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2 years ago

Can someone answer these ?!!

Mathematics

2 answers:

olya-2409 [2.1K]2 years ago

8 0

Answer:

1 shaded to 5 unshaded

Step-by-step explanation:

Counting the number of shaded parts is 1.

Counting the number of unshaded parts is 5.

Therefore, 1 shaded for 5 unshaded.

Oduvanchick [21]2 years ago

3 0

Answer:

1 shaded to 5 unshaded

Step-by-step explanation:

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B=3,c=7 Please help Iam struggling

Afina-wow [57]

Please provide more information

8 0

2 years ago

A computer store has certain software on sale at 3 for $20.00, with a limit of 3 at the sale price. Additional

alexgriva [62]

Answer:

a) y = 9.95(x-3) + 20.00

b) Cost of 5 software packages = $39.90

Step-by-step explanation:

Given that:

Cost of 3 software at sale = $20

Cost of regular = $9.95

Let,

x be the number of software purchased.

We will subtract 3 from x because they cost $20.

y be the total cost.

y = 9.95(x-3)+ 20.00

Putting x=5 in expression

y = 9.95(5-3) + 20.00

y = 2(9.95) + 20.00

y = 19.90 + 20.00

y = $39.90

Hence,

a) y = 9.95(x-3) + 20.00

b) Cost of 5 software packages = $39.90

7 0

2 years ago

We are planning a reception for our daughter. The hall that we are renting charges a $90 cleanup fee. In addition to the cleanup

iris [78.8K]

Answer:

The equation is y = 29x + 90. So, if you plug in 3,100.00 for y.

3,100.00 = 29x + 90

3,100.00 - 90 = 29x

3,010.00 = 29x

3,010.00/29 = 29x/29

x = 103 guests

4 0

2 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

1/5 x 2 1/5 will be 2 1/5.

4vir4ik [10]

Answer:less than

Step-by-step explanation:

1/5 x 2 1/5 = 11/25

2 1/5 as a mixed number is 11/5

so 11/25 is less than 11/5 therefore

1/5 x 2 1/5 < 2 1/5

5 0

2 years ago