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3 years ago

15

Could someone please let me know these answers , thank you :)

2 answers:

KATRIN_1 [288]3 years ago

8 0

Answer:

1: CD and BC

2: CE and AC

3: AE , AB , BD, DE

4: AD and BE

There you go!

Soloha48 [4]3 years ago

7 0

Answer:

1) COB and DOC

2) EOC and AOC

3) AOB, AOE, EOD, and BOD

4) BOE and AOD

Step-by-step explanation:

An acute angle less than 90 degrees

An obtuse more than 90 degrees

A right angle is 90 degrees

A straight angle is a 180 degrees

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Simplify Fraction improper 13/8

Soloha48 [4]

Answer:

Exact Form:

13 /8

Decimal Form:

1.625

Mixed Number Form:

1 5/ 8

8 0

2 years ago

Solve 2(x+3)3/2=54 x = <br>show your work

hjlf

Answer:

x=15

Step-by-step explanation:

Step 1: Multiply both sides by 2.

Step 2: Simplify both sides of the equation.

6x+18=108

Step 3: Subtract 18 from both sides.

6x+18−18=108−18

6x=90

Step 4: Divide both sides by 6.

x=15

6 0

2 years ago

Use substitution. What is the solution to the system of equations? Use the drop-down menus to explain your answer.

Hoochie [10]

Answer:

The system of equations has one solution. The two equations are intersecting lines.

Step-by-step explanation:

if you want to use the substitution method, you would have to plug in y for 12x+2 in the second equation.

2(12x+2)=x+4

next, you have to distribute

24x+4=x+4

subtract 24x from each side

4=-23x+4

subtract 4 from each side

0=23x

divide each side by 23

x=0

now, plug in 0 to one equation.

y=12(0)+2

y=2

the lines intersect at one point, (0,2)

7 0

2 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

2 years ago

Solve (x-3)^2=5 plssss <br> thxs

Andreyy89

Answer:

=3±√5

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers