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2 years ago

Find AB if BC = 4, BD = 5, and AD = 2. show your work.

Mathematics

1 answer:

miskamm [114]2 years ago

7 0

Solving this system of equations without a CAS: $4=a+c$, $5=d+ac+b$, $4=ad+bc$, $4=bd$ 2

Let $A,B,C,D$ be the vertices of a four sided polygon taken in anti clockwise. Given $|AB|=|BC|=3,|AD|=|CD|=4,|BD|=5$ , Find $|AC|$ 1

Find $ad - bc$ if $ac = 18$ and $bd = 50$ 2

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3 years ago

I need to know the measurement of the fourth angle

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Answer:

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2 years ago

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Given the linear programming problem, use the method of corners to determine where the minimum occurs and give the minimum value

algol [13]

Answer:

Minimum value of function $C=x+10y$ is 63 occurs at point (3,6).

Step-by-step explanation:

To minimize :

$C=x+10y$

Subject to constraints:

$x\leq 3---(1)\\y\leq 9---(2)\\x+y\geq 9----(3)\\x\geq 0\\y\geq 0$

Eq (1) is in blue in figure attached and region satisfying (1) is on left of blue line

Eq (2) is in green in figure attached and region satisfying (2) is below the green line

Considering $x+y\geq 9$ , corresponding coordinates point to draw line are (0,9) and (9,0).

Eq (3) makes line in orange in figure attached and region satisfying (3) is above the orange line

Feasible region is in triangle ABC with common points A(0,9), B(3,9) and C(3,6)

Now calculate the value of function to be minimized at each of these points.

$C=x+10y$

at A(0,9)

$C=0+10(9)\\C=90$

at B(3,9)

$C=3+10(9)\\C=93$

at C(3,6)

$C=3+10(6)\\C=63$

Minimum value of function $C=x+10y$ is 63 occurs at point C (3,6).

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3 years ago

Which table contains points that lie on a line with a slope of −2?

o-na [289]

C, or the top right table. I don’t exactly feel like explaining unless you really want me to... in that case lmk.

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2 years ago

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At a high school, students can choose between three art electives, four history electives, and five computer electives. Each stu

V125BC [204]

Answer:

$\frac{^3C_1\times ^4C_1}{^{12}C_2}$

Step-by-step explanation:

Given,

Art electives = 3,

History electives = 4,

Computer electives = 5,

Total number of electives = 3 + 4 + 5 = 12,

Since, if a student chooses an art elective and a history elective,

So, the total combination of choosing an art elective and a history elective = $^3C_1\times ^4C_1$

Also, the total combination of choosing any 2 subjects out of 12 subjects = $^{12}C_2$

Hence, the probability that a student chooses an art elective and a history elective = $\frac{\text{Total combination of choosing an art elective and a history}}{\text{ Total combination of choosing any 2 subjects}}$

$=\frac{^3C_1\times ^4C_1}{^{12}C_2}$

Which is the required expression.

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3 years ago

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