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3 years ago

13

(2.3x-3)-(8x-9.1) please help me solve this its for a test

1 answer:

Tasya [4]3 years ago

4 0

Answer:

−5.7x + 6.1

Step-by-step explanation:

Does the answer need to be simplyfied?? if yes you have the right answer on top

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60 points Need help Asap

Anna [14]

Answer:

45 units ^2

Step-by-step explanation:

make sure to divide the triangle and sqaure up when finding the area of this polygon..

hope this helps!!

8 0

3 years ago

Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter

Vaselesa [24]

We have been given a function $f(x)=2x^3+3x^2-336x$ . We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

$f'(x)=2(3)x^{2}+3(2)x^1-336$

$f'(x)=6x^{2}+6x-336$

$6x^{2}+6x-336=0$

$x^{2}+x-56=0$

$x^{2}+8x-7x-56=0$

$(x+8)-7(x+8)=0$

$(x+8)(x-7)=0$

$x=-8,x=7$

So $x=-8,7$ are critical points and these will divide our function in 3 intervals $(-\infty,-8)U(-8,7)U(7,\infty)$ .

Now we will find derivative over each interval as:

$f'(x)=(x+8)(x-7)$

$f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16$

Since $f'(9)>0$ , therefore, function is increasing on interval $(-\infty,-8)$ .

$f'(x)=(x+8)(x-7)$

$f'(1)=(1+8)(1-7)=(9)(-6)=-54$

Since $f'(1)$ , therefore, function is decreasing on interval $(-8,7)$ .

Let us check for the derivative at $x=8$ .

$f'(x)=(x+8)(x-7)$

$f'(8)=(8+8)(8-7)=(16)(1)=16$

Since $f'(8)>0$ , therefore, function is increasing on interval $(7,\infty)$ .

(b) Since $x=-8,7$ are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

$f(-8)=2(-8)^3+3(-8)^2-336(-8)$

$f(-8)=1856$

$f(7)=2(7)^3+3(7)^2-336(7)$

$f(7)=-1519$

Therefore, $(-8,1856)$ is a local maximum and $(7,-1519)$ is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

$f''(x)=6(2)x^{1}+6$

$f''(x)=12x+6$

$12x+6=0$

$12x=-6$

$\frac{12x}{12}=-\frac{6}{12}$

$x=-\frac{1}{2}$

Therefore, $x=-\frac{1}{2}$ is an inflection point of given function.

3 0

3 years ago

Can someone please help me on this???I keep messing it up.

Alchen [17]

Answer:

The answer is B

Step-by-step explanation

As it shows in the table each week the amount adds to 75$. in the table it only shows 4 weeks amount. and the question asks you whats the amount going to be in the 7th week . the amount will be 825.

4 0

3 years ago

Let X1, X2 and X3 be three independent random variables that are uniformly distributed between 50 and 100.

sergejj [24]

Answer:

a) the probability that the minimum of the three is between 75 and 90 is 0.00072

b) the probability that the second smallest of the three is between 75 and 90 is 0.396

Step-by-step explanation:

Given that;

fx(x) = { 1/5 ; 50 < x < 100

0, otherwise}

Fx(x) = { x-50 / 50 ; 50 < x < 100

1 ; x > 100

a)

n = 3

F(1) (x) = nf(x) ( 1-F(x)^n-1

= 3 × 1/50 ( 1 - ((x-50)/50)²

= 3/50 (( 100 - x)/50)²

=3/50³ ( 100 - x)²

Therefore P ( 75 < (x) < 90) = ⁹⁰∫₇₅ 3/50³ ( 100 - x)² dx

= 3/50³ [ -2 (100 - x ]₇₅⁹⁰

= (3 ( -20 + 50)) / 50₃

= 9 / 12500 = 0.00072

b)

f(k) (x) = nf(x) ( ⁿ⁻¹_k₋ ₁) ( F(x) )^k-1 ; ( 1 - F(x) )^n-k

Now for n = 3, k = 2

f(2) (x) = 3f(x) × 2 × (x-50 / 50) ( 1 - (x-50 / 50))

= 6 × 1/50 × ( x-50 / 50) ( 100-x / 50)

= 6/50³ ( 150x - x² - 5000 )

therefore

P( 75 < x2 < 90 ) = 6/50³ ⁹⁰∫₇₅ ( 150x - x² - 5000 ) dx

= 99 / 250 = 0.396

3 0

3 years ago

Share $70 in a ratio of 4:1?

laiz [17]

I am pretty sure the answer is 56:14. Please correct me if I am incorrect.

6 0

3 years ago

Read 2 more answers