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3 years ago

13

PLEASE HELP ME ITS DIE IN A FEW

1 answer:

Alina [70]3 years ago

8 0

It won’t let me see the photo just one second

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What is the y-intercept of the graph of the line whose equation is y=3/4x+1

Tanzania [10]

Answer:

The correct answer is "1".

Step-by-step explanation:

You need only set x to zero, and solve for y:

$y = \frac{3}{4} x + 1\\y = \frac{3}{4}0 + 1\\y = 1$

4 0

3 years ago

Please help me to answer this

Igoryamba

A. 1 = 3/3 = 4/4 = 52/52 = x/x

B. 5 = 10/2 = 15/3 = 7.5/1.5 = 120/24 = 5a/a

All the ratios must equal each other

5 0

3 years ago

A carton of large eggs way about 1.5 pounds. If a carton holds a dozen eggs, how many pounds does each egg weigh

nata0808 [166]

0.125 or 1/8 of a pound is your answer

3 0

3 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]

<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

3 years ago

The height of a certain banner is equal to one third of its length. If the banner is 5 ft tall, write the equation that can be u

pav-90 [236]

Answer:

$L=3h$

Step-by-step explanation:

Let height of a banner be $h$ , then we know that this is $\frac{1}{3}$ of the length $L$ of the banner; so we have

$h=\frac{1}{3} L$ .

Now we can rearrange this equation to solve for the length of the banner given it's height:

$L=3h$

So there we have it!

If the banner is 5 ft tall, then it's length will be:

$L=3*5=15ft$

5 0

3 years ago