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irinina [24]
2 years ago
15

What does x equal with this equation x/6-33= -27

Mathematics
2 answers:
lesya692 [45]2 years ago
8 0

Answer:

The first thing that we have to do is add 33 to both sides to try to reduce the amount of variables on the left hand side.

\frac{x}{6} - 33 = -27

\frac{x}{6} - 33 + 33 = -27 + 33

\frac{x}{6} = 6

Now that we have simplified the left hand side even more we only have one part left to get x by itself which is to multiply both sides by 6.

\frac{x}{6} = 6

\frac{x}{6} * 6 = 6 * 6

x = 36

Therefore, our final answer for what the value of x equals is 36.

<u><em>Hope this helps!  Let me know if you have any questions</em></u>

Dafna11 [192]2 years ago
5 0
X/6-33=-27
-27+33=6
x/6=6
6x6=36
X=36
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Given: Sides lengths of the triangle : 16 units, 10 units, 8 units.

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Then,

s=\dfrac{a+b+c}{2}=\dfrac{16+10+8}{2}=17

Using Heron's formula:-

\text{Area of triangle}=\sqrt{17(17-16)(17-10)(17-8)}\\\\\Rightarrow\ \text{Area of triangle}=\sqrt{1071}=32.7261363439\approx33\text{ square units}

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0.2 is 1/10 of 2 true or false
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true

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Circle A has center of (4,5) and a radius of 3 and circle B has a center of (1,7) and a radius of 9. What steps will help show t
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Moving the circles to the same center would make them concentric.  Then, by multiplying the radius of Circle A by 3, the circle is dilated to the exact size of circle B, since its radius is 9.  This shows that Circle B is proportional to Circle A, thus they are similar.  

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A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20
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Answer:

The smallest possible value for the third term of the geometric progression is 1.

Step-by-step explanation:

Given : A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting numbers form a geometric progression.

To find : What is the smallest possible value for the third term of the geometric progression?  

Solution :

The arithmetic progression is given by a,a+d,a+2d

First term a=9

Second term - 2 is added to second term i.e. a+d+2=9+d+2=11+d

Third term - 20 is added to the third term i.e.a+2d+20=9+2d+20=29+2d

The geometric progression is given by a,ar,ar^2

First term a=9

Second term -ar=11+d

Third term ar^2=29+2d

r is the common ratio which is second term divided by first term,

So, r=\frac{ar}{a}=\frac{11+d}{9}

or  third term divided by second term,

So, r=\frac{ar^2}{ar}=\frac{29+2d}{11+d}

Equating both the r,

\frac{11+d}{9}=\frac{29+2d}{11+d}

Cross multiply,

(11+d)\times (11+d)=(29+2d)(9)

11^2+11d+11d+d^2=261+18d

121+22d+d^2-261-18d=0

d^2+4d-140=0

Solving by middle term split,

d^2+14d-10d-140=0

d(d+14)-10(d+14)=0

(d+14)(d-10)=0

d=-14,10

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Third term ar^2=29+2d

When d=-14,

Third term ar^2=29+2(-14)=29-28=1

When d=10,

Third term ar^2=29+2(10)=29+20=49

Therefore, The smallest possible value for the third term of the geometric progression is 1.

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3 years ago
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