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3 years ago

HELP PLEASE I REALLY NEED IT

Mathematics

1 answer:

Svetlanka [38]3 years ago

5 0

Answer:

Step-by-step explanation:

1/2 bh

1/2 (6)(6) multiply

1/2 (36) divide

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Which recursive formula correctly models the values shown in the table?

Vedmedyk [2.9K]

The right answers are

1. C, 1

2. B, an+1 = an x 3, a1 =5

3. C, an = -6 x 0.5^n-1

4. D, an + 2/3 x (1/2)^n-1

5. D, an+1 = an x (-2), a1 = 5

I just took the test and got 100% so these should be correct

4 0

3 years ago

Can someone help me PQR is similar to XYZ

enot [183]

Answer:

D. 126

Step-by-step explanation:

30 ÷ 5 = 6

5 * 6 = 30

6 * 6 = 36

10 * 6 = 60

60 + 30 + 36 = 126

6 0

3 years ago

Read 2 more answers

Sam buys 4 tropical fish. Each fish is 5/8 of an inch long. Write an equation that represents 4 x 5/8 as a multiple of a unit fr

UkoKoshka [18]

4 x 5/8= 20/8= 10/4=5/2 inches lng

3 0

3 years ago

Help pls :))))))))))))

Aneli [31]

Answer:

yes, the triangle are congruent by SAS

6 0

2 years ago

At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

Answer-

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

Solution-

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$

Now, equating this to 0

$(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x}) =0$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}-(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}=(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{1}{2}}=48e^{2x}$

$\Rightarrow (1+16e^{2x})}=48^2e^{2x}=2304e^{2x}$

$\Rightarrow 2304e^{2x}-16e^{2x}-1=0$

Solving this eq,

we get $x= \frac{1}{2304e^4-16e^2}$

∴ At $x= \frac{1}{2304e^4-16e^2}$ the curvature is maximum.

6 0

3 years ago