrt(2973)/8) + (73 y)/8 + (sqrt(2973) λ)/8 + λ^2

Complete the square on the right hand side:

(y^2 + sqrt(2973)/16 + λ)^2 = (y sqrt(2 λ + 115/8 + sqrt(2973)/8) + 73/(16 sqrt(2 λ + 115/8 + sqrt(2973)/8)))^2 + (4 (2 λ + 115/8 + sqrt(2973)/8) (λ^2 + (sqrt(2973) λ)/8) - 5329/64)/(4 (2 λ + 115/8 + sqrt(2973)/8))

To express the right hand side as a square, find a value of λ such that the last term is 0.

This means 4 (2 λ + 115/8 + sqrt(2973)/8) (λ^2 + (sqrt(2973) λ)/8) - 5329/64 = 1/64 (512 λ^3 + 96 sqrt(2973) λ^2 + 3680 λ^2 + 460 sqrt(2973) λ + 11892 λ - 5329) = 0.

Thus the root λ = 1/48 (-3 sqrt(2973) - 115) + 1/12 (-i sqrt(3) + 1) ((3 i sqrt(10705335) - 8327)/2)^(1/3) + (173 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(10705335) - 8327)^(1/3)) allows the right hand side to be expressed as a square.

(This value will be substituted later):

(y^2 + sqrt(2973)/16 + λ)^2 = (y sqrt(2 λ + 115/8 + sqrt(2973)/8) + 73/(16 sqrt(2 λ + 115/8 + sqrt(2973)/8)))^2

Take the square root of both sides:

y^2 + sqrt(2973)/16 + λ = y sqrt(2 λ + 115/8 + sqrt(2973)/8) + 73/(16 sqrt(2 λ + 115/8 + sqrt(2973)/8)) or y^2 + sqrt(2973)/16 + λ = -y sqrt(2 λ + 115/8 + sqrt(2973)/8) - 73/(16 sqrt(2 λ + 115/8 + sqrt(2973)/8))

Solve using the quadratic formula:

y = 1/8 (sqrt(2) sqrt(16 λ + 115 + sqrt(2973)) + sqrt(2) sqrt((10252 - 32 sqrt(2973) λ - 256 λ^2 + 292 sqrt(2) sqrt(16 λ + 115 + sqrt(2973)))/(16 λ + 115 + sqrt(2973)))) or y = 1/8 (sqrt(2) sqrt(16 λ + 115 + sqrt(2973)) - sqrt(2) sqrt((10252 - 32 sqrt(2973) λ - 256 λ^2 + 292 sqrt(2) sqrt(16 λ + 115 + sqrt(2973)))/(16 λ + 115 + sqrt(2973)))) or y = 1/8 (sqrt(2) sqrt((10252 - 32 sqrt(2973) λ - 256 λ^2 - 292 sqrt(2) sqrt(16 λ + 115 + sqrt(2973)))/(16 λ + 115 + sqrt(2973))) - sqrt(2) sqrt(16 λ + 115 + sqrt(2973))) or y = 1/8 (-sqrt(2) sqrt(16 λ + 115 + sqrt(2973)) - sqrt(2) sqrt((10252 - 32 sqrt(2973) λ - 256 λ^2 - 292 sqrt(2) sqrt(16 λ + 115 + sqrt(2973)))/(16 λ + 115 + sqrt(2973)))) where λ = 1/48 (-3 sqrt(2973) - 115) + 1/12 (-i sqrt(3) + 1) ((3 i sqrt(10705335) - 8327)/2)^(1/3) + (173 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(10705335) - 8327)^(1/3))

Substitute λ = 1/48 (-3 sqrt(2973) - 115) + 1/12 (-i sqrt(3) + 1) ((3 i sqrt(10705335) - 8327)/2)^(1/3) + (173 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(10705335) - 8327)^(1/3)) and approximate:

y = -3.23079 or y = -1.40272 or y = 0.642002 or y = 3.99151

Substitute back for y = x - 1/4:

x - 1/4 = -3.23079 or y = -1.40272 or y = 0.642002 or y = 3.99151

Add 1/4 to both sides:

x = -2.98079 or y = -1.40272 or y = 0.642002 or y = 3.99151

Substitute back for y = x - 1/4:

x = -2.98079 or x - 1/4 = -1.40272 or y = 0.642002 or y = 3.99151

Add 1/4 to both sides:

x = -2.98079 or x = -1.15272 or y = 0.642002 or y = 3.99151

Substitute back for y = x - 1/4:

x = -2.98079 or x = -1.15272 or x - 1/4 = 0.642002 or y = 3.99151

Add 1/4 to both sides:

x = -2.98079 or x = -1.15272 or x = 0.892002 or y = 3.99151

Substitute back for y = x - 1/4:

x = -2.98079 or x = -1.15272 or x = 0.892002 or x - 1/4 = 3.99151

Add 1/4 to both sides:

Answer: x = -2.98079 or x = -1.15272 or x = 0.892002 or x = 4.24151

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If you wish to warm 40 kg of water by 18 ∘C for your bath, find what the quantity of heat is needed. Express your answer in calo
Harman [31]

Answer:

Quantity of heat needed (Q) = 722.753 × 10³

Step-by-step explanation:

According to question,

Mass of water (m) = 40 kg

Change in temperature ( ΔT) = 18°c

specific heat capacity of water = 4200 j kg^-1 k^-1

The specific heat capacity is the amount of heat required to change the temperature of 1 kg of substance to 1 degree celcius or 1 kelvin .

So, Heat (Q) = m×s×ΔT

Or,          Q = 40 kg × 4200 × 18

or,           Q = 3024 × 10³ joule

Hence, Quantity of heat needed (Q) = 3024 × 10³ joule    

In calories 4.184 joule = 1 calorie

So, 3024 × 10³ joule = 722.753 × 10³

6 0
3 years ago
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