Find the equation of the line through point (4,−7) and parallel to y=−23x+32.

Answers

Answer 1

Answer:

Answer:

y = -23x + 85

Step-by-step explanation:

for an equation to be parallel to another, they must have the same slope. So plugging in (4, -7) we get -7 = -23(4) + P. P being the new y intercept we must find. Now simple algebra add -23(4) = -92 to both sides and we get 85 = P. Plug in to y = mx + b and we get

y = -23x + 85

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16= -5+z/4
a. 69
b. -1
c. 9
d. 59

Answers

If you would like to solve the equation 16 = (- 5 + z) / 4, you can do this using the following steps:

16 = (- 5 + z) / 4 /*4
16 * 4 = - 5 + z
16 * 4 + 5 = z
z = 69

The correct result would be a. 69.

$16 = (-5+z)/(4)$

Multiply both sides by 4

$16* 4 = (-5+z)/(4) * 4$

$64 = -5+z$

Rearrange the terms

$64 = z - 5$

Add 5 on both sides.

$64 + 5 = z - 5 + 5$

$\boxed{\bf{z = 69}}$

Your final answer is a. 69.

For which system of equations is (2, 2) a solution? A.–3x + 3y = 0
x + 6y = 10B.–2x + 5y = –6
4x – 2y = 4C.5x – 2y = –6
3x – 4y = 2D.2x + 3y = 10
4x + 5y = 18

Answers

$A.)\n\n-3x + 3y = 0\nx + 6y = 10 \ / *3\n\n-3x + 3y = 0\n3x + 18y = 30\n+------\n21y =30 \ / :21\n \ny=(30)/(21)=(10)/(7)\n\nnot \ true$

$B.)\n\n-2x + 5y = -6 \ / \cdot 2\n4x - 2y = 4\n \n-4x + 10y = -12 \n4x - 2y = 4 \n+-------- \n8y=-8\ /:8\n \ny=-1\n \n not \ true$

$C.\n\n5x - 2y = -6 \ / \cdot 2\n3x - 4y = 2 \n \n 10x - 4y = -12 \n 3x- 4y = 2\n+------\n13x=-11 \ / :13\n \nx=-(11)/(13)\n\n not \ true$

$D.)\n\n2x + 3y = 10\ / \cdot (-2)\n4x + 5y = 18\n\n-4x -6y =-20\n4x + 5y = 18\n+-------\n-y = -2 \ / \cdot (-1)\n \ny=2$

$2x + 3*2 = 10\n \n2x=10-6 \n \n2x=4 \ / :2 \n \n x=2 \n \n true$

$x=2\ \ \ and\ \ \ y=2\n\nA.\ \ \ \ \ \ x + 6y = 2+6\cdot2=2+12=14 \neq 10\n\nB.\ \ \ \ \ -2x + 5y =-2\cdot2+5\cdot2=-4+10=6 \neq -6\n\nC.\ \ \ \ \ \ \ 5x-2y =5\cdot2-2\cdot2=10-4=6 \neq -6\n\nD.\ \ \ \ \ \ \ 2x + 3y =2\cdot2+3\cdot2=4+3= 10\nand\ \ \ \ \ \ 4x + 5y = 4\cdot2+5\cdot2=8+10=18\n\nAns.\ (2,\ 2)\ is\ a\ solution\ for\ system \ D$

Find the magnitude of the force needed to accelerate a 220g mass with a⃗ = -0.145m/s2 i^+0.430m/s2 j^.

Answers

Themagnitude of theforce needed to accelerate a 220g mass with a⃗ = -0.145m/s2 i^+0.430m/s2 j^. is 108.6 degrees and is located in the secondquadrant of the Cartesian. I am hoping that this answer has answered all yourqueries about this specific question.

Slope -2/3 , x-intercept at 3

Answers

Answer:

Y=(-3)x+6

Step-by-step explanation:

Situation: a 25 gram sample of a substance that’s used for drug research has a K-value of 0.1205N0(zero)= initial mass(at time t=0)
N= mass at time t
K= a positive constant that depends on the substance itself and on the units used to measure time
t= time, in days
Find the substances half-life, in days. Round your answer to the nearest tenths????

Answers

The decay model is given as

$N=N_0e^(-Kt)$

The substance's half-life is the time it takes for the substance to decay to half its original amount. In other words, it's the time $t$ such that $N=\frac{N_0}2$ . Substituting into the model, you have

$\frac12=e^(-Kt)$

You can solve for $t$ now.

$\frac12=e^(-Kt)$
$\implies\ln\frac12=\ln e^(-Kt)$
$\implies-\ln2=-Kt\ln e$
$\implies\frac{\ln2}K=t\approx5.7523\text{ days}$

Prove:
sin^2x (sec^2x + csc^2x)=sec^2x

Answers

sin^2x (sec^2x + csc^2x) = sec^2x

I would convert the functions in the parentheses to their reciprocals.

sin^2x (1/cos^2x + 1/sin^2x) = sec^2x

Now distribute the sine.

sin^2x/cos^2x + sin^2x/sin^2x = sec^2x

Remember that sine divided by cosine is always tangent.

tan^2x + sin^2x/sin^2x = sec^2x

The remaining fraction is simply 1.

tan^2x + 1 = sec^2x

Use the Pythagorean identity to add the left side.

sec^2x = sec^2x

Q.E.D.

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