Find the first term and the common difference of the arithmetic sequence whose 6th term is 30 and 12th term
is 54.

Answer:

(x-3)(5x+4)

x=3 x=-4/5

Answer:

(5x + 4)(x - 3)

Step-by-step explanation:

Hello!

Factor:

Think: What two numbers add up to -11 but multiply to (5)(-12)?

Answer: -15 and 4

Continue:

The factored expression is (5x + 4)(x - 3)

Answer:

8r+8pg-pq

Step-by-step explanation:

The subtractable pg cancels out one of the 9 pg's. So 9 pg-1 pg= 8 pg

Hope this helps!

Answer: x>12

so i think x is 16.

Answer:

a) -8x³+x²+6x

d) 16x²-9

Step-by-step explanation:

a) -2x(x+4x²)+3(x²+2x)

Expand each bracket:

-2x(x+4x²)

As the -2x is on the outside of the bracket, you have to times everything inside the bracket by -2x.

-2x times x equals -2x²

-2x times 4x² equals -8x³

Then we expand the other bracket:

3(x²+2x)

3 times x² equals 3x²

3 times 2x equals 6x

We then put all of it together:

-2x²-8x³+3x²+6x

Collect like terms:

-8x³+x²+6x

b) (4x-3)(4x+3)

We will use the FOIL method:

F-First

O-outer

I-Inner

L-Last

Times the first two terms in each bracket:

4x times 4x equals 16x²

Times the outer terms in the bracket:

4x times 3 equals 12x

Times the inside terms in the bracket:

-3 times 4x equals -12x

Times the last terms in the bracket:

-3 times 3 equals -9

Put it together:

16x²+12x-12x-9

The 12x and -12x cancel out to leave 16x²-9

Hope this helps :)

We know at centroid medians bisect each other in the ratio 2:1.

[tex]\\ \sf\longmapsto TY=2x[/tex]

[tex]\\ \sf\longmapsto 2x=30[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{30}{2}[/tex]

[tex]\\ \sf\longmapsto x=15[/tex]

Answer:

A

Step-by-step explanation:

On the median TW the distance from the vertex to the centroid is twice the distance from the centroid to the midpoint , then

YW = [tex]\frac{1}{2}[/tex] × TY = [tex]\frac{1}{2}[/tex] × 30 = 15

Answer:

[tex]4 \sqrt{122} \sqrt{12} \\ (4 \times 2) \times ( \sqrt{12} \times \sqrt{12} ) \\ (4 \times 2) \times 12 \\ 8 \times 12 \\ 96[/tex]

Answer:

535 students passed and 107 students failed

Step-by-step explanation:

Create a system of equations where p is the number of students who passed and f is the number of students who failed:

p + f = 642

p = 5f

Solve by substitution by plugging in 5f as p into the first equation, then solving for f:

p + f = 642

5f + f = 642

6f = 642

f = 107

So, 107 students failed.

Find how many students passed by multiplying this by 5:

107(5)

= 535

535 students passed and 107 students failed.

Answer:

x=4sqrt3 a=4 b=3  ,y=8sqrt3 c=8 d=3

Step-by-step explanation:

because this is a 30-60-90 triangle, it is easy to find the side lengths. the longer leg is sqrt(3) times the shorter leg so x= 12/sqrt(3) or 4sqrt(3). the hypotenuse is 2 times the shorter leg so y= 8sqrt(3)

Answer:

9--7

Step-by-step explanation:

Answer:

This type of study is called a simulation

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

(y^(4))^(1/5)=y^(4/5)

Answer:

–10m4n3 + 8m2n6 + 3m4n3 – 2m2n6 – 6m2n6 = -7m4n3

⇒It is a monomial with a degree of 7 is correct

Step-by-step explanation:

Step-by-step explanation:

5.

[tex](5 + 4 \sqrt{7} ){x}^{2} + (4 - 2 \sqrt{7} ) x- 1 = 0[/tex]

Simplify both radicals.

[tex](5 + \sqrt{112) {x}^{2} } + (4 - \sqrt{28} )x - 1 = 0[/tex]

Apply Quadratic Formula

First. find the discramnint.

[tex](4 - \sqrt{28} ) {}^{2} - 4(5 + \sqrt{112} )( - 1) = 64[/tex]

Now find the divisor 2a.

[tex]2(5 + \sqrt{112} ) = 10 + 8 \sqrt{7} [/tex]

Then,take the square root of the discrimant.

[tex] \sqrt{64} = 8[/tex]

Finally, add -b.

[tex] - (4 + 2 \sqrt{7} )[/tex]

So our possible root is

[tex] - (4 + 2 \sqrt{7} ) + \frac{8}{10 + 8 \sqrt{7} } [/tex]

Which simplified gives us

[tex] \frac{ 4 + 2 \sqrt{7} }{10 + 8 \sqrt{7} } [/tex]

Rationalize the denominator.

[tex] \frac{4 + 2 \sqrt{7} }{10 + 8 \sqrt{7} } \times \frac{10 - 8 \sqrt{7} }{10 - 8 \sqrt{7} } = \frac{ - 72 - 12 \sqrt{7} }{ - 348} [/tex]

Which simplified gives us

[tex] \frac{6 + \sqrt{7} }{29} [/tex].

6. The answer is 2.

9514 1404 393

Answer:

  5. x = (6 +√7)/29; a=6, b=1, c=29

  6. x = 2

Step-by-step explanation:

The quadratic formula can be used, where a=(5+4√7), b=(4-2√7), c=-1.

  [tex]x=\dfrac{-b+\sqrt{b^2-4ac}}{2a}=\dfrac{-(4-2\sqrt{7})+\sqrt{(4-2\sqrt{7})^2-4(5+4\sqrt{7}})(-1)}{2(5+4\sqrt{7})}\\\\=\dfrac{-4+2\sqrt{7}+\sqrt{16-16\sqrt{7}+28+20+16\sqrt{7}}}{10+8\sqrt{7}}=\dfrac{4+2\sqrt{7}}{2(5+4\sqrt{7})}\\\\=\dfrac{(2+\sqrt{7})(5-4\sqrt{7})}{(5+4\sqrt{7})(5-4\sqrt{7})}=\dfrac{10-3\sqrt{7}-28}{25-112}=\boxed{\dfrac{6+\sqrt{7}}{29}}[/tex]

__

Use the substitution z=3^x to put the equation in the form ...

  z² -3z -54 = 0

  (z -9)(z +6) = 0 . . . . . factor

  z = 9 or -6 . . . . . . . . value of z that make the factors zero

Only the positive solution is useful, since 3^x cannot be negative.

  z = 9 = 3^2 = 3^x . . . . use the value of z to find x

  x = 2

Answer:

AED 972

Step-by-step explanation:

Area of the wall = 9² = 81 m²

each m² costs AED 12

so 81 m² will cost 12×81 = AED 972

Answer:

x = 1

y=1.5

Step-by-step explanation:

3*1+2*1.5=6

The values of x and y in equation 6=3x+2y is for x is 1 and for y is 1.5.

We have given that,

x and y are integers and 0 < x < y.

6 = /3x+2y.

x=1 then

A statement of an order relationship greater than, greater than or equal to, less than, or less than or equal to between two numbers or algebraic expressions.

6=3+2y

6-3=2y

3/2=y

y=1.5

3*1+2*1.5=6

Therefore we get the values of x and y is for x is 1 and for y is 1.5.

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Answer:

SOH-CAH-TOA

[tex]\sin \left(x\right)=\frac{6}{\sqrt{49+36}}[/tex] = 40.60°

SOH = SIN = OPP/HYP

SIN(Θ) = 6/[tex]\sqrt{49+36 }[/tex]

Step-by-step explanation:

Answer: -1

Step-by-step explanation:

12x^2-7x-12 = (4x+3)(3x-4)

4x+3=0. X = -3/4

3x-4=0. X = 4/3

(-3/4) (4/3) = -1

Answer:

b is the correct answer

Step-by-step explanation:

9514 1404 393

Answer:

  c = 14

  no extraneous solutions

Step-by-step explanation:

You can subtract the right-side expression, combine fractions, and set the numerator to zero.

  [tex]\dfrac{c-4}{c-2}-\left(\dfrac{c-2}{c+2}-\dfrac{1}{2-c}\right)=0\\\\\dfrac{c-4}{c-2}-\dfrac{1}{c-2}-\dfrac{c-2}{c+2}=0\\\\\dfrac{(c-5)(c+2)-(c-2)^2}{(c-2)(c+2)}=0\\\\\dfrac{(c^2-3c-10)-(c^2-4c +4)}{(c-2)(c+2)}=0\\\\\dfrac{c-14}{(c-2)(c+2)}=0\\\\\boxed{c=14}[/tex]

__

Check

  (14 -4)/(14 -2) = (14 -2)/(14 +2) -1/(2 -14) . . . . substitute for c

  10/12 = 12/16 -1/-12

  5/6 = 3/4 +1/12 . . . . true

There is one solution (c=14) and it is a solution to the original equation. There are no extraneous solutions.

Answer:

not enough info

Step-by-step explanation:

The simplification of the expression is  [tex]\frac{15\sqrt{2} }{4}[/tex]

First, find the factors of the number that ahs square root

Given,

= [tex]\frac{5}{\sqrt{2} } + \frac{9}{\sqrt{6} } - \frac{2}{\sqrt{50} } + \sqrt{32}[/tex]

Multiply the numerators by the surd of the denominators

= [tex]\frac{5 *\sqrt{2} }{\sqrt{2}*\sqrt{2 } } + \frac{9 *\sqrt{8} }{\sqrt{8}* \sqrt{8} } - \frac{2 *\sqrt{50} }{\sqrt{50 * \sqrt{50} } } + \sqrt{32}[/tex]

Multiply through and find their square root

= [tex]\frac{5\sqrt{2} }{2} + \frac{18\sqrt{2} }{8 } - \frac{10\sqrt{2} }{50} + 16\sqrt{2}[/tex]

To simply, we have

= [tex]\frac{5\sqrt{2} }{2}+ \frac{9\sqrt{2} }{4} + \frac{1\sqrt{2} }{5} + 16\sqrt{2}[/tex]

Find the LCM

= [tex]\frac{10\sqrt{2} + 45\sqrt{2}+ 4\sqrt{2} + 16\sqrt{2} }{20}[/tex]

Add through

= [tex]\frac{75\sqrt{2} }{20}[/tex]

= [tex]\frac{15\sqrt{2} }{4}[/tex]

Thus, the simplification of the expression is  [tex]\frac{15\sqrt{2} }{4}[/tex]

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Answer:

It's Micheal because he started with 6 and added 15 and he divided them into 3 piles

Answer:

147.3

Step-by-step explanation:

this is an hendecagon there is 11 sides the total interior measure is 1,620 if you divide that by each side which would be 11 you get 147.272727 which would ultimately round up to be 147.3

Answer:

13. 10

14. 51

Step-by-step explanation:

Answer:

13. 10

14. 51

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Given:

The population, P, of six towns with time t in years are given by the following exponential equations:

(i) [tex]P=1000(1.08)^t[/tex]

(ii) [tex]P = 600 (1.12)^2[/tex]

(iii) [tex]P =2500 (0.9)^t[/tex]

(iv) [tex]P=1200 (1.185)^t[/tex]

(v) [tex]P=800 (0.78)^t[/tex]

(vi) [tex]P=2000 (0.99)^t[/tex]

To find:

The town whose population is decreasing the fastest.

Solution:

The general form of an exponential function is:

[tex]P(t)=ab^t[/tex]

Where, a is the initial value, b is the growth or decay factor.

If b>1, then the function is increasing and if 0<b<1, then the function is decreasing.

The values of b for six towns are 1.08, 1.12, 0.9, 1.185, 0.78, 0.99 respectively. The minimum value of b is 0.78, so the population of (v) town  [tex]P=800 (0.78)^t[/tex] is decreasing the fastest.

Therefore, the correct option is b.

Answer:

Not a function

Step-by-step explanation:

we see on this graph,

(1,2) and (1,7)

we have two of the same x points, when put on a graph, that would be having 2 of the same inputs.

which always means it's not a function.