ASAP!!! FOR BRAINlIEST!!

Answer:

I don't really know what you mean so I would say to just look on the tag but if you gave more info then I could help you out

Answer:

The answer is 6

Step-by-step explanation:

This is due to the nth term sequence going up by 6

[tex] {n}^{2} - n - 56 = [/tex]

[tex](n - 8)(n + 7)[/tex]

_____________________________________________

[tex]a {x}^{2} + bx + c = 0[/tex]

[tex]delta = {b}^{2} - 4ac[/tex]

[tex]x(1) = \frac{ - b + \sqrt{delta} }{2a} \\ [/tex]

[tex]x(2) = \frac{ - b - \sqrt{delta} }{2a} \\ [/tex]

_____________________________________________

[tex] {n}^{2} - n - 56 = 0[/tex]

[tex]delta = ({ - 1})^{2} - 4(1)( - 56)[/tex]

[tex]delta = 1 + 224[/tex]

[tex]delta = 225[/tex]

Thus ;

[tex]n(1) = \frac{ - ( -1) + \sqrt{225} }{2 \times 1} \\ [/tex]

[tex]n(1) = \frac{1 + 15}{2} \\ [/tex]

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

[tex]n(1) = 8[/tex]

[tex]n(2) = \frac{ - ( - 1) - \sqrt{225} }{2 \times 1} \\ [/tex]

[tex]n(2) = \frac{ 1 - 15}{2} \\ [/tex]

[tex]n(2) = - \frac{14}{2} \\ [/tex]

[tex]n(2) = - 7[/tex]

[tex]n(1) = 8[/tex]

[tex]n(1) - 8 = 0 \: \: \: (\alpha )[/tex]

[tex]n(2) = - 7[/tex]

[tex]n(2) + 7 = 0 \: \: \: \: \: ( \beta )[/tex]

_____________________________________________

[tex] \alpha \times \beta = [/tex]

[tex](n - 8)(n + 7)[/tex]

Answer:

x = 3 +√43, x = 3 - √43

Step-by-step explanation:

x^2 - 6x - 34 = 0

move the 34 across

x^2 - 6x = 34

take -6 and divide by two, then square it

(-6/2)^2 = 9

add the product to your equation

x^2 - 6 + 9 = 34 + 9

solve

(x - 3)^2 = 43

x - 3 = ± √43

x = 3 ± √43

Answer: [tex]\dfrac{200}{S}[/tex] represents the transport cost per student in the expression.

Step-by-step explanation:

Given: For trip,

Admission fee per student = $10

Transport cost (for all students0 = $200

Here, S represents the number of students  went on the field trip.

Transport cost per student = (Total cost) ÷ Total student

So, [tex]\dfrac{200}{S}[/tex] represents the transport cost per student.

Hence,  [tex]\dfrac{200}{S}[/tex] represents the transport cost per student in the expression.

Answer:

5) y=2x^2+16x+32

Step-by-step explanation:

So the following equations on your paper are in vertex form. We want to turn these into standard form.

Remember, standard form is ax^2+bx+c and vertex form is y=a(x-h)^2+k

Lets expand our first problem, y=2(x+4)^2

y=2(x+4)(x+4)

y=2(x^2+8x+16)

y=2x^2+16x+32

Our answer for question 5 is y=2x^2+16x+32

If you need help with the other questions just add a comment.

Answer:

#5

#6

#7

#8

Hope this helps~! Brainliest is 'preciated if it does!

she cut the sum in (3+5+4) in 12 parts

J had 3/12 of the sum S

L had 5/12 of the sum S

and R had 4/12 of the sum S

4/12 x S = 320

S = 320 x 12/4

S = 960

$960 shared

Answer:

r=2

Step-by-step explanation:

r + 8 r + 11 = 29

( 1 + 8 ) r + 11 = 29 9 r + 11 = 29

Now we can isolate and solve for r while always keeping the equation balanced: First, subtract 11 from each side of the equation:

9 r + 11 − 11 = 29 − 11

9 r + 0 = 18

9 r = 18

Now we can divide each side of the equation by 9 to get

r : 9 over 9 = 18 under 9

1 r = 2

r = 2

Answer: r = 2

Steps:  

r + 8r + 11 = 29

9r + 11 = 29

9r + 11 - 11 = 29 - 11

9r = 18

r = 2

plz mark brainliest

Answer:

Rational

Step-by-step explanation:

An irrational number is any number that cannot be turned into a fraction. 7 can be turned into a fraction.

The required expression is v(t) = 80,000 - $6000t

Cost of the truck =  $80,000

Depreciation rate = $6000/year

Cost price is that price for buyer which he pays to seller for an object or product.

Required expression-

m = $6000
t = time in years
b = original price
v(t) = mt + b
v(t) = -$6000t + $80000

negative sign for depreciation

Thus, the required expression is v(t) = 80,000 - $6000t

Learn more about cost price here:
https://brainly.com/question/11027396

#SPJ2

Answer:

A XD  (um i dont know what else to put sof frpijfpoemd-eod)

PLEASE CHECK THE ATTACHED FILE

We are given that ;

[tex]{\quad \longrightarrow \displaystyle \sf \lim_{x\to 0}\bigg\{1+\dfrac{p(x)}{x^2}\bigg\}=2}[/tex]

Where p(x) is a polynomial of degree 4 , it will help us later, but let's do some manipulations first ;

Can be further written as ;

[tex]{:\implies \quad \displaystyle \sf 1+\lim_{x\to 0}\bigg\{\dfrac{p(x)}{x^2}\bigg\}=2}[/tex]

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 0}\bigg\{\dfrac{p(x)}{x^2}\bigg\}=1}[/tex]

So, here p(x) is polynomial of degree 4, so it will be a biquadratic polynomial, so we will write p(x) in the form of general biquadratic polynomial, so p(x) = ax⁴ + bx³ + cx² + dx + e

Now, first find p(x)/x² ;

[tex]{:\implies \quad \sf \dfrac{p(x)}{x^2}=\dfrac{ax^{4}+bx^{3}+cx^{2}+dx+e}{x^2}}[/tex]

[tex]{:\implies \quad \sf \dfrac{p(x)}{x^2}=ax^{2}+bx+c+\dfrac{d}{x}+\dfrac{e}{x^2}}[/tex]

So now, as [tex]{\bf{x\to 0}}[/tex] , for any values of a, b, c, d and e, the RHS will approach ∞ iff d ≠ e ≠ 0, as the denominator of d and e will be approaching 0 and so the whole limit will be ∞ , but we want the limit to be approaching 1, so when if d = e = 0, the denominator of d and e will be approaching 0 (not absolutely 0), and if d = e = 0, we will have the limit be approaching ax²+ bx + c for x approaching 0 being the limit 1 , and for any values of a, b and c . So now we have ;

[tex]{:\implies \quad \displaystyle \bf \lim_{x\to 0}\dfrac{p(x)}{x^{2}}=\begin{cases}\bf \infty \:, iff\: d\neq e\neq 0 \\ \\ \bf 1\:, iff\: d=e=0\end{cases}}[/tex]

So, now we had to consider the second case, for which the limit is approaching 1, for d = e = 0, so the limand here will just be ax² + bx + c

Now, we so have ;

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 0}ax^{2}+bx+c=1}[/tex]

Putting the limit we will have ;

[tex]{:\implies \quad \sf c=1}[/tex]

So, now as p(x) have extremum at 1 and 2, so p'(x) = 0, for x = 1, 2 , so now finding p'(x)

[tex]{:\implies \quad \sf p(x)=ax^{4}+bx^{3}+x^{2}\quad \qquad \{\because c=1\: and\: d=e=0\}}[/tex]

So, differentiating both sides wr.t.x ;

[tex]{:\implies \quad \sf p^{\prime}(x)=4ax^{3}+3bx^{2}+2x}[/tex]

Now, p'(1) and p'(2) must be 0

[tex]{:\implies \quad \sf p^{\prime}(1)=4a+3b+2=0}[/tex]

[tex]{:\implies \quad \sf p^{\prime}(2)=32a+12b+4=0}[/tex]

So, now we have ;

[tex]{\quad \longrightarrow \displaystyle \begin{cases}\bf 4a+3b=-2 \\ \\ \bf 32a+12b=-4\end{cases}}[/tex]

On multiplying first equation by 8 on both sides we can thus obtain ;

[tex]{\quad \longrightarrow \displaystyle \begin{cases}\bf 32a+24b=-16 \\ \\ \bf 32a+12b=-4\end{cases}}[/tex]

On solving both the equations we will be having ;

[tex]{\quad \longrightarrow \displaystyle \begin{cases}\bf a=\dfrac{1}{4} \\ \\ \bf b=-1\end{cases}}[/tex]

So , now as d = e = 0, c = 1, a = (1/4), b = -1, so putting all the values in p(x) we can obtain p(x) as ;

[tex]{:\implies \quad \sf p(x)=\dfrac{1}{4}(x)^{4}-(x)^{3}+x^{2}}[/tex]

Now, at x = 2 ;

[tex]{:\implies \quad \sf p(2)=\dfrac{1}{4}(2)^{4}-(2)^{3}+2^{2}}[/tex]

[tex]{:\implies \quad \sf p(2)=4-8+4=8-8}[/tex]

[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{p(2)=0}}}[/tex]

This is the required answer

Answer:

= 1= RHS

Step-by-step explanation:

shee i got u bud ok LHS-

(1+cot^2A)(1+cosA)(1-cosA)

=Cosec^2A(1-cos^2A)

=1/sin^2A×(sin^2A)

= 1= RHS

Answer:

The slope is 0 or there is no slope.

Step-by-step explanation:

Only a y-intercept of -4.

Answer:

0

Step-by-step explanation:

The slope is defined as rise/run. Since y does not increase at all as x increases, the slope is 0.

It is 5,760.

Stay safe & humble <3

Answer:

x = -4

Step-by-step explanation:

Step 1: Write equation

-x + 3 = 7

Step 2: Solve for x

Step 3: Check

Plug in x to verify it's a solution.

-(-4) + 3 = 7

4 + 3 = 7

7 = 7

Answer:

B. -3

Step-by-step explanation:

From (-1,4) it goes down 3 (rise) units then 1 unit (run) to the right. Simply, its sloping to the left, so negative.

Solving using slope formula:

y2-y1/x2-x1

4--2/ -1-1

6/-2

3/-1

-3

Answer:  B

Step-by-step explanation:

y2-y1/x2-x1

-2-4/1-(-1)

-6/1+1

-6/2

-3

Answer: 4 Hours

Step-by-step explanation:

volume of cylinder = 3.14×1.5×1.5×4

1.34×2.25×4=

1.34×9=

12.06 in³

volume of cone =1/3×3.14×2×2×6

1/3×3.14×4×6

1/3×3.14×24

3.14×8

=25.12 in³

therefore he need to buy the cone shaped cup

Answer:

63

Step-by-step explanation:

5(13)-2 is 65-2 or 63.

Answer:

  244 sq. ft.

Step-by-step explanation:

The net for a triangular prism consists of a central rectangle and two triangles. The length of the central rectangle is equal to the perimeter of the triangular faces.

__

The area of the central rectangle of the net is ...

  A = LW = (12 +7 +10 ft)(6 ft) = 174 ft²

The two triangles are right triangles, so the area of each is half the product of the leg lengths.

  A = 1/2(bh) = 1/2(10 ft)(7 ft) = 35 ft²

Then the sum of the rectangle and two triangles is ...

  surface area = 174 ft² +2 × 35 ft² = 244 ft²

Answer:

48 degrees

Step-by-step explanation:

Since triangles WZY and WXY are congruent by HL, angles WYZ and WYX are congruent by CPCTC.

Answer:

2x+12.5

Step-by-step explanation:

0.5(x4-9) +17

=0.5(4x-9)+12

=2x-4.5+17

=2x+12.5

Answer:

A. If the dividend is greater than the divisor, the quotient will be greater than 1.

Step-by-step explanation:

Checking all options

A. If the dividend is greater than the divisor, the quotient will be greater than 1.

That is,

4 ÷ 3 = 1.33

Where,

4 is the dividend

3 is the divisor

1.33 is the quotient

THIS IS TRUE

B.If the dividend is less than the divisor, the quotient will be greater than 1

That is,

3 ÷ 4 = 0.75

Where,

3 is the dividend

4 is the divisor

0.75 is the quotient

NOT TRUE

C. If the dividend is equal to the divisor, the quotient will be less than 1

That is,

3 ÷ 3 = 1

Where,

3 is the dividend

3 is the divisor

1 is the quotient

NOT TRUE

D. If the dividend is greater than the divisor, the quotient will be less than 1.

That is,

4 ÷ 3 = 1.33

Where,

4 is the dividend

3 is the divisor

1.33 is the quotient

NOT TRUE

Therefore, option A is TRUE

Answer:

120.7016

Step-by-step explanation:

surface area of sphere=4πr²

solution

=4*3.14*3.1*3.1

=120.7016