solve for the missing sidebof the triangle shown in the figure​

Answer:

8.

a) f'x means you find the derivative.

2 * d/dx x^2 -b * d/dx x + d/dx c

use power rule x^2 = 2x^1

2*2x = 4x.  the derivative of the differentiation variable, x is 1 and the derivative of a constant, c is 0

4x-b+0

4x-b is our derivative

(I am still figuring out b and c, I will edit this answer and put the solution for b and c.)

Step-by-step explanation:

Answer:

Step-by-step explanation:

120 : 30

ivans get £90 more

Answer:

3

Step-by-step explanation:

f(x) =2x+3

Let x=0

f(0) =2*0+3

Multiply

f(0) =0+3

Add

f(0) =3

Answer:

3

step-by-step explanation

f ( x ) = 2 x + 3

f ( 0 ) = 2 × 0 + 3 .. ( f ( x = 0 ) - given )

multiply

f ( 0 ) = 0 + 3

Add the numbers

f ( 0 ) = 3

Answer:

The correct option is (b).

Step-by-step explanation:

The formula for the side of a cube of surface area SA is as follows :

[tex]s=\sqrt{\dfrac{SA}{6}}[/tex]

When SA = 180 m²

[tex]s=\sqrt{\dfrac{180}{6}}\\\\s=\sqrt{30}[/tex]

When SA = 120 m²

[tex]s=\sqrt{\dfrac{120}{6}}\\\\s=\sqrt{20}\\\\=2\sqrt5[/tex]

Difference,

[tex]=30-2\sqrt5[/tex]

So, the correct option is (b).

Answer:

a

Step-by-step explanation:

Answer:

39.52

Step-by-step explanation:

Answer:

Table multiplication:

7.6 times

5.2 =

15.2

38.0

—-

39.52

Answer:

a. 0.1536

b. 0.9728

Step-by-step explanation:

The probability that a component fails, P(Y) = 0.2

The number of components in the system = 4

The number of components required for the subsystem to operate = 2

a. By binomial theorem, we have;

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) is given as follows;

P(Y = 2) = [tex]\dbinom{4}{2}[/tex] × 0.2² × 0.8² = 0.1536

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) = 0.1536

b. The probability that the system last longer than 1,000 hours, P(O) = The probability that no component fails + The probability that only one component fails + The probability that two component fails leaving two working

Therefore, we have;

P(O) = P(Y = 0) + P(Y = 1) + P(Y = 2)

P(Y = 0) = [tex]\dbinom{4}{0}[/tex] × 0.2⁰ × 0.8⁴ = 0.4096

P(Y = 1) = [tex]\dbinom{4}{1}[/tex] × 0.2¹ × 0.8³ = 0.4096

P(Y = 2) = [tex]\dbinom{4}{2}[/tex] × 0.2² × 0.8² = 0.1536

∴ P(O) = 0.4096 + 0.4096 + 0.1536 = 0.9728

The probability that the subsystem operates longer than 1,000 hours = 0.9728

Answer:

6th term = 25

Step-by-step explanation:

3n + 7

3 x (6) + 7

18 + 7 = 25

If this helps you, please mark brainliest!

Have a nice day!

Answer:

[tex]25[/tex]

Step-by-step explanation:

[tex] Tn_{n} = 3n + 7 \\Tn _{6} = 3 n + 7 \\ = 3 \times 6 + 7 \\ = 18 + 7 \\ = 25[/tex]

Hope this helps you

Have a nice day!

Answer/Step-by-step explanation:

The diagram of the circular track is missing, and so also its dimensions.

However, let's assume the dimensions of the circular track given is diameter (d) = 20 meters or radius (r) = 10 meters.

Since it's a circular track, the circumference of the track would give us the number of meters she runs in 1 lap.

Circumference = πd

d = 20 m (we are assuming the diameter is 20 meters)

π = 3.14

Circumference of circular track = 3.14 × 20 = 62.8 m.

This means that 1 lap = 62.8 m that she would have to run.

Therefore,

6 laps would be = 6 × 62.8 = 376.8 m

Therefore, if she completes 6 laps around the circular track that has a diameter of 20 m, she will have to run about 376.8 m.

Answer:

4.61 years

Step-by-step explanation:

hope it helped!

Answer:

500000 cm2

Step-by-step explanation:

Answer:

The surface charge density is [tex]7.8\times10^{-8} C/m^2[/tex].

Step-by-step explanation:

separation, d = 85 mm

Work, W = 6 x 10^-3 J

charge , Q = 8µC

The potential difference is given by

W = q V

[tex]V=\frac{6\times 10^{-3}}{8\times 10^{-6}}=750 V[/tex]

Let the charge on he capacitor is q.

[tex]q = CV\\\\q = \frac{\varepsilon oA}{d}\times V\\\\\frac{q}{A} = \frac{8.85\times 10^{-12}\times750}{0.085} =7.8\times10^{-8} C/m^2[/tex]

Question options:

A. He should report them directly on form 1040

B. He should report them on form 8949 and then on schedule D

C. He should report them on schedule D

D. He is not required to report them until he sells the underlying securities

Answer:

B. He should report them on form 8949 and then on schedule D

Explanation:

John has shares which have capital gains from a mutual fund and a brokerage account. In order to report his taxes, he would need to use the Schedule D(form 1040) for his mutual fund capital gains and the form 8949 for his brokerage capital gains. The brokerage capital gains is then transferred to schedule D.

Answer:

3 * 45 = 135 which is NOT 180

45, 45, 90 would work

Step-by-step explanation:

hope it helps!

Answer:

87 minutes

Step-by-step explanation:

Let the total number of minutes = m

Our equation is given as:

$20.21 = $14.99 + 0.06m

20.21 = 14.99 + 0.06m

Collect like terms

0.06m= 20.21 - 14.99

0.06m = 5.22

m = 5.22/0.06

m = 87

Therefore, the total number rod minutes used is 87 minutes

Answer:

hypotenuse = 7.3

Step-by-step explanation:

the length two legs of the given triangle are 7 and 2 respectively.

using pythagoras theorem

a^2 + b^2 = c^2

7^2 + 2^2 = c^2

49 + 4 = c^2

53 = c^2

[tex]\sqrt{53}[/tex] = c

7.3 = c

Answer:

Option B

Step-by-step explanation:

By applying the converse theorem of corresponding angles,

"If corresponding angles formed between two parallel lines and the transversal line are equal then both the lines will be parallel"

Angle between line B and Y = 90°

Angle between line B and Z = 90°

Therefore, corresponding angles are equal.

By applying converse theorem, line Y and line Z will be parallel.

Option B will be the answer.

Answer:

Option A = 1/15 cubic meters

Step-by-step explanation:

Formule to find volume of rectangular prism:

Volume = width × height × length

V = w×h×l

V = 1/3 × 1/4 × 4/5

V = 1/15 cubic meters

Answer:

11

Step-by-step explanation:

2(x + 1) - 3

Let x= 6

2(6+1) -3

Parentheses first

2(7) -3

Then multiply

14-3

Then subtract

11

Answer:

The answer is D

Step-by-step explanation:

-x is n

Y=0

Answer:

D

Step-by-step explanation:

Basically....... graph shown in picture

Answer:

timely, timely,measurable

Step-by-step explanation:

Answer:

specific, timely, measurable

Step-by-step explanation:

i took the test on plato

Answer: 280 km

Step-by-step explanation:

[tex]3\dfrac{1}{2} \: hours = 3.5 \: hours[/tex]

S = V × t

V = 80 km/h

t = 3.5 h

S = 80 × 3.5 = 280 km

Answer:

[tex]{ \tt{ \frac{( \frac{x + y}{x - y}) }{( \frac{y + x}{y - x}) } }} \\ \\ { \tt{ = \frac{x + y}{x - y} \times \frac{y - x}{y + x} }} \\ \\ { \tt{ = \frac{-(x- y)}{x - y} }}[/tex]

Answer: = -1

Answer:

3 < x

Step-by-step explanation:

3(8 – 4x) < 6(x – 5)

Divide each side by 3

3/3(8 – 4x) < 6/3(x – 5)

(8 – 4x) < 2(x – 5)

Distribute

8-4x < 2x-10

Add 4x to each side

8-4x+4x < 2x-10+4x

8 < 6x-10

Add 10 to each side

8+10 < 6x-10+10

18 < 6x

Divide by 6

18/6 < 6x/6

3 < x

Answer:

Associative property

Step-by-step explanation:

The associative property is a math rule that says that the way in which factors are grouped in a multiplication problem does not change the product.

Hope this helps

Answer:

[tex] \displaystyle - \frac{1}{2} [/tex]

Step-by-step explanation:

we would like to compute the following limit:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{1}{ \ln(x + \sqrt{ {x}^{2} + 1} ) } - \frac{1}{ \ln(x + 1) } \right) [/tex]

if we substitute 0 directly we would end up with:

[tex] \displaystyle\frac{1}{0} - \frac{1}{0} [/tex]

which is an indeterminate form! therefore we need an alternate way to compute the limit to do so simplify the expression and that yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right) [/tex]

now notice that after simplifying we ended up with a rational expression in that case to compute the limit we can consider using L'hopital rule which states that

[tex] \rm \displaystyle \lim _{x \to c} \left( \frac{f(x)}{g(x)} \right) = \lim _{x \to c} \left( \frac{f'(x)}{g'(x)} \right) [/tex]

thus apply L'hopital rule which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \dfrac{d}{dx} \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right) [/tex]

use difference and Product derivation rule to differentiate the numerator and the denominator respectively which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \frac{1}{x + 1} - \frac{1}{ \sqrt{x + 1} } }{ \frac{ \ln(x + 1)}{ \sqrt{ {x}^{2} + 1 } } + \frac{ \ln(x + \sqrt{x ^{2} + 1 } }{x + 1} } \right) [/tex]

simplify which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \sqrt{ {x}^{2} + 1 } - x - 1 }{ (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right) [/tex]

unfortunately! it's still an indeterminate form if we substitute 0 for x therefore apply L'hopital rule once again which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \sqrt{ {x}^{2} + 1 } - x - 1 }{ \dfrac{d}{dx} (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right) [/tex]

use difference and sum derivation rule to differentiate the numerator and the denominator respectively and that is yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \frac{x}{ \sqrt{ {x}^{2} + 1 } } - 1}{ \ln(x + 1) + 2 + \frac{x \ln(x + \sqrt{ {x}^{2} + 1 } ) }{ \sqrt{ {x}^{2} + 1 } } } \right) [/tex]

thank god! now it's not an indeterminate form if we substitute 0 for x thus do so which yields:

[tex] \displaystyle \frac{ \frac{0}{ \sqrt{ {0}^{2} + 1 } } - 1}{ \ln(0 + 1) + 2 + \frac{0 \ln(0 + \sqrt{ {0}^{2} + 1 } ) }{ \sqrt{ {0}^{2} + 1 } } } [/tex]

simplify which yields:

[tex] \displaystyle - \frac{1}{2} [/tex]

finally, we are done!

9514 1404 393

Answer:

  -1/2

Step-by-step explanation:

Evaluating the expression directly at x=0 gives ...

  [tex]\dfrac{1}{\ln(\sqrt{1})}-\dfrac{1}{\ln(1)}=\dfrac{1}{0}-\dfrac{1}{0}\qquad\text{an indeterminate form}[/tex]

Using the linear approximations of the log and root functions, we can put this in a form that can be evaluated at x=0.

The approximations of interest are ...

  [tex]\ln(x+1)\approx x\quad\text{for x near 0}\\\\\sqrt{x+1}\approx \dfrac{x}{2}+1\quad\text{for x near 0}[/tex]

__

Then as x nears zero, the limit we seek is reasonably approximated by the limit ...

  [tex]\displaystyle\lim_{x\to0}\left(\dfrac{1}{x+\dfrac{x^2}{2}}-\dfrac{1}{x}\right)=\lim_{x\to0}\left(\dfrac{x-(x+\dfrac{x^2}{2})}{x(x+\dfrac{x^2}{2})}\right)\\\\=\lim_{x\to0}\dfrac{-\dfrac{x^2}{2}}{x^2(1+\dfrac{x}{2})}=\lim_{x\to0}\dfrac{-1}{2+x}=\boxed{-\dfrac{1}{2}}[/tex]

_____

I find a graphing calculator can often give a good clue as to the limit of a function.