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2 years ago

Hi please help me with this and complete all the directions pls

Mathematics

2 answers:

7nadin3 [17]2 years ago

8 0

Answer:

(3m+5)(m^2+4)

Step-by-step explanation:

add and subtract the second term to the expression and factor by grouping

Brilliant_brown [7]2 years ago

8 0

3m (3*3*3)+5m (5*5)+12m+20

(The * mean times btw if you didn’t know just in case)

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3 years ago

Solve for X. Each figure is a parallelogram.

vichka [17]

Answer:

1. x = 8

2. x = 3

3. x = 25

4. x = 6

5. x = 11

6. x = 3

7. x = 6

8. x = 10

Step-by-step explanation:

Since the opposite sides of a parallelogram are congruent, the length of segment QR is equal to the length of segment PS.

10 = -6 + 2x

Add 6 to each side

10 + 6 = -6 + 6 + 2x

Simplify each side

16 = 2x

Divide both sides by 2

16/2 = 2x/2

Simplify

x = 8

to check if x is correct:

you already know that segment QR is 10, so your answer needs to be equal to 10

so sement PS is -6x + 2x

-6 + 2x

= -6 + 2(8)

= -6 + 16

= 10

so PS = 10

2. 34x + 5 = 35x + 2

Step 1: Subtract 35x from both sides.

34x + 5 − 35x = 35x + 2 − 35x

−x + 5 = 2

Step 2: Subtract 5 from both sides.

−x + 5 − 5 = 2 − 5

−x = −3

Step 3: Divide both sides by -1.

-x/-1 = -3/-1

Step 4: Simplify

x = 3

115 = 2x + 65

Step 1: Flip the equation.

2x + 65 = 115

Step 2: Subtract 65 from both sides.

2x + 65 − 65 = 115 − 65

2x = 50

Step 3: Divide both sides by 2.

2x/2 = 50/2

x = 25

84 + 16x = 180

84 - 84 + 16x = 180 - 84

16x = 96

16x/16 = 96/16

x = 6

2(3x − 10) = 46

Step 1: Simplify both sides of the equation.

2(3x − 10) = 46

(2)(3x) + (2)(−10 )= 46(Distribute)

6x + −20 = 46

6x − 20 = 46

Step 2: Add 20 to both sides.

6x − 20 + 20 = 46 + 20

6x = 66

Step 3: Divide both sides by 6.

6x/6 = 66/6

x = 11

6. RH = 1/2 FH or RH = 2 × FH

Step 1: Simplify both sides of the equation.

(2)(12)=13x−2

24=13x+−2

24=13x−2

Step 2: Flip the equation.

13x−2=24

Step 3: Add 2 to both sides.

13x−2+2=24+2

13x=26

Step 4: Divide both sides by 13.

13x/13 = 26/13

x = 3

2x+2=3x−4

Step 1: Subtract 3x from both sides.

2x+2−3x=3x−4−3x

−x+2=−4

Step 2: Subtract 2 from both sides.

−x+2−2=−4−2

−x=−6

Step 3: Divide both sides by -1.

-x/-1 = -6/-1

x = 6

2(2x−6)=28

Step 1: Simplify both sides of the equation.

2(2x−6)=28

(2)(2x)+(2)(−6)=28(Distribute)

4x+−12=28

4x−12=28

Step 2: Add 12 to both sides.

4x−12+12=28+12

4x=40

Step 3: Divide both sides by 4.

4x/4 = 40/4

x = 10

7 0

2 years ago

Consider the following function.

Kryger [21]

Answer:

See below

Step-by-step explanation:

I assume the function is $f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, $x=0$ is the only vertical asymptote.

B) Set the first derivative equal to 0 and solve:

$f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

$f'(x)=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-5x+8$

$5x=8$

$x=\frac{8}{5}$

Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:

$f'(2)=-\frac{5}{2^2}+\frac{8}{2^3}=-\frac{5}{4}+\frac{8}{8}=-\frac{5}{4}+1=-\frac{1}{4}$

$f'(1)=-\frac{5}{1^2}+\frac{8}{1^3}=-\frac{5}{1}+\frac{8}{1}=-5+8=3$

Therefore, the function increases on the interval $(0,\frac{8}{5})$ and decreases on the interval $(-\infty,0),(\frac{8}{5},\infty)$ .

C) Since we determined that the slope is 0 when $x=\frac{8}{5}$ from the first derivative, plugging it into the original function tells us where the extrema are. Therefore, $f(\frac{8}{5})=1+\frac{5}{\frac{8}{5}}-\frac{4}{\frac{8}{5}^2 }=\frac{41}{16}$ , meaning there's an extreme at the point $(\frac{8}{5},\frac{41}{16})$ , but is it a maximum or minimum? To answer that, we will plug in $x=\frac{8}{5}$ into the second derivative which is $f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$ . If $f''(x)>0$ , then it's a minimum. If $f''(x)$ , then it's a maximum. If $f''(x)=0$ , the test fails. So, $f''(\frac{8}{5})=\frac{10}{\frac{8}{5}^3}-\frac{24}{\frac{8}{5}^4}=-\frac{625}{512}$ , which means $(\frac{8}{5},\frac{41}{16})$ is a local maximum.

D) Now set the second derivative equal to 0 and solve:

$f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$

$0=\frac{10}{x^3}-\frac{24}{x^4}$

$0=10x-24$

$-10x=-24$

$x=\frac{24}{10}$

$x=\frac{12}{5}$

We then test where $f''(x)$ is negative or positive by plugging in test values. I will use -1 and 3 to test this:

$f''(-1)=\frac{10}{(-1)^3}-\frac{24}{(-1)^4}=-34$ , so the function is concave down on the interval $(-\infty,0)\cup(0,\frac{12}{5})$

$f''(3)=\frac{10}{3^3}-\frac{24}{3^4}=\frac{2}{27}>0$ , so the function is concave up on the interval $(\frac{12}{5},\infty)$

The inflection point is where concavity changes, which can be determined by plugging in $x=\frac{12}{5}$ into the original function, which would be $f(\frac{12}{5})=1+\frac{5}{\frac{12}{5}}+\frac{4}{\frac{12}{5}^2 }=\frac{43}{18}$ , or $(\frac{12}{5},\frac{43}{18})$ .

E) See attached graph

5 0

2 years ago

Hi please help me with this and complete all the directions pls​

Hi please help me with this and complete all the directions pls