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3 years ago

If the measure of angle 4 is (11x) and angle 3 is (4x) what is the measure of angle 3 in degrees

Mathematics

1 answer:

Xelga [282]3 years ago

3 0

Answer:

idk i suc dwa

Step-by-step explanation:

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PLEASE HELP, GOOD ANSWERS GET BRAINLIEST. +40 POINTS WRONG ANSWERS GET REPORTED

MA_775_DIABLO [31]

1. Ans:(A) 123

Given function: $f(x) = 8x^2 + 11x$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2 + 11x)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2) + \frac{d}{dx}(11x)$
=> $\frac{d}{dx} f(x) = 2*8(x^{2-1}) + 11$
=> $\frac{d}{dx} f(x) = 16x + 11$

Now at x = 7:
$\frac{d}{dx} f(7) = 16(7) + 11$

=> $\frac{d}{dx} f(7) = 123$

2. Ans:(B) 3

Given function: $f(x) =3x + 8$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(3x + 8)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(8)$
=> $\frac{d}{dx} f(x) = 3*1 + 0$
=> $\frac{d}{dx} f(x) = 3$

Now at x = 4:
$\frac{d}{dx} f(4) = 3$ (as constant)

=>Ans: $\frac{d}{dx} f(4) =$ 3

3. Ans:(D) -5

Given function: $f(x) = \frac{5}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{5}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(5x^{-1})$
=> $\frac{d}{dx} f(x) = 5*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = -5x^{-2}$

Now at x = -1:
$\frac{d}{dx} f(-1) = -5(-1)^{-2}$

=> $\frac{d}{dx} f(-1) = -5 *\frac{1}{(-1)^{2}}$
=> Ans: $\frac{d}{dx} f(-1) = -5$

4. Ans:(C) 7 divided by 9

Given function: $f(x) = \frac{-7}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{-7}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(-7x^{-1})$
=> $\frac{d}{dx} f(x) = -7*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = 7x^{-2}$

Now at x = -3:
$\frac{d}{dx} f(-3) = 7(-3)^{-2}$

=> $\frac{d}{dx} f(-3) = 7 *\frac{1}{(-3)^{2}}$
=> Ans: $\frac{d}{dx} f(-3) = \frac{7}{9}$

5. Ans:(C) -8

Given function:
$f(x) = x^2 - 8$

Now if we apply limit:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 - 8)$

=> $\lim_{x \to 0} f(x) = (0)^2 - 8$
=> Ans: $\lim_{x \to 0} f(x) = - 8$

6. Ans:(C) 9

Given function:
$f(x) = x^2 + 3x - 1$

Now if we apply limit:
$\lim_{x \to 2} f(x) = \lim_{x \to 2} (x^2 + 3x - 1)$

=> $\lim_{x \to 2} f(x) = (2)^2 + 3(2) - 1$
=> Ans: $\lim_{x \to 2} f(x) = 4 + 6 - 1 = 9$

7. Ans:(D) doesn't exist.

Given function: $f(x) = -6 + \frac{x}{x^4}$
In this case, even if we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

Check:
$f(x) = -6 + \frac{x}{x^4} \\ f(x) = -6 + \frac{1}{x^3} \\ f(x) = \frac{-6x^3 + 1}{x^3} \\ Rationalize: \\ f(x) = \frac{-6x^3 + 1}{x^3} * \frac{x^{-3}}{x^{-3}} \\ f(x) = \frac{-6x^{3-3} + x^{-3}}{x^0} \\ f(x) = -6 + \frac{1}{x^3} \\ Same$

If you apply the limit, answer would be infinity.

8. Ans:(A) Doesn't Exist.

Given function: $f(x) = 9 + \frac{x}{x^3}$
Same as Question 7
If we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

9, 10.
Please attach the graphs. I shall amend the answer. :)

11. Ans:(A) Doesn't exist.

First We need to find out: $\lim_{x \to 9} f(x)$ where,
$f(x) = \left \{ {{x+9, ~~~~~x \textless 9} \atop {9- x,~~~~~x \geq 9}} \right.$

If both sides are equal on applying limit then limit does exist.

Let check:
If x $\textless$ 9: answer would be 9+9 = 18
If x $\geq$ 9: answer would be 9-9 = 0

Since both are not equal, as $18 \neq 0$ , hence limit doesn't exist.

12. Ans:(B) Limit doesn't exist.

Find out: $\lim_{x \to 1} f(x)$ where,

$f(x) = \left \{ {{1-x, ~~~~~x \textless 1} \atop {x+7,~~~~~x \textgreater 1} } \right. \\ and \\ f(x) = 8, ~~~~~ x=1$

If all of above three are equal upon applying limit, then limit exists.

When x < 1 -> 1-1 = 0
When x = 1 -> 8
When x > 1 -> 7 + 1 = 8

ALL of the THREE must be equal. As they are not equal. 0 $\neq$ 8; hence, limit doesn't exist.

13. Ans:(D) -∞; x = 9

f(x) = 1/(x-9).

Table:

x f(x)=1/(x-9)

----------------------------------------

8.9 -10

8.99 -100

8.999 -1000

8.9999 -10000

9.0 -∞

Below the graph is attached! As you can see in the graph that at x=9, the curve approaches but NEVER exactly touches the x=9 line. Also the curve is in downward direction when you approach from the left. Hence, -∞, x =9 (correct)

14. Ans: -6

s(t) = -2 - 6t

Inst. velocity = $\frac{ds(t)}{dt}$

Therefore,

$\frac{ds(t)}{dt} = \frac{ds(t)}{dt}(-2-6t) \\ \frac{ds(t)}{dt} = 0 - 6 = -6$

At t=2,

Inst. velocity = -6

15. Ans: +∞, x =7

f(x) = 1/(x-7)^2.

Table:

x f(x)= 1/(x-7)^2

--------------------------

6.9 +100

6.99 +10000

6.999 +1000000

6.9999 +100000000

7.0 +∞

Below the graph is attached! As you can see in the graph that at x=7, the curve approaches but NEVER exactly touches the x=7 line. The curve is in upward direction if approached from left or right. Hence, +∞, x =7 (correct)

-i

7 0

3 years ago

Read 2 more answers

Help me please !!! And thank you

Neko [114]

Answer:

804.4

Step-by-step explanation:

the formula for the volume of a cube is the side length cubed,

so:

(9.3)(9.3)(9.3) = 804.357

then round it to 804.4

hope this helped!

4 0

3 years ago

If (x) = 3x - 5 and g(x) = x + 3, find (f - g)(x).

Temka [501]

Answer:

2x -8

Step-by-step explanation:

f (x) = 3x - 5

g(x) = x + 3,

(f - g)(x) = 3x - 5 - ( x+3)

Distribute the minus sign

= 3x-5 -x-3

Combine like terms

= 2x -8

8 0

3 years ago

Read 2 more answers

HELP: I need help on multiple questions for a quiz before tomorrow night. If all is answered, you will get 40+ points and marked

Vinil7 [7]

Answer:

1. $33,000

2. $6077.53

a) 2.72 seconds

b) 0.96 seconds

4. 189 square meters

a) Average Rate of Change = 4 pounds per week, or -4

b) 160 pounds (after 5 weeks)

Independent Variables = p, A, and H

Dependent Variable = T

141 beats per minute (rounded to nearest whole number)

Step-by-step explanation:

The value of a tractor V(t) decreases over time, t. Its value is given by:

$V(t)=55000-5500t$

To find value of tractor after 4 years, we would need to substitute "4" into t and calculate. It is shown below:

$V(t)=55000-5500t\\V(4)=55000-5500(4)\\V(4) =33000$

The value of the tractor after 4 years is $33,000

The value of investment that is growing each year, is given by:

$f(n)=5000(1.05)^{n}$

Where

5000 is the initial investment (deposit)

1.05 means a 5% growth rate per year

n is the time in years

We want the investment's value after 4 years, so n would be 4. Substituting we get our answer:

$f(n)=5000(1.05)^{n}\\f(4)=5000(1.05)^{4}\\f(4)=6077.53$

To the nearest cent, the value of the investment would be:

$6077.53

The time it takes of pendulum to make one swing is given by the formula:

$T(L)=2\pi \sqrt{\frac{L}{32}}$

Where L is the length of the pendulum in feet

If L = 6ft, the time it will take is:

$T(6)=2\pi \sqrt{\frac{6}{32}}\\T(6)=2.72$

About 2.72 seconds

Now, the length is 9 inches, we convert it to feet first:

9/12 = 0.75 feet

So, the time it will take:

$T(0.75)=2\pi \sqrt{\frac{0.75}{32}}\\T(0.75)=0.96$

So, it will take about 0.96 seconds

The area of weed after t days can be modeled by:

$A(t)=0.006 \pi t^2$

Where t is number of days

Now, we want to find area after 100 days, so let t = 100, we get:

$A(t)=0.006 \pi t^2\\A(100)=0.006 \pi (100)^2\\A(100)=188.5$

Rounded to nearest sq. m, we have the area to be:

189 square meters

The avg. rate of change is basically how much the program is advertising that someone can loose in a week. It says "4 pounds per week". Since decrease, we give the value of "4" and negative sign So:

Average Rate of Change = 4 pounds per week, or -4

Initial weight is 180 pounds, we know 4 pounds is decreased every week when going through the program course. So, after 5 weeks,

5 * 4 = 20 pounds will be less

So, he will be:

180 - 20 = 160 pounds (after 5 weeks)

The Karvonen formula is given as:

$T=p(220-A)+H(1-p)$

Where

T is target heart rate (in bpm)

p is the percent intensity (expressed as decimal)

A is the age (in yrs)

H is the resting heart rate (in bpm)

We need to identity the independent and dependent variables. Now, lets that a simple example:

y = 2x

Here,

x is the independent variable

y is dependent on x, so y in dependent

Similarly, if you look at the formula, you can see:

p, A, and H are all independent

T depends on them

So,

p, A, H are independent variables

T is the dependent variable

Independent Variables = p, A, and H

Dependent Variable = T

We are given

A = 35

H = 60

p = 65% = 65/100 = 0.65

Now, we want T, lets substitute and find:

$T=0.65(220-35)+60(1-0.65)\\T=141.25$

The target heart rate should be: 141 beats per minute (rounded to nearest whole number)

3 0

4 years ago

Which algebra expression represents this phase ?

lapo4ka [179]

Answer:

Step-by-step explanation:

Product is what you get when you multiply

5 0

3 years ago

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