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2 years ago

Calculate the perimeter of a triangle with the following measurements: 200 mm, 40 cm and 400 dm Please HELP

Mathematics

1 answer:

melisa1 [442]2 years ago

5 0

Answer:

640 dm

Step-by-step explanation:

200+40+400=640dm

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Rapunzel sees a flying saucer. The radius of the flying saucer is 18 feet. What is the area of the flying saucer using 3.14 roun

Vladimir79 [104]

Answer:

$1017.4ft^{2}$

Step-by-step explanation:

Area of a circle = $\pi r^{2}$

However, the question wants to use 3.14 instead of $\pi$ .

Since the radius is given, which is 18 feet, we can plug and chug.

$A = 3.14*(18ft)^{2} \\A = 3.14*324ft^{2}\\A = 1017.36ft^{2}$

However, the question wants us to round it to the nearest tenth.

Therefore, the answer is $1017.4ft^{2}$

4 0

3 years ago

Find the total surface area and volume of a cylinder with a radius of 2 and a height of 4.

Margarita [4]

Answer:

75.4 units³

Step-by-step explanation:

SA = 2×pi×r×(r+h)

= 2×pi×2×(2+4)

= 24pi

= 75.3982236861

= 75.4 units²

4 0

3 years ago

Pls help me

Vladimir [108]

<h3>Learning Task 1</h3>

Correct responses;

$\displaystyle 1. \hspace{0.3 cm} 7^{-1} = \frac{1}{7}$

2. (14·a·b·c)⁰ = 1

$\displaystyle 3. \hspace{0.3 cm} 10^{-9} = \frac{1}{10^9}$

4. 5·(x·y)⁰ = 5

5. 0¹⁵ = 0

$\displaystyle 6. \hspace{0.3 cm} \frac{24 \cdot x^8 \cdot y^4}{6 \cdot x^5 \cdot y^4} = 4 \cdot x^3$

$\displaystyle 7. \hspace{0.3 cm} \left(\frac{5 \cdot x^0}{y} \right)^{-1} = \frac{y}{5}$

$8. \hspace{0.3 cm}\displaystyle \frac{120 \cdot a^5 \cdot b^6 \cdot c^{-5} }{12 \cdot a^{-2} \cdot b^0 \cdot c^{2}} = \frac{10 \cdot a^7 \cdot b^6}{c^7}$

$\displaystyle 9. \hspace{0.3 cm} \frac{(4 \cdot x)^0 \cdot y^{-5} \cdot z^{-2} }{(234 \cdot x \cdot y \cdot z)^0} = \frac{1}{y^{5} \cdot z^2}$

$\displaystyle 10. \hspace{0.3 cm} \left(\frac{(5 \cdot x \cdot y)^0}{10} \right)^{-2} = 100$

$11. \displaystyle \hspace{0.3 cm} \left(\frac{(a \cdot b)}{9} \right)^{-1} = \frac{9}{a \cdot b}$

$\displaystyle 12. \hspace{0.3 cm} \frac{9}{x^{-2}} = 9 \cdot x^2$

$13. \displaystyle \hspace{0.3 cm} \frac{12 \cdot b^{-3}}{a^{-4}} = \frac{12 \cdot a^4}{b^3}$

$14. \displaystyle \hspace{0.3 cm} \frac{1}{a^{-5 \cdot n}} =a^{5 \cdot n}$

$15. \displaystyle \hspace{0.3 cm} \left(\frac{1}{2} \right)^{-5} = 32$

<h3>Methods by which the above expressions are simplified</h3>

The given expressions expressed into non zero and non negative exponents are;

$\displaystyle 1. \hspace{0.3 cm} \mathbf{7^{-1}} = \underline{\frac{1}{7}}$

2. (14·a·b·c)⁰ = 14⁰ × a⁰ × b⁰ × c⁰ = 1 × 1 × 1 × 1 =<u> 1</u>

$\displaystyle 3. \hspace{0.3 cm} \mathbf{ 10^{-9}} = \underline{ \frac{1}{10^9}}$

4. 5·(x·y)⁰ = 5 × x⁰ × y⁰ = 5 × 1 × 1 = <u>5</u>

5. 0¹⁵ = <u>0</u>

$\displaystyle 6. \hspace{0.3 cm} \mathbf{\frac{24 \cdot x^8 \cdot y^4}{6 \cdot x^5 \cdot y^4}} = \frac{ 4 \times 6 \times x^5 \times x^3 \times y^4}{6 \times x^5 \times y^4} = \underline{4 \cdot x^3}$

$\displaystyle 7. \hspace{0.3 cm} \mathbf{\left(\frac{5 \cdot x^0}{y} \right)^{-1} }= \frac{1}{\frac{5 \times 1}{y} } = \underline{\frac{y}{5}}$

$8. \hspace{0.3 cm}\displaystyle \mathbf{\frac{120 \cdot a^5 \cdot b^6 \cdot c^{-5} }{12 \cdot a^{-2} \cdot b^0 \cdot c^{2}}} = 10 \cdot a^{5 - (-2)} \cdot b^{6 - 0} \cdot c^{-5 -2} = \underline{\frac{10 \cdot a^7 \cdot b^6}{c^7}}$

$\displaystyle 9. \hspace{0.3 cm} \mathbf{\frac{(4 \cdot x)^0 \cdot y^{-5} \cdot z^{-2} }{(234 \cdot x \cdot y \cdot z)^0}} = y^{-5} \cdot z^{-2} = \underline{\frac{1}{y^{5} \cdot z^2}}$

$\displaystyle 10. \hspace{0.3 cm} \mathbf{\left(\frac{(5 \cdot x \cdot y)^0}{10} \right)^{-2}} = \left(\frac{1}{10} \right)^{-2} = \frac{1}{\left(\frac{1}{10} \right)^2 } = \frac{1}{\frac{1}{100} } = \underline{100}$

$11. \displaystyle \hspace{0.3 cm} \mathbf{\left(\frac{(a \cdot b)}{9} \right)^{-1}} = \frac{1}{\left(\frac{(a \cdot b)}{9} \right) } = \underline{\frac{9}{a \cdot b}}$

$\displaystyle 12. \hspace{0.3 cm} \mathbf{\frac{9}{x^{-2}}} = \frac{9}{\frac{1}{x^2} } = \underline{9 \cdot x^2}$

$13. \displaystyle \hspace{0.3 cm} \mathbf{\frac{12 \cdot b^{-3}}{a^{-4}}} = 12 \cdot \frac{b^{-3}}{a^{-4}} =12 \cdot \frac{1}{\frac{b^3}{a^4} } = 12 \cdot \frac{a^4}{b^3} = \underline{\frac{12 \cdot a^4}{b^3}}$

$14. \displaystyle \hspace{0.3 cm} \mathbf{\frac{1}{a^{-5 \cdot n}}} = \frac{1}{\frac{1}{a^{5 \cdot n}} } = \underline{a^{5 \cdot n}}$

$15. \displaystyle \hspace{0.3 cm} \mathbf{\left(\frac{1}{2} \right)^{-5}} = \frac{1}{\left(\frac{1}{2} \right)^5} = 2^5 = \underline{32}$

Learn more about the laws of indices here:

brainly.com/question/8959311

6 0

1 year ago

Femi keeps track of a number of gallons of gasoline he uses for each of several trips. Part A: Complete the table. The vehicle c

Katena32 [7]

The table is attached.

A) The gasoline consume grows as the distance traveled increases. This means that the two quantities are directly proportional.

Therefore, the proportionality constant is given by:
k = gasoline / distance
   = 2 / 40
   = 0.05
We could have used also:
k = 3 / 60
   = 0.05

In order to find the gasoline consumed, you need to multiply the distance by the proportionality constant:
30 × 0.05 = 1.5
55 × 0.05 = 2.75

In order to find the distance traveled, you need to divide the gasoline consumed by the proportionality constant:
3.5 ÷ 0.05 = 70

B) The function of the proportionality found is:
y = 0.05·x
where:
x = distance
y = gasoline

Therefore:
y = 0.05·110 = 5.5

Femi for a trip of 110 miles expects to use 5.5 gallons of gasoline.

7 0

2 years ago

What is the length of BC, AC, ED, DK, &’ EK?

Tpy6a [65]

For the truth

am not sure

Answer:

BC = 12, AC = 12, ED = 8, DK = 14, EK = 12

7 0

2 years ago