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2 years ago

What are the solutions to the system of equations? y=x2−7x+12y=−x+7

Mathematics

1 answer:

kap26 [50]2 years ago

3 0

Answer:

x = 1, y = 6

x = 5, y = 2.

Step-by-step explanation:

y=x2−7x+12

y=−x+7

Substitute for y in the first equation:

- x + 7 = x^2 - 7x + 12

x^2 - 7x + x + 12 - 7 = 0

x^2 - 6x + 5 = 0

(x - 1)(x - 5) = 0

x = 1, 5.

When x = 1, y = -1 + 7 = 6.

when x = 5, y = -5+7 = 2.

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BRAINLIEST BRAINLIEST!!! EASY MATH! Do with explanation!!

Anuta_ua [19.1K]

Answer:

Answer is given below with explanations.,

Step-by-step explanation:

$1)formula \: for \: linear \: equation \: is \: \\ ax + by + c = 0 \\ 2) \: solution \: of \: the \: equation \: y = 9x - 3 \\ (2,15) \: is \: the \: solution \: \\ on \: substitutuing \: the \: points \\ 15 = 9(2) - 3 \\ 15 = 18 - 3 \\ 15 = 15 \\ hence \: proved\\ 3) \: given \:( k,7) \: is \: a \: solution\\ \: of \: y = - 4x + 1 \\ on \: substitutuing \: the \: points \\ 7 = - 4k + 1 \\ 4k = - 7 + 1 \\ 4k = - 6 \\ k = \frac{ - 6}{4} \\ k = \frac{ - 2}{3}$

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3 years ago

The arc of the parabola y = x^2 from (3,9) to (4,16) is rotated about the y-axis. Find the area of the resulting surface. Please

Hitman42 [59]

Answer:

156.5

Step-by-step explanation:

Thinking process:

The area can be calculated using the formula:

$S = \int\limits^4_3 {2\pi x\sqrt{1+}(2x)^{2} } \, dx$

We let the substitution take place.

Therefore, we let $u = 1 + 4x^{2}$

Thus, $du = 8dx.$

So,

$xdx = \frac{1}{8}du$

Also, the interval of the integration changes to [ 37, 65]

Thus,

$S = \int\limits^4_3 {2\pi \sqrt{1+4x^{2} } } \, dx \\= \int\limits^65_35 {2\pi \sqrt{\frac{1}{8}u } } \, dx$

= $\frac{1}{6} [ 65^{\frac{3}{2}-37^{\frac{3}{2} } }$

= 156.5 units²

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3 years ago

PLEASE HELP ILL GIVE BRAINLIEST

dusya [7]

$(-2+\sqrt{-5})^2\implies (-2+\sqrt{-1\cdot 5})^2\implies (-2+\sqrt{-1}\sqrt{5})^2\implies (-2+i\sqrt{5})^2 \\\\\\ (-2+i\sqrt{5})(-2+i\sqrt{5})\implies +4-2i\sqrt{5}-2i\sqrt{5}+(i\sqrt{5})^2 \\\\\\ 4-4i\sqrt{5}+[i^2(\sqrt{5})^2]\implies 4-4i\sqrt{5}+[-1\cdot 5] \\\\\\ 4-4i\sqrt{5}-5\implies -1-4i\sqrt{5}$

let's recall that the conjugate of any pair a + b is simply the same pair with a different sign, namely a - b and the reverse is also true, let's also recall that i² = -1.

$\cfrac{6-7i}{1-2i}\implies \stackrel{\textit{multiplying both sides by the denominator's conjugate}}{\cfrac{6-7i}{1-2i}\cdot \cfrac{1+2i}{1+2i}\implies \cfrac{(6-7i)(1+2i)}{\underset{\textit{difference of squares}}{(1-2i)(1+2i)}}} \\\\\\ \cfrac{(6-7i)(1+2i)}{1^2-(2i)^2}\implies \cfrac{6-12i-7i-14i^2}{1-(2^2i^2)}\implies \cfrac{6-19i-14(-1)}{1-[4(-1)]} \\\\\\ \cfrac{6-19i+14}{1-(-4)}\implies \cfrac{20-19i}{1+4}\implies \cfrac{20-19i}{5}\implies \cfrac{20}{5}-\cfrac{19i}{5}\implies 4-\cfrac{19i}{5}$