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3 years ago

What's an integer for 300ft above sea level

Mathematics

2 answers:

butalik [34]3 years ago

8 0

Since its above sea level its 300+

jeyben [28]3 years ago

3 0

Above is postive. So the answer is +300

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What is the product of −2/5 and −0.8? Select all that apply.

stepan [7]

Answer:

0.32

Step-by-step explanation:

You can convert -2/5 to -0.4. Then multiply them together to get 0.32. Remember that multiplying two negatives makes a positive. Hope this helps :)

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2 years ago

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Drag each tile to the correct box. Each tile shows an investor's tax bracket along with the tax-free yield of a bond the investo

shusha [124]

The tax bracket and tax-free yield will be (18%, 3%) < (32%, 3%) < (32% , 4%) < (22% , 5%) < (24% , 6%) .

<h3>Taxable equivalent yield based problem:</h3>

The taxable equivalent yield will be:

= Tax-free yield / (100 - Tax bracket)

Taxable equivalent yield = 3 / (100 - 18) = 0.03659

Taxable equivalent yield = 6 / (100 - 24) = 0.07895

Taxable equivalent yield = 3 / (100 - 32) = 0.04412

Taxable equivalent yield = 5 / (100 - 22) = 0.06410

Taxable equivalent yield = 4 / (100 - 32) = 0.05882

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5 0

1 year ago

Pls help me with this :)<br><br><br> .

san4es73 [151]

Answer:

35 is the answer thank u 35 is a real answer

3 0

2 years ago

What is the slope of 3,5 and -3,-5

olya-2409 [2.1K]

The slope is 5/3 five over 3. You youse the formula y2 - y1 over x2-x1. Plug into calculator,and then reduce.

5 0

3 years ago

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

2 years ago