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3 years ago

How do I I figured out what 1,476 divided by 12

Mathematics

1 answer:

azamat3 years ago

6 0

Answer: 123 THIS IS CORRECT

Step-by-step explanation: 1,476 divided by 12 = 123

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Which equation represents a linear function that has a slope of 4/5<br> and a y-intercept of –6?

vredina [299]

Answer:

$\large\boxed{y=\dfrac{4}{5}x-6}$

Step-by-step explanation:

The slope-intercept form of equation of a linear function:

$y=mx+b$

m - slope

b - y-intercept

We have the slope m = 4/5 and the y-intercept b = -6. Substitute:

$y=\dfrac{4}{5}x-6$

3 0

3 years ago

Read 2 more answers

Tanner is picking three varieties of apples at an orchard: Gala, Fuji, and Jazz. Each type of apple has a different cost per pou

Makovka662 [10]

Answer:

2g + 2f = 5.16

7g + 9f = 21.24

Step-by-step explanation:

Let

Gala = g ; Fuji = f ; jazz = j

g + f + j = 5.16 - - - (1)

j = g + f - - - (2)

From (1):

g + f + g + f = 5.16

2g + 2f = 5.16 - - - (3)

Total pound picked by Tanner = 12

3 Pounds of g ; 5 pounds of f

j = 12 - (5 +3) = 4 pounds

3g + 5f + 4j = 21.24 - - - (4)

3g + 5f + 4(g + f) = 21.24

3g + 5f + 4g + 4f = 21.24

7g + 9f = 21.24 - - - - - (5)

6 0

2 years ago

3/4 greater or less than 5/7

Mars2501 [29]

3/4 is greater
hope it helps

4 0

2 years ago

Read 2 more answers

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

Please help I’m stuck on this question

yan [13]

It helps demonstrate it since it forms a right triangle in the middle with 2 base lengths and a hypotenuse. The areas of the squares can be plugged into the pythagorean theorem in order to find the answer

3 0

2 years ago