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Varvara68 [4.7K]
3 years ago
15

Plz help I was in the hospital and I am behind in math now. How do I show my work for this?!

Mathematics
1 answer:
denis-greek [22]3 years ago
8 0

Answer:

-16

Step-by-step explanation:

28-(\sqrt{44} )^2\\28 - (44)\\-16

When you square a square root, the number inside is the answer.

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The area of a square photo is 9 square inches how long is each side of the photo
Marta_Voda [28]
Let the side of each side of photo is a inch.

So area of photo = (side)^2 = a^2

But as given area of photo is 9 square inch.

So a^2 = 9

a =  \sqrt{9}
          = 3

So side of photo is 3 inch.
3 0
3 years ago
Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.
Leni [432]

Answer:

a)

g(2.9) \approx -6.6

g(3.1) \approx -3.4

b)

The values are too small since g'' is positive for both values of x in. I'm speaking of the x values, 2.9 and 3.1.

Step-by-step explanation:

a)

The point-slope of a line is:

y-y_1=m(x-x_1)

where m is the slope and (x_1,y_1) is a point on that line.

We want to find the equation of the tangent line of the curve g at the point (3,-5) on g.

So we know (x_1,y_1)=(3,-5).

To find m, we must calculate the derivative of g at x=3:

m=g'(3)=(3)^2+7=9+7=16.

So the equation of the tangent line to curve g at (3,-5) is:

y-(-5)=16(x-3).

I'm going to solve this for y.

y-(-5)=16(x-3)

y+5=16(x-3)

Subtract 5 on both sides:

y=16(x-3)-5

What this means is for values x near x=3 is that:

g(x) \approx 16(x-3)-5.

Let's evaluate this approximation function for g(2.9).

g(2.9) \approx 16(2.9-3)-5

g(2.9) \approx 16(-.1)-5

g(2.9) \approx -1.6-5

g(2.9) \approx -6.6

Let's evaluate this approximation function for g(3.1).

g(3.1) \approx 16(3.1-3)-5

g(3.1) \approx 16(.1)-5

g(3.1) \approx 1.6-5

g(3.1) \approx -3.4

b) To determine if these are over approximations or under approximations I will require the second derivative.

If g'' is positive, then it leads to underestimation (since the curve is concave up at that number).

If g'' is negative, then it leads to overestimation (since the curve is concave down at that number).

g'(x)=x^2+7

g''(x)=2x+0

g''(x)=2x

2x is positive for x>0.

2x is negative for x.

That is, g''(2.9)>0 \text{ and } g''(3.1)>0.

So 2x is positive for both values of x which means that the values we found in part (a) are underestimations.

6 0
3 years ago
Part A: Factor x2y2 + 6xy2 + 8y2. Show your work. (4 points) Part B: Factor x2 + 8x + 16. Show your work. (3 points) Part C: Fac
skad [1K]

Answer:

Part A : y²(x + 2)(x + 4)

Part B: (x + 4) (x + 4)

Part C: (x + 4) (x - 4)

Step-by-step explanation:

Part A: Factor x²y²+ 6xy²+ 8y²

x²y²+ 6xy²+ 8y²

y² is very common across the quadratic equation , hence

= y² (x² + 6x + 8)

= (y²) (x² + 6x + 8)

= (y²) (x² + 2x +4x + 8)

= (y²) (x² + 2x)+(4x + 8)

= (y²) (x(x + 2)+ 4(x + 2))

= y²(x+2)(x+4)

Part B: Factor x² + 8x + 16

x² + 8x + 16

= x² + 4x + 4x + 16

= (x² + 4x) + (4x + 16)

= x( x + 4) + 4(x + 4)

= (x + 4) (x + 4)

Part C: Factor x² − 16

= x² − 16

= x² + 0x − 16

= x² + 4x - 4x - 16

= (x² + 4x) - (4x - 16)

= x (x + 4) - 4(x + 4)

= (x + 4) (x - 4)

6 0
3 years ago
What expression is equivalent to cos(1.4x)-cos(.6x)
andreev551 [17]
The answer should be:
cos (0.8 x)

BRAINLIEST PLS!!!
6 0
3 years ago
In the exercise, X is a binomial variable with n = 5 and p = 0.2. Compute the given probability. Check your answer using technol
Ronch [10]

Answer:

0.0512

Step-by-step explanation:

We are told in the question that X isa Binomial variable. Hence we solve this question, using the formula for Binomial probability.

The formula for Binomial Probability = P(x) = (n!/(n - x) x!) × p^x × q ^n - x

In the above question,

x = 3

n = 5

p = probability for success = 0.2

q = probability for failures = 1 - 0.2 = 0.8

P(x) = (n!/(n - x) x!) × p^x × q ^n - x

P(3) = (5!/(5 - 3) 3!) × 0.2^3 × 0.8^5-3

P(3) =( 5!/2! × 3!) × 0.2³ × 0.8²

P(3) = 10 × 0.008 × 0.64

P(3) = 0.0512

6 0
3 years ago
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