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Ugo [173]
4 years ago
15

a circle has a radius of 2 cm. find the length s of the arc intercepted by a central angle of 1.9 radians.​

Mathematics
1 answer:
Luba_88 [7]4 years ago
3 0

Answer:

3.8 cm

Step-by-step explanation:

The arc length formula is s = r·Ф, where Ф is the central angle in radians.

Here, s = (2 cm)(1.9 rad) = 3.8 cm

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A polynomial is to be constructed that will touch the x axis at most 7 times. What is the minimum degree of the polynomial?
tia_tia [17]
7 times, means 7 x intercepts, means 7th degree

answer is 7th degree
7 0
3 years ago
PLEASE HELP ME I need help on this ASAP!!
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Answer:

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Step-by-step explanation:

5 0
3 years ago
Arnie is mixing red and yellow paints to make two different shades of orange. To make 1 cup of dark orange​ paint, he needs 7 ou
hjlf

Answer:

Arnie has enough red paint

Step-by-step explanation:

step 1

Find out the number of ounces of red paint needed to make 3 cups of dark orange paint

Let

x -----> number of ounces of red paint

y -----> number of ounces of yellow paint

z ----> cups of dark orange

we know that

\frac{x}{z}=\frac{7}{1}\ \frac{ounces\ red\ paint}{cups\ dark\ orange}

z

x

=

1

7

cups dark orange

ounces red paint

so

For z=3 cups

substitute and solve for x

\frac{x}{3}=\frac{7}{1}

3

x

=

1

7

x=21\ ounces\ of\ red\ paintx=21 ounces of red paint

step 2

Find out the number of ounces of red paint needed to make 3 cups of light orange paint

Let

x -----> number of ounces of red paint

y -----> number of ounces of yellow paint

z ----> cups of light orange

we know that

\frac{x}{z}=\frac{3}{2}\ \frac{ounces\ red\ paint}{cups\ light\ orange}

z

x

=

2

3

cups light orange

ounces red paint

so

For z=3 cups

substitute and solve for x

\frac{x}{3}=\frac{3}{2}

3

x

=

2

3

x=4.5\ ounces\ of\ red\ paintx=4.5 ounces of red paint

step 3

Adds the ounces of red paint

21+4.5=25.5\ ounces\ of\ red\ paint21+4.5=25.5 ounces of red paint

25.5\ ounces < 32\ ounces25.5 ounces<32 ounces

therefore

Arnie has enough red paint

7 0
3 years ago
sonya has a bag with 4 green marbles, 7 orange marbles and 9 blue marbles. She randomly selects one marble, does not replace it
wel
What’s the answer choices?
8 0
3 years ago
Over the closed interval [3,8] for which function can the extreme value theorem be applied?
vodka [1.7K]

The extreme value theorem can be applied on an interval if the function is continuous in the entire interval. Testing the continuity for each function, we get that the correct option is the third option.

Continuity:

A function f is continuous at an interval if all points in the interval are in the domain of the function, and, for each point of x^{\ast}, the limit exists and:

\lim_{x \rightarrow x^{\ast}} = f(x^{\ast})

First function:

At x = 4, the denominator is 0, and so, the extreme value theorem cannot be applied.

Second function:

At x = 5, the denominator is 0, and so, the extreme value theorem cannot be applied.

Third function:

The only point there can be a discontinuity is at x = 4, where the definition of the function changes. First we have to find the lateral limits, and if they are equal, the limits exist:

To the left(-), it is less than 4, so we take the definition for x < 4.

To the right(+), it is more than 4, so we take the definition for x >= 4.

\lim_{x \rightarrow 4^{-}} h(x) = \lim_{x \rightarrow 4} \frac{9x}{10-x} = \frac{9*4}{10-4} = \frac{36}{6} = 6

\lim_{x \rightarrow 4^{+}} h(x) = \lim_{x \rightarrow 4} x + 2 = 4 + 2 = 6

The lateral limits are equal, so the limit exists, and it's value is 6.

The definition at x = 4 is h(x) = x + 2, so h(4) = 4 + 2 = 6.

Since \lim_{x \rightarrow 4} = h(4), the function is continuous over the entire interval, and this is the correct answer.

Fourth function:

There can be a discontinuity at x = 5, so we test the limits:

\lim_{x \rightarrow 5^{-}} h(x) = \lim_{x \rightarrow 5} -x = -5

\lim_{x \rightarrow 5^{+}} h(x) = \lim_{x \rightarrow 5} x^2 - 20 = 5^2 - 20 = 25 - 20 = 5

Different limits, so the limit does not exist and the function is not continuous at x = 5, and the extreme value theorem cannot be applied.

For more on the extreme value theorem, you can check brainly.com/question/15585098

6 0
3 years ago
Read 2 more answers
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