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2 years ago

-9y + 8(4z + 3y) -6x PLZ HELP

Mathematics

2 answers:

hichkok12 [17]2 years ago

4 0

Answer: 32z+15y-6x

Step by step:

-9y+8(4z+3y)-6x

-9y+32z+24y-6x

15y+32z-6x

32z+15y-6x

miss Akunina [59]2 years ago

3 0

Answer:

15y + 32z - 6x

Step-by-step explanation:

-9y + 8(4z + 3y) -6x

=> -9y + 32z + 24y - 6x

=> 15y + 32z - 6x

Hoped this helped.

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Which of the following is a solution to the inequality below? 12 - x/11 > 8

n200080 [17]

Answer:

12 - x/11 > 8

12 -8 - x/11 > 8 -8

4 -x/11 > 0

4 -x/11 + x/11> 0 + x/11

4 > x/11

44 > x

x is less than 44

Step-by-step explanation:

7 0

2 years ago

identify all of the central angles. please help me asap. i have to have this done soon! :( thann you so much!

vova2212 [387]

Answer:

see explanation

Step-by-step explanation:

Central angles are angles at the centre of the circle subtended by 2 radii

Angles that meet this criteria are

∠AOB, ∠BOC, ∠AOC

8 0

3 years ago

Determine the slope of the line that has the following coordinates: (1.5, 6) (3.75, -12)

mrs_skeptik [129]

Answer:

= -18/2.25

Step-by-step explanation:

We can find the slope by using the following

m = (y2-y1)/(x2-x1)

= (-12-6)/(3.75 - 1.5)

= -18/2.25

7 0

2 years ago

Read 2 more answers

! PLEASE HELP QUICKLY !

Travka [436]

Answer:

see below

Step-by-step explanation:

y= 1/2x - 2

To find the x intercept, set y =0 and solve for x

0= 1/2x - 2

Add 2 to each side

2 = 1/2x -2+2

2 = 1/2x

Multiply by 2

2*2 = 1/2x *2

4 =x

The x intercept is 4

3 0

2 years ago

The sum of two terms of gp is 6 and that of first four terms is 15/2.Find the sum of first six terms.

Gnoma [55]

Given:

The sum of two terms of GP is 6 and that of first four terms is $\dfrac{15}{2}.$

To find:

The sum of first six terms.

Solution:

We have,

$S_2=6$

$S_4=\dfrac{15}{2}$

Sum of first n terms of a GP is

$S_n=\dfrac{a(1-r^n)}{1-r}$ ...(i)

Putting n=2, we get

$S_2=\dfrac{a(1-r^2)}{1-r}$

$6=\dfrac{a(1-r)(1+r)}{1-r}$

$6=a(1+r)$ ...(ii)

Putting n=4, we get

$S_4=\dfrac{a(1-r^4)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1-r^2)(1+r^2)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1+r)(1-r)(1+r^2)}{1-r}$

$\dfrac{15}{2}=6(1+r^2)$ (Using (ii))

Divide both sides by 6.

$\dfrac{15}{12}=(1+r^2)$

$\dfrac{5}{4}-1=r^2$

$\dfrac{5-4}{4}=r^2$

$\dfrac{1}{4}=r^2$

Taking square root on both sides, we get

$\pm \sqrt{\dfrac{1}{4}}=r$

$\pm \dfrac{1}{2}=r$

$\pm 0.5=r$

Case 1: If r is positive, then using (ii) we get

$6=a(1+0.5)$

$6=a(1.5)$

$\dfrac{6}{1.5}=a$

$4=a$

The sum of first 6 terms is

$S_6=\dfrac{4(1-(0.5)^6)}{(1-0.5)}$

$S_6=\dfrac{4(1-0.015625)}{0.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Case 2: If r is negative, then using (ii) we get

$6=a(1-0.5)$

$6=a(0.5)$

$\dfrac{6}{0.5}=a$

$12=a$

The sum of first 6 terms is

$S_6=\dfrac{12(1-(-0.5)^6)}{(1+0.5)}$

$S_6=\dfrac{12(1-0.015625)}{1.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Therefore, the sum of the first six terms is 7.875.

5 0

2 years ago