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2 years ago

Economic Leading Indicators are based on historical results. A.True B False

Mathematics

1 answer:

deff fn [24]2 years ago

4 0

I think it’s A I’m not 100 percent sure tho sorry if I’m wrong

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Which choices are equivalent to the fraction below? Check all that apply.<br> 20/24

laiz [17]

Answer:

5/6

Step-by-step explanation:

20 divide by 2 is 10

24 divide by 2 is 12

then have 10/12

10 divide by 2 is 5

12 divide by 2 is 6

answer is 5/6

7 0

2 years ago

Gary earns $42,990 per year he is paid weekly he currently has a $456 per month car loan payment and he pays $1277 per year for

Ber [7]

Answer:

Yes

Step-by-step explanation:

There are about 52 weeks in a year, 42,990 ÷ 52 = 826.73076 ,

1277 ÷ 12 = 106.41666

826.73076 - 106.41666 = 720.3141

720.3141 - 456 = 264.3141

8 0

3 years ago

PLEASE HELP!!! WHOEVER GETS THE ANSWER RIGHT GETS BRAINLIEST!!!

masha68 [24]

There is 4 X so you use 360- (11+1) which is 348 and now divide by 4 which is now 87

<h2>Thank you for asking</h2><h2 /><h2>From Cambridge</h2><h2 />

7 0

2 years ago

Read 2 more answers

Given that the points (-3, 2) and (1, 2) are vertices of a rectangle, what two sets of coordinates could form the other two vert

Lesechka [4]

Answer:

Step-by-step-by-step

explanation:

4 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