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DaniilM [7]
3 years ago
15

PLEASE HELP ASAP FOR 10 POINTS

Mathematics
1 answer:
victus00 [196]3 years ago
7 0
A^2 +a-6. You use the method of foiling where you multiply the first two terms, the the outside terms, the inside terms, then the last terms. You would do a•a, then a•-2 the a•3 then 3•-2. You would get a^2 -2a +3a -6. Then combine like terms to get your answer.
You might be interested in
4x + 5 is greater than 11
erastova [34]

Answer:

if your asking if x is greater than 11 it will be false

Step-by-step explanation:

8 0
4 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
4 years ago
I got this far in my working out and am already stuck. have i gone wrong?
Kipish [7]

Yes, you made an algebraic error. By multiplying 18 x 4, you messed up on order of operations.

(x+9)(x+2)*4

becomes:

(x^2+11x+18)*4

= 4x^2 + 44x + 72


anyway, this is how I would do this problem:

(x+9)(x+2)(4) = 912

(x+9)(x+2) = 228

x^2 + 11x + 18 = 228

x^2 + 11x - 220 = 0

(x+21)(x-10) = 0

x = -21 or 10

The dimensions can't be negative, so we only use x = 10

The dimensions are then 10+9, 10+2, 4

or 19, 12, 4

6 0
3 years ago
2х - Зу = 32<br> x + 4у = -20<br><br><br> solve the system of equations
ANTONII [103]
Answer: x=68/11 and y=-72/11


Steps:

2(x+4y=-20)

2x+8y=-40

2x - 3y = 32
- 2x + 8y = -40

-11y=72

y=-72/11


x + 4(-72/11) = -20

x - 288/11 = -20

x=68/11
5 0
3 years ago
Can someone PLEASE help me on this Geometry question!!!
grandymaker [24]
What is your geometry question?
8 0
3 years ago
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