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3 years ago

Evaluate -8y - 4x When x = -6. y = 9

Mathematics

2 answers:

sergij07 [2.7K]3 years ago

6 0

-8(9)= -72; -4(-6)= 24; -72+24= -48

Diano4ka-milaya [45]3 years ago

6 0

Answer:

-48

Step-by-step explanation:

We are given the expression:

$-8y-4x$

and asked to evaluated when x= -6 and y=9. Therefore, we must substitute -6 in for x and 9 in for y.

$-8(9)-4(-6)$

Solve according to PEMDAS: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.

Multiply -8 and 9 ⇔ -8*9= -72

$-72-4(-6)$

Multiply -4 and -6 ⇔ -4*-6=24

$-72+24$

Add 24 to -72.

$-48$

The expression -8y-4x when evaluated at x=-6 and y=9 is equal to -48.

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Let's set W as the width of the room and L as the length of the den. We know that the den is 10 ft longer than it is wide so that means:

W + 10 = L

Let's call this equation 1.

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So now we have 2 equations and 2 unknowns. Let's take equation 1 and solve for W:

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Now we can substitute this value into equation 2.

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This looks very close to a quadratic equation. In fact, if we subtract 144 from both sides, we get the quadratic equation of:

L2 - 10L - 144 = 0

Now we can factor this equation into:

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That means the two answers for L are -8 ft and 18 ft.

We can now substitute these values for L into equation 1 to solve for W.

W = L - 10

W = 18 - 10 or W = -8 - 10

So our 2 values for W are 8 ft and -18 ft.

Let's just make sure we did this correctly. We know that the area is 144 so (-8*-18) and (18*8) should equal 144 and it does! That means we did it correctly. Give that a den exists, it does not make sense that it has negative values, so the one real answer for this problem is that the width is 8 ft and the length is 18 ft.

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Step-by-step explanation:

The parametric equation given is;

x = 1 + ln t and y = t² + 2 at (1, 3)

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