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alekssr [168]
3 years ago
5

1/3 is equal to x/216 find x

Mathematics
2 answers:
kirill [66]3 years ago
7 0

Answer:

72

Step-by-step explanation:

proportions

aleksandr82 [10.1K]3 years ago
7 0

Answer:

x = 72

Step-by-step explanation:

1/3 = x/216

cross multiply

216/3 = x

x = 72

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0.367 centimeters squared

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What is the answer to 1.1x + 1.2x - 5.4 = -10
lord [1]
1.1x+1.2x-5.4=-10

11/10x+12/10x-54/10=10

11x/2.5+2^2-1*3/5x-3^3/5=-10

(11x)+2(2*3x)+2(-3^3)/2*5=-10

11x+2(6x)+2(-27)/2*5=-10

11x+2*6x-2*27/2*5=-10

11x+12x-54/2*5=-10

23x-54/2*5=-10

23x-54=-100

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x=-2*23/23

x=-2
8 0
3 years ago
Read 2 more answers
Please help me with this question??m
Keith_Richards [23]

Answer:

75

Step-by-step explanation:

f(1) = 7

f(n) = 3f(n-1) + 3

So what you are trying to do here is find the recursive value (that's what it is called) for f(n). Computers love this sort of thing, but we humans have to do it slowly and carefully.

So let's try f(2)

That means that f(2) = 3*f(n-1) + 3

but if f(2) is used it means that you have to know f(2-1) which is just f(1) and we know that.

so f(2) = 3*f(1)+3

f(2) = 3*7 + 3

f(2) = 21 + 3

f(2) = 24

Now do it again. We now know F(2), so we should be able to find f(3)

f(3) = 3*f(3 - 1) + 3  

f(3) = 3*f(2) + 3      We know that f(2) = 24

f(3) = 3* 24 + 3

f(3) = 72 + 3

f(3) = 75

7 0
2 years ago
Can someone pls pls PLS, help me out? ASAP!! (Geometry)<br> “Complete the proof”
Maru [420]

1) \overline{AB} \perp \overline{BD}, \overline{AB} \cong \overline{DB}, \overline{AC} \cong \overline{DE} (given)

2) \angle ABD is a right angle (perpendicular lines form right angles)

3) \triangle ABD and \triangle DEB are right triangles (a triangle with a right angle is a right angle)

4) \triangle ABD \cong \triangle DEB (HL)

5) \angle ACB \cong \angle DEB (CPCTC)

8 0
2 years ago
Help! If you know this can you tell me how to do it?
aleksandr82 [10.1K]

Answer:

c

Step-by-step explanation:

Here's how this works:

Get everything together into one fraction by finding the LCD and doing the math.  The LCD is sin(x) cos(x).  Multiplying that in to each term looks like this:

[sin(x)cos(x)]\frac{sin(x)}{cos(x)}+[sin(x)cos(x)]\frac{cos(x)}{sin(x)} =?

In the first term, the cos(x)'s cancel out, and in the second term the sin(x)'s cancel out, leaving:

\frac{sin^2(x)}{sin(x)cos(x)}+\frac{cos^2(x)}{sin(x)cos(x)}=?

Put everything over the common denominator now:

\frac{sin^2(x)+cos^2(x)}{sin(x)cos(x)}=?

Since sin^2(x)+cos^2(x)=1, we will make that substitution:

\frac{1}{sin(x)cos(x)}

We could separate that fraction into 2:

\frac{1}{sin(x)}×\frac{1}{cos(x)}

\frac{1}{sin(x)}=csc(x)  and  \frac{1}{cos(x)}=sec(x)

Therefore, the simplification is

sec(x)csc(x)

5 0
3 years ago
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