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Alinara [238K]
3 years ago
5

Using paper folding to construct a line perpendicular to a given line through a point fold the paper through the point so that t

he given line segment lies:
A. Perpendicular to the paper's edge
B. On the point
C. At an angle
D. Upon Itself
Mathematics
2 answers:
Sonja [21]3 years ago
5 0
Correct answer: D. Upon Itself
A line segment has an angle that is exactly 180°. As perpendicular lines are lines meeting at the right angle, or 90°. If we then use paper folding to construct a line perpendicular to a given line, the given line should be directly above each other once folded.
Pavlova-9 [17]3 years ago
3 0

Upon Itself ~~~~~~~~~~~~~~~~~ APEX

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What is the simplified form of the following expression? Assume x greater-than-or-equal-to 0 and y greater-than-or-equal-to 0 2
finlep [7]

To solve such questions we need to know more about expression.

<h2 /><h2>Expression </h2>

In mathematics, an expression is defined as a set of numbers, variables, and mathematical operations that are formed according to rules which are dependent on the context.  

<h2 /><h2 /><h2>Given to us,</h2>

2\sqrt[4]{16x}-2\sqrt[4]{2y}+3\sqrt[4]{81x}-4\sqrt[4]{32y}+5\sqrt[4]{x}-4\sqrt[4]{32y}+5\sqrt[4]{x}-16\sqrt[4]{2y}+13\sqrt[4]{4}-10\sqrt[4]{2y}+35\sqrt[4]{x}-18\sqrt[4]{2y}

rearranging we get,

=2\sqrt[4]{16x}+3\sqrt[4]{81x}+5\sqrt[4]{x}+5\sqrt[4]{x}+13\sqrt[4]{4}+35\sqrt[4]{x}-2\sqrt[4]{2y}-4\sqrt[4]{32y}-4\sqrt[4]{32y}-16\sqrt[4]{2y}-10\sqrt[4]{2y}-18\sqrt[4]{2y}

we know that,

\sqrt[4]{81}=3\\\sqrt[4]{16}=2\\\sqrt[4]{32}=2\sqrt[4]{2}\\

therefore,

=4\sqrt[4]{x}+9\sqrt[4]{x}+5\sqrt[4]{x}+5\sqrt[4]{x}+13\sqrt[4]{x}+35\sqrt[4]{x}-2\sqrt[4]{2y}-8\sqrt[4]{2y}-8\sqrt[4]{2y}-16\sqrt[4]{2y}-10\sqrt[4]{2y}-18\sqrt[4]{2y}

taking \sqrt[4]{x} and \sqrt[4]{2y} as common

=[(\sqrt[4]{x})(4+9+5+5+13+35)]-[(2+8+8+16+10+18)(\sqrt[4]{2y})]

Further simplifying

=71\sqrt[4]{x}-62\sqrt[4]{2y}

Learn more about the expression:

brainly.com/question/13947055?referrer=searchResults

7 0
3 years ago
Read 2 more answers
keith spent 60% of jis birthday money at the mall. If he spent $24 at the mall, how much money did Keith receive for his birthda
rodikova [14]
24 x 100 = 2400

2400 divided by 60 = 40

Keith got $40 for his birthday.
8 0
3 years ago
Read 2 more answers
How do you do pictures
VikaD [51]

Answer:

click the add attachments file and upload

Step-by-step explanation:


6 0
3 years ago
let x = the amoun of raw sugar in tons a procesing plant is a sugar refinery process in one day . suppose x can be model as expo
anygoal [31]

Answer:

The answer is below

Step-by-step explanation:

A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently,a.Find the probability that any given plant processes more than 5 tons of raw sugar on a given day.b.Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day.c.How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05?

Answer: The mean (μ) of the plants is 4 tons. The probability density function of an exponential distribution is given by:

f(x)=\lambda e^{-\lambda x}\\But\ \lambda= 1/\mu=1/4 = 0.25\\Therefore:\\f(x)=0.25e^{-0.25x}\\

a) P(x > 5) = \int\limits^\infty_5 {f(x)} \, dx =\int\limits^\infty_5 {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_5=e^{-1.25}=0.2865

b) Probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day can be solved when considered as a binomial.

That is P(2 of the three plant use more than five tons) = C(3,2) × [P(x > 5)]² × (1-P(x > 5)) = 3(0.2865²)(1-0.2865) = 0.1757

c) Let b be the amount of raw sugar should be stocked for the plant each day.

P(x > a) = \int\limits^\infty_a {f(x)} \, dx =\int\limits^\infty_a {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_a=e^{-0.25a}

But P(x > a) = 0.05

Therefore:

e^{-0.25a}=0.05\\ln[e^{-0.25a}]=ln(0.05)\\-0.25a=-2.9957\\a=11.98

a  ≅ 12

6 0
3 years ago
The following figure shows a circle with centre O
Semmy [17]

Step-by-step explanation:

This is tough! However, using a little bit of logic and playing around with formulas, we can solve this!

We know that an inscribed angle = (1/2) intercepted arc. The arc is directly correlated to the area with 16 degrees.  Therefore, r=(1/2)(16)=8. An inscribed angle counts as long as its sides intersect the circle to form an intercepted arc, and this is not an option given, so I can understand why you're having trouble. I would recommend asking your teacher about this.

7 0
3 years ago
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