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NikAS [45]
3 years ago
10

Find the X intercept and the Y intercept of the graph of the equation. 9/8x+8y=18

Mathematics
2 answers:
aleksandr82 [10.1K]3 years ago
8 0

Answer:

\large\boxed{x-intercept=16\to(16,\ 0)}\\\boxed{y-intercept=\dfrac{9}{4}\to\left(0,\ \dfrac{9}{4}\right)}

Step-by-step explanation:

\dfrac{9}{8}x+8y=18\\\\x-intercept\ \text{is for}\ y=0:\\\\\dfrac{9}{8}x+8(0)=18\\\\\dfrac{9}{8}x+0=18\\\\\dfrac{9}{8}x=18\qquad\text{multiply both sides by}\ \dfrac{8}{9}\\\\\dfrac{8\!\!\!\!\diagup^1}{9\!\!\!\!\diagup_1}\cdot\dfrac{9\!\!\!\!\diagup^1}{8\!\!\!\!\diagup_1}x=\dfrac{8}{9\!\!\!\!\diagup_1}\cdot18\!\!\!\!\!\diagup^2\\\\x=16\\\\y-intercept\ \text{is for}\ x=0:\\\\\dfrac{9}{8}(0)+8y=18\\\\0+8y=18\\\\8y=18\qquad\text{divide both sides by 8}\\\\y=\dfrac{18}{8}\\\\y=\dfrac{9}{4}

joja [24]3 years ago
7 0

Answer:

(16, 0) and (0, \frac{9}{4})

Step-by-step explanation:

Given

\frac{9}{8} x + 8y = 18

Multiply through by 8 to eliminate the fraction

9x + 64y = 144

When the graph crosses the x- axis the y- coordinate of the point is zero.

Let y = 0 in the equation and solve for x

9x +0 = 144

9x = 144 ( divide both sides by 9 )

x = 16 ← x - intercept ⇒ (16, 0 )

When the graph crosses the y- axis the x- coordinate of the point is zero

let x = 0 in the equation and solve for y

0 + 64y = 144

64y = 144 ( divide both sides by 64 )

y = \frac{144}{64} = \frac{9}{4} ← y- intercept ⇒ (0, \frac{9}{4} )

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mote1985 [20]

Answer:

x is greater than or equal to -5 (the second graph)

Step-by-step explanation:

-5c+2<=27

-5x<=25

x>=-5

8 0
3 years ago
What is 5au + 24av - 5bu - 30bv​
Blizzard [7]

Answer:

59

Step-by-step explanation:

5au+24av-5bu-30bv

rearrange it

5au+5bu-24av-30bv

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5a+5b-24a-30b

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5a+24a-5b-30b

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29- -30=59

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5 0
2 years ago
Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |
dsp73

Looks like f(x,y)=x^2+y^2-x^2y+7.

f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1

f_y=2y-x^2=0\implies2y=x^2

  • If x=0, then y=0 - critical point at (0, 0).
  • If y=1, then x=\pm\sqrt2 - two critical points at (-\sqrt2,1) and (\sqrt2,1)

The latter two critical points occur outside of D since |\pm\sqrt2|>1 so we ignore those points.

The Hessian matrix for this function is

H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}

The value of its determinant at (0, 0) is \det H(0,0)=4>0, which means a minimum occurs at the point, and we have f(0,0)=7.

Now consider each boundary:

  • If x=1, then

f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4

which has 3 extreme values over the interval -1\le y\le1 of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

  • If x=-1, then

f(-1,y)=8-y+y^2

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

  • If y=1, then

f(x,1)=8

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

  • If y=-1, then

f(x,-1)=2x^2+8

which has 3 extreme values, but the previous cases already include them.

Hence f(x,y) has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

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