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1 year ago

Angles H and F are corresponding angles True or False?

Mathematics

2 answers:

lutik1710 [3]1 year ago

5 0

True! they are corresponding angles.

Crazy boy [7]1 year ago

3 0

TRUE my bro its true

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4.6.PS-12 Question Help Frank is going to plant n vegetable seeds in one garden and 3n + 2 vegetable seeds in another. How many

miskamm [114]

Answer:

Step-by-step explanation:

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2 years ago

Which of the following graphs could be the graph of the function f(x)=-0.08x(x^2-11x+18)?

MariettaO [177]

Answer:

option C

Step-by-step explanation:

f(x)=-0.08x(x^2-11x+18)

Factor the parenthesis and find out the x intercepts

(x^2-11x+18)

We find out two factors whose product is 18 and sum is -11

-9 and -2 gives us product 18

-9 -2 gives us sum -11

(x^2-11x+18) =(x-9)(x-2)

f(x)=-0.08x(x^2-11x+18)

f(x)= -0.08x(x-9)(x-2)

set f(x) =0 and solve for x

-0.08x(x-9)(x-2) =0

-0.08x =0, so x=0

x-9 = 0 , so x=9

x-2 =0 , so x=2

x intercepts are (0,0) (2,0) and (9,0)

x intercepts are not repeating

so the graph crosses x -axis at x=0, 2,9

option C is correct

The graph is attached below4

5 0

2 years ago

Read 2 more answers

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]

<u>Answer-</u>

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

<u>Solution-</u>

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

2 years ago

Hurry up plz it’s due in like 5 mins

777dan777 [17]

uhm i think its 15 cause you just add them all i guess

7 0

2 years ago

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Mark is pitching a tent he ties a rope from a nail to the top of the tent what is the length of the rope between the nail and th

Leto [7]

I believe that it is 7

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3 years ago