Ask question

Ask question

All categories

2 years ago

15

Find the minimum or maximum value of the function g(x)=−4x^2+16x−15

2 answers:

lara31 [8.8K]2 years ago

8 0

Answer:

(-2, -31)

Extra Letters.

valkas [14]2 years ago

3 0

I got (2,1)

but the maximum/minimum will always be the y

so the why in that case is 1

You might be interested in

this week is a struggle for me i may have to have a tutor . but here is my problem 2x-3+3x= -28 what is th value of x?

pantera1 [17]

Answer:

x = -5

Step-by-step explanation:

2x-3+3x= -28

combine like terms

5x -3 = -28

add 3 to each side

5x-3+3 = -28+3

5x = -25

divide each side by 5

5x/5 = -25/5

x = -5

8 0

3 years ago

What is the probability that the sample mean life span of 64 calculators from Caseo will exceed the sample mean life span of 81

musickatia [10]

Answer:

it would be 64-81 than answer devide by 6.

8 0

2 years ago

What is the Distribution property of multiply addends of number,the add the products

KonstantinChe [14]

The address the products

5 0

3 years ago

Which of the following is the portion of a line that starts at one point and goes off in a particular direction to infinity?

NeTakaya

Answer: Vertex
Explanation:

4 0

2 years ago

Read 2 more answers

Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro

Ugo [173]

Answer:

$Side\ B = 6.0$

$\alpha = 56.3$

$\theta = 93.7$

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with $\alpha, \beta, \theta$

[See Attachment for Triangle]

$A = 10cm$

$C = 12cm$

$\beta = 30$

What the question is to calculate the third length (Side B) and the other 2 angles ( $\alpha\ and\ \theta$ )

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

$B^2 = A^2 + C^2 - 2ABCos\beta$

Substitute: $A = 10$ , $C =12$ ; $\beta = 30$

$B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30$

$B^2 = 100 + 144 - 240*0.86602540378$

$B^2 = 100 + 144 - 207.846096907$

$B^2 = 36.153903093$

Take Square root of both sides

$\sqrt{B^2} = \sqrt{36.153903093}$

$B = \sqrt{36.153903093}$

$B = 6.0128115797$

$B = 6.0$ <em>(Approximated)</em>

Calculating Angle $\alpha$

$A^2 = B^2 + C^2 - 2BCCos\alpha$

Substitute: $A = 10$ , $C =12$ ; $B = 6$

$10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 180 - 144 *Cos\alpha$

Subtract 180 from both sides

$100 - 180 = 180 - 180 - 144 *Cos\alpha$

$-80 = - 144 *Cos\alpha$

Divide both sides by -144

$\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}$

$\frac{-80}{-144} = Cos\alpha$

$0.5555556 = Cos\alpha$

Take arccos of both sides

$Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)$

$Cos^{-1}(0.5555556) = \alpha$

$56.25098078 = \alpha$

$\alpha = 56.3$ <em>(Approximated)</em>

Calculating $\theta$

Sum of angles in a triangle = 180

Hence;

$\alpha + \beta + \theta = 180$

$30 + 56.3 + \theta = 180$

$86.3 + \theta = 180$

Make $\theta$ the subject of formula

$\theta = 180 - 86.3$

$\theta = 93.7$

5 0

3 years ago