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jasenka [17]
3 years ago
10

Please help with a practice sheet my parents gave me.

Mathematics
2 answers:
Anestetic [448]3 years ago
5 0

Hope it helps

~ʆᵒŕ∂ཇꜱꜹⱽẻⱮë

KonstantinChe [14]3 years ago
3 0

Answer:

  1. Volume of cylinder =πr²h=π×4²×5=80π m³
  2. Volume of cuboid=l*b*h=7*3*2=42in³
  3. Volume of cone =(⅓πr²h)+(⅓πr²h)⅓π×(6/2)²(12+8)=60πft³

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3 years ago
Choose the correct classification of 2x4 − 8x5 − 2x2 + 5.
Dennis_Churaev [7]
I would say to the 5th degree because that's the highest exponent there but not 100% sure but I'm pretty sure
Sorry I can't help you more but I hope this helps ome
4 0
3 years ago
Read 2 more answers
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
2 years ago
tobey says the equations 3x+2y=270 and y=4/5x+20 intersect at the point (50,60).Is this true?how do you know? Be clear and compl
topjm [15]
Plug in the point to find out!
Step 1: Fill in the x and y values. x=50 and y=60
3x+2y=270 —> 3(50)+2(60)=270
y=4/5x+20 —> 60=4/5(50)+20

Step 2: solve the filled in equations.
3(50)+2(60)=270 —> 150+120=270
60=4/5(50)+20 —> 60-20=4/5(50)+20-20 —> 40=4/5(50) —> 40=0.8(50) —> 40=40

Answer: It is equivalent because it works for both equations, so it does intersect the pint (50,60).
8 0
3 years ago
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