6x - 2 < 5x + 7
I DID THIS AND GOT -9 BUT THE ANSWER IS 9 WHAT DID I DO WRONG

Answer:

(a).

[tex]S_{n} = \frac{n}{2} (4n + 20) \\ \\S _{n - 1} = \frac{(n - 1)}{2} (4n - 4 + 20) \\ \\ S _{n - 1} = \frac{(n - 1)}{2} (4n + 16) \\ \\S _{n - 1} = \frac{(n - 1)(4n + 16)}{2} \\ \\ { \boxed{S _{n - 1} = {2 {n}^{2} + 6n - 8}}} \\ [/tex]

(b).

from general equation:

[tex]S _{n - 1} = \frac{(n - 1)}{2} (4n + 16)[/tex]

first term is 4n

common difference:

[tex]16 = \{(n - 1) - 1 \}d \\ 16 = (n - 2)d \\ d = \frac{16}{n - 2} [/tex]

f(x) = 4x

g(x) =

[tex] \sqrt{x^{2} \: - \: 1 } [/tex]

h(x) =

[tex] \sqrt{16x \: - \: 1} [/tex]

We want to check if h(x) = g(f(x))

So g(f(x)) =

[tex] \sqrt{(4x) {}^{2} - \: 1} [/tex]

Simplified;

g(f(x)) =

[tex] \sqrt{16x {}^{2} \: - \: 1 } [/tex]

But h(x) =

[tex] \sqrt{16x \: - \: 1} [/tex]

Hence g(f(x)) is not equal to h(x)

Answer:

Equation is y=2x+12

The image is a picture of the graph(it's not the best but it's something.

Step-by-step explanation:

compound inequality: a statement formed by two.......

intersection: elements that are in both set A and B

union: elements that are in either set A or B

Answer:

D

Step-by-step explanation:

6(x + 8) =                 Plug in x with 4

6(4 + 8) =

6(12) =

72

After setting [tex]u=\sqrt{1+e^{2x}}[/tex], partially solving for x in terms of u gives

[tex]u = sqrt{1+e^{2x}} \implies u^2 = 1 + e^{2x} \implies e^{2x} = u^2 - 1[/tex]

Then taking differentials, you get

[tex]2 e^{2x} \,\mathrm dx = 2u \, \mathrm du \implies \mathrm dx = \dfrac{u}{u^2-1}\,\mathrm du[/tex]

Replacing everything in the original integral then gives

[tex]\displaystyle \int \frac9{\sqrt{1+e^{2x}}}\,\mathrm dx = \int \frac9u \times \frac u{u^2-1}\,\mathrm du = 9 \int \frac{\mathrm du}{u^2-1}[/tex]

Split up the integrand into partial fractions:

[tex]\dfrac1{u^2-1} = \dfrac a{u-1} + \dfrac b{u+1} \\\\ 1 = a(u+1) + b(u-1) = (a+b)u + a-b \\\\ \implies \begin{cases}a+b=0\\a-b=1\end{cases} \implies a=\dfrac12,b=-\dfrac12[/tex]

so that

[tex]\displaystyle 9 \int \frac{\mathrm du}{u^2-1} = \frac92 \int \left(\frac1{u-1} - \frac1{u+1}\right) \,\mathrm du \\\\ = \frac92 \left(\ln|u-1| - \ln|u+1|\right) + C \\\\ = \frac92 \ln\left|\frac{u-1}{u+1}\right| + C \\\\ = \frac92 \ln\left(\frac{\sqrt{1+e^{2x}}-1}{\sqrt{1+e^{2x}}+1}\right) + C[/tex]

Step-by-step explanation:

11.

[tex] \frac{x - 4}{ \frac{x}{4} - \frac{4}{x} } [/tex]

[tex] \frac{x - 4}{ \frac{ {x}^{2} }{4x} - \frac{16}{4x} } [/tex]

[tex] \frac{x - 4}{ \frac{ {x}^{2} - 16 }{4x} } [/tex]

[tex] \frac{x - 4}{ \frac{(x + 4)(x - 4)}{4x} } [/tex]

[tex] \frac{4x}{x + 4} [/tex]

12.

[tex] \frac{ \frac{ {x}^{2} - 1 }{x} }{ \frac{(x - 1) {}^{2} }{x} } [/tex]

[tex] \frac{ {x}^{3} - x}{ {x}^{3} - 2 {x}^{2} + x } [/tex]

[tex] \frac{x( {x} - 1)(x + 1) }{x( {x} - 1) {}(x - 1) } [/tex]

[tex] \frac{x + 1}{x - 1} [/tex]

13.

[tex] \frac{ \frac{ {t}^{2} }{ \sqrt{ {t}^{2} + 1} } - \sqrt{ {t}^{2} + 1} }{ {t}^{2} } [/tex]

[tex] \frac{ \frac{ {t}^{2} }{ \sqrt{ {t}^{2} + 1 } } - \frac{ {t}^{2} + 1}{ \sqrt{ {t}^{2} + 1 } } }{ {t}^{2} } [/tex]

[tex] \frac{ \frac{1}{ \sqrt{ {t}^{2} + 1 } } }{ {t}^{2} } [/tex]

[tex] \frac{ - \frac{ \sqrt{ {t}^{2} + 1 } }{ {t}^{2} + 1 } }{ {t}^{2} } [/tex]

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

14.

[tex] \frac{3x { }^{ \frac{1}{3} } - x {}^{ \frac{ - 2}{3} } }{3x {}^{ - \frac{2}{3} } } [/tex]

[tex] \frac{3x - 1}{3} [/tex]

answer:

(a).

Equation of circle is x² + y² - 25 = 0

(b).

(-5, 0) » yes

(√7, 1) » no

(-3, √21) » no

(0, 7) » no

Step-by-step explanation:

(a).

If centred at origin, centre is (0, 0)

General equation of circle:

[tex]{ \boxed{ \bf{ {x}^{2} + {y}^{2} + 2gx + 2fy + c = 0}}}[/tex]

but g and f are 0:

[tex] {x}^{2} + {y}^{2} + c = 0 \\ but : \\ c = {g}^{2} + {f}^{2} - {r}^{2} \\ c = { - 25} [/tex]

Answer:

Angle DBC = 51

Step-by-step explanation:

Angle ABC = angle ABD + angle DBC

We are given that angle ABC = 122 and angle ABD = 71

So 122 = 71 + angle DBC

* Solve for angle DBC *

Subtract 71 from both sides

122 - 71 = 71 - 71 + angle DBC

51 = angle DBC

Answer:

Washington had 115 hits and Sanchez had 111 hits.

Step-by-step explanation:

2s + 4 = 226

2s = 222

s = 111

s + 4 = 115

====================================================

Explanation:

Refer to the diagram below. The rectangle MATH has the diagonal MT that cuts the rectangle into two identical right triangles.

We then use the pythagorean theorem to find the length of the hypotenuse MT.

a^2 + b^2 = c^2

10^2 + 24^2 = c^2

100 + 576 = c^2

676 = c^2

c^2 = 676

c = sqrt(676)

c = 26

Segment MT is 26 cm long.

This applies to the other diagonal AH as well.

Answer:

(-1,-5)

Step-by-step explanation:

To understand where this point was before the transformation you must first understand the transformation itself. Shifting left is a horizontal translation, this means that it will affect the x-coordinate. The x-coordinate is the first digit in the pair. Therefore, in the case of (3,5), 3 would be the x-coordinate. Additionally, moving left is moving towards the negative, so it is the same as subtracting. Thus, to find where you started, work backward. Do this by adding 4 to the x-coordinate. This means the beginning coordinate was (-1,-5).

Answer:

x=5

JK = 40 units

Step-by-step explanation:

JM=MK since the have the little red line which shows they are equal

7x+5 = 8x

Subtract 7x from each side

7x+5 -7x= 8x-7x

5 = x

JK = 7x+5 = 7(5)+5 = 35+5 = 40

9514 1404 393

Answer:

  see attached

Step-by-step explanation:

The x-coordinate is the same in each pair of points, so the line segment will be a vertical line segment 3 units long.

Answer:

Step-by-step explanation:

Answer: 0.0623

Step-by-step explanation:

The probability of correctly answering 8 questions if you guess randomly on each question is 0.062348.

It is given that multiple choice test contains 25 questions with 5 answer choices.

To find probability of correctly answering.

The extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.

Given that:

The probability of each correct answer is  [tex]p_{s}[/tex][tex]=\frac{1}{5}[/tex]

The probability of 8 successful answers in 25 independent trials for a binomial probability distribution is:

p(k|n)=[tex]\frac{n!}{(n-k)!*n!}[/tex][tex]p_{s}^{k} (1-p_{s})^{n-k} \\[/tex]

p(8|25)=[tex]\frac{25!}{(25-8)!*8!}[/tex][tex]{\frac{1}{5} }^{k} (1-1/5)^{25-8} \\[/tex]

p(8|25)=0.062348

So, the probability of correctly answering 8 questions if you guess randomly on each question is 0.062348.

Learn more about probability here:

https://brainly.com/question/24385262

#SPJ2

Step-by-step explanation:

we first find the gradient

when y1=-4, x1=-5 and y2=1, x2=4

gradient={(y2-y1)÷(x2-x1)}

gradient={1-(-4)}÷{4-(-5)}

gradient=(1+4)÷(4+5)

gradient=5÷9 OR 0.5555

To find the equation interm of x

we substitute for the gradient in the formula

y-y1=gradient (x-x1)

y-(-4)=(5÷9)(x-×-5)

y+4=(5÷9)(x+5)

y+4=(5x+25)÷9 Cross multiply ✖️

9(y+4)=(5x+25)(1)

9y+36=5x+25 collect like terms

9y=5x+25-36

9y=5x-11 divide both sides by 9

y=(5x-11)÷9.......ans

Answer:

17.2

Step-by-step explanation:

30.1÷1.75=17.2

plssss mark brainliesttr

Answer:

m∠AKD = 46°

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given,

Measure of <AKG = 133°

So,

<AKG = <AKD + <DKG (As both together combine to form one <AKG as shown)

=> 133° = <AKD + 87°

=> <AKD = 133° - 87°

=> <AKD = 46° (Ans)

the answer would be

A. 53-53i !! :)

it was also the correct answer on the test ,, goodluck !!

Answer = B. -17 -53i

Answer:

X=25

Step-by-step explanation:

This is because X has to be greater than 23. Because you are taking away 35 and the answer is 2, you have to add 23 and 2, and that gives you 25!

Answer:

p = k(d)/u^2

Step-by-step explanation:

BRAINLIEST, PLEASE!

Well this could be incredibly easy, by just saying [tex]m=0[/tex] and [tex]n=-3x[/tex].

This is a completely valid case but nevertheless seems too easy. The requirements are probably that n and m are integers or natural numbers but since that was not specified such probable requirement ought not to be followed.

Hope this helps :)

Answer:

[tex]\sup(S) = 3[/tex].

[tex]\displaystyle \inf(S) = \frac{1}{3}[/tex].

Step-by-step explanation:

When factored, [tex]3\,x^{2} - 10\, x + 3[/tex] is equivalent to [tex](3\, x - 1)\, (x - 3)[/tex].

[tex]3\, x^{2} - 10\, x + 3 < 0[/tex] whenever [tex]\displaystyle x \in \left(\frac{1}{3},\, 3\right)[/tex].

Typically, the supremum and infimum of open intervals are the two endpoints. In this question, [tex]\sup(S) = 3[/tex] whereas [tex]\displaystyle \inf(S) = \frac{1}{3}[/tex].

Below is a proof of the claim that [tex]\sup(S) = 3[/tex]. The proof for [tex]\displaystyle \inf(S) = \frac{1}{3}[/tex] is similar.

In simple words, the supremum of a set is the smallest upper bound of that set. (An upper bound of a set is greater than any element of the set.)

It is easy to see that [tex]3[/tex] is an upper bound of [tex]S[/tex]:

To see that [tex]3[/tex] is the smallest upper bound of [tex]S[/tex], assume by contradiction that there exists some [tex]\epsilon > 0[/tex] for which [tex](3 - \epsilon)[/tex] (which is smaller than [tex]3\![/tex]) is also an upper bound of [tex]S\![/tex].

The next step is to show that [tex](3 - \epsilon)[/tex] could not be a lower bound of [tex]S[/tex].

There are two situations to consider:

Either way, it would be necessary to find (or construct) an element [tex]z[/tex] of [tex]S[/tex] such that [tex]z > 3 - \epsilon[/tex].

For the first situation, it would be necessary that [tex]\displaystyle 3 - \epsilon \le \frac{1}{3}[/tex], such that [tex]\displaystyle \epsilon \ge \frac{8}{3}[/tex]. Let [tex]z := 1[/tex] (or any other number between [tex](1/3)[/tex] and [tex]3[/tex].)

With the first situation [tex]\displaystyle \epsilon \ge \frac{8}{3}[/tex] accounted for, the second situation may assume that [tex]\displaystyle 0 < \epsilon < \frac{8}{3}[/tex].

Claim that  [tex]\displaystyle z:= \left(3 - \frac{\epsilon}{2}\right)[/tex] (which is strictly greater than [tex](3 - \epsilon)[/tex]) is also an element of [tex]S[/tex].

Either way, [tex](3 - \epsilon)[/tex] would not be an upper bound of [tex]S[/tex]. Contradiction.

Hence, [tex]3[/tex] is indeed the smallest upper bound of [tex]S[/tex]. By definition, [tex]\sup(S) = 3[/tex].

The proof for [tex]\displaystyle \inf(S) = \frac{1}{3}[/tex] is similar and is omitted because of the character limit.

Answer:

C. $1.73

Step-by-step explanation:

$5.03

$3.30

---------

$1.73

Answer:

between 15.5 and 16

Step-by-step explanation:

which std

Answer:

Try h=(1-6πs)/6π

Step-by-step explanation:

That's the same answer in another format