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Umnica [9.8K]
3 years ago
5

Find the recursive formula 32, 232, 432, 632.....

Mathematics
1 answer:
ss7ja [257]3 years ago
6 0
Looks alot like the previous term +200
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Which of the lists of letters all have line symmetry?
Sunny_sXe [5.5K]

Answer:

The first row

Step-by-step explanation:

Letters A,B,C,D

6 0
3 years ago
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Suppose y varies directly with x, and y = 15 when x = 3. What is y when x = 12?
sveticcg [70]

Answer:

y = 60 when x = 12

Step-by-step explanation:

Direct variation is

y = kx

15 = k3

Divide by 3

15/3 = k

5 =k

y = 5x

Let x = 12

y = 5*12

y = 60

5 0
3 years ago
What is this answer???
wlad13 [49]

Answer:

∠ ABC = 116°

Step-by-step explanation:

Since the angles are supplementary, they sum to 180° , that is

8x - 36 + 4x - 12 = 180

12x - 48 = 180 ( add 48 to both sides )

12x = 228 ( divide both sides by 12 )

x = 19

Then

∠ ABC = 8x - 36 = 8(19) - 36 = 152 - 36 = 116°

5 0
3 years ago
Find the value of x <br>a.2 <br>b.5<br>c.2.5<br>d.4​
nikdorinn [45]
I should say 5 is the most clearest answer to me
6 0
3 years ago
Read 2 more answers
How can i differentiate this equation?
Dmitry_Shevchenko [17]

\bf y=\cfrac{2x^2-10x}{\sqrt{x}}\implies y=\cfrac{2x^2-10x}{x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2}x^{-\frac{1}{2}} \right)}{\left( x^{\frac{1}{2}} \right)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2\sqrt{x}} \right)}{\left( x^{\frac{1}{2}} \right)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x}


\bf\cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{ \frac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2\sqrt{x}}}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2x\sqrt{x}}


\bf \cfrac{dy}{dx}=\cfrac{(4x-10)2x~~-~~(2x^2-10x)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~(2x^2-10x)}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~2x^2+10x}{2x\sqrt{x}} \implies \cfrac{dy}{dx}=\cfrac{6x^2-10x}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{2x(3x-5)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{3x-5}{\sqrt{x}}

8 0
3 years ago
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