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2 years ago

6

Solve the following equation for X, if Y = 3 and Z = -1. 3XY- 5XZ^2+Y=19

1 answer:

ArbitrLikvidat [17]2 years ago

5 0

Y=7-4*[4-3^3]-7888=421

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PLZ HELP ASAP TANGENTS OF CIRCLES

bezimeni [28]

First, we are given that the inscribed angle of arc CB which is angle D is equal to 65°. This is half of the measure of the arc which is equal to the measure of the central angle, ∠O.

m∠O = 2 (65°) = 130°

Also, the measure of the angles where the tangent lines and the radii meet are equal to 90°. The sum of the measures of the angle of a quadrilateral ACOB is equal to 360°.

m∠O + m∠C + m∠B + m∠A = 360°

Substituting the known values,
130° + 90° + 90° + m∠A = 360°

The value of m∠A is equal to 50°.

<em>Answer: 50°</em>

7 0

3 years ago

This is K12 in sixth grade

zmey [24]

Answer:

13

Step-by-step explanation:

divide 150 by 12

the answer is 12.5

round 12.5

;)

8 0

3 years ago

Read 2 more answers

melisa1 [442]

Answer:

$\boxed{\sf 10}$

Step-by-step explanation:

The additive number of any number is the number when added to the number gives a result of zero.

So, if we add 10 to -10 we get a result of zero.

=> -10+10

=> Zero

5 0

3 years ago

The total monthly profit for a firm is P(x)=6400x−18x^2− (1/3)x^3−40000 dollars, where x is the number of units sold. A maximum

wlad13 [49]

Answer:

Maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

Step-by-step explanation:

We are given the following information: $P(x) = 6400x - 18x^2 - \frac{x^3}{3} - 40000$ , where P(x) is the profit function.

We will use double derivative test to find maximum profit.

Differentiating P(x) with respect to x and equating to zero, we get,

$\displaystyle\frac{d(P(x))}{dx} = 6400 - 36x - x^2$

Equating it to zero we get,

$x^2 + 36x - 6400 = 0$

We use the quadratic formula to find the values of x:

$x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$ , where a, b and c are coefficients of $x^2, x^1 , x^0$ respectively.

Putting these value we get x = -100, 64

Now, again differentiating

$\displaystyle\frac{d^2(P(x))}{dx^2} = -36 - 2x$

At x = 64, $\displaystyle\frac{d^2(P(x))}{dx^2} < 0$

Hence, maxima occurs at x = 64.

Therefore, maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

6 0

3 years ago

F(x) = x^2+3; g(x) = square root of x-2 find f(g(x))

Anettt [7]

F(g(x)) = x + 1 . Just replace square root of x-2 to x in f(x)

5 0

3 years ago