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1 year ago

Can you guys help me please

Mathematics

1 answer:

Bogdan [553]1 year ago

6 0

The shortest length of fabric, in inches, that Kirk could use to make the tablecloth without putting any separate pieces of fabric together is 30 inches

<h3>How to solve for the shortest length of fabric required to make the table cloth</h3>

<u>Given data</u>

diameter of the table = 3 feet

hang down dimension = 5 inches

sew down dimension = 1 inch

width of rectangular fabric = 60 inches

The question seeks for area of the table cloth, for a circular table area of a circle = pi * r^2

required radius of cloth

= radius of the table + hang down dimension + sew down dimension

= 3 feet / 2 + 5 inches + 1 inch

= 1.5 * 12 inches + 5 inches + 1 inch

= 18 + 5 + 1

= 24 inches

Area of required cloth

= pi * 24^2

= 1809.557 square inches

Area of fabric = Area of required cloth

length * wide = 1809.557 square inches

length * 60 = 1809.557 square inches

length = 1809.557 / 60 square inches

length = 30.159 inches ≈ 30 inches

Therefore option H which is 30 inches is the answer

Read more on areas here: brainly.com/question/25292087

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Lady_Fox [76]

Answer:

B. - 0.383

Step-by-step explanation:

Since cotangent of angle is:

cot λ = cos λ / sin γ

And

cos λ = x/1, sin λ = y / 1

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cot λ = x/y = - 0.358/0.934 = - 0.383 (rounded)

Correct choice is B

8 0

2 years ago

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Step-by-step explanation:

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6 0

4 years ago

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stira [4]

You know that there is a right angle which equals 90 degrees, you take 90 + 67 +37 which equals 194. You have 2 x's so it should equal 2x. Then you take 194 divided by 2, and you should have the answer to what x equals

4 0

4 years ago

(15 Points)

expeople1 [14]

ANSWER 1

Note that,

$f(u)=tan^{-1}(u)$

is the same as

$f(u)=arctan(u)$

We apply the product rule.

$f(x)=x^2tan^{-1}(x)$

So we keep the second function and differentiate the first,plus we keep the first function and differentiate the second.

$f'(x)=(x^2)'tan^{-1}(x)+x^2(tan^{-1}(x))'$

Recall that,

$f(u)=tan^{-1}(u)$

Then,

$f'(u)=\frac{1}{1+u^2}} \times u'$

This implies that,

$f'(x)=2xtan^{-1}(x)+\frac{x^2}{x^2+1}$

ANSWER 2

We apply the product rule and the chain rules of differentiation here.

$f(x)=xsin^{-1}(1-x^2)$

$f'(x)=x'sin^{-1}(1-x^2)+x(sin^{-1}(1-x^2))'$

Recall that,

$f(u)=sin^{-1}(u)$

Then,

$f'(u)=\frac{1}{\sqrt{1-u^2}} \times u'$

This implies that,

$f'(x)=sin^{-1}(1-x^2)+x \times \frac{1}{\sqrt{1-(1-x^2)^2}}\times (-2x)$

$f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-(1-2x^2+x^4)}}$

$f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-1+2x^2-x^4}}$

$f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{2x^2-x^4}}$

5 0

4 years ago

Sarah invested £12000 in a unit trust 5 years ago, the value of the unit trust increased by 7% per annum for each of the last 3

trasher [3.6K]

Answer:

The current price of the unit trust = £13,831.72

Step-by-step explanation:

Since it increased 7% per annum in last three years and decreased by 3% per annum before that, it implies that the unit trust decreased by 3% per annum for the first 2 years and then increased by 7% per annum for the next 3 years in the total 5 year period

The invested after 2 years = 12,000*(1-0.03)^2 = £11,290.8

This amount then grows by 7% for the next 3 years making it = 11,290.80*(1+0.07)^3 = £13,831.7155 = £13,831.72 (Rounded to 2 decimals)

The current price of the unit trust = £13,831.72 (Rounded to 2 decimals)

6 0

4 years ago