Question 3 of 10

Determine the maximum r-value of the polar equation r =3+3 cos 0

AURORKA [14]

[r] =6

Solve for r by simplifying both sides of the equation, then isolating the variable.

<em> </em>I hope this makes sense

8 0

3 years ago

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A student who takes a multiple-choice test by reading the stem of each item, generating the correct response before looking at t

Ludmilka [50]

Answer:

(A) a heuristic

Explanation:

A heuristic:It is a reasoning strategy to find answers, make judgement about any something."

It is possible to choose between the options given, we have the following interpretation as; Interpretation: "A heuristic: finding answers is a thinking technique, evaluating something." Mechanism: In the heuristic approach the method is to find solutions or answers to a question by choosing the right and optimal compositions.

5 0

3 years ago

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

5 0

3 years ago

During the storm a tree fell over into a river what might happen to the tree

mart [117]

The tree might get swept away by the current and it will disappear when it catches on something

4 0

3 years ago

One thousand grams of seawater would consist of how many grams of dissolved substances? 65 635 35 965

Naya [18.7K]

<span>One thousand grams of seawater has 35 grams of dissolved substances ... on the average. While the salinity of the Earth's oceans and seas varies, the average salinity of seawater rests at 3.5%. Consider one liter or sea or ocean water. One liter has 1,000 milliliters (mL) in it. To find 3.5% of 1,000, we would multiply with the decimal place adjusted for percentages: 1000 x .035 = 35. Therefore, for every 1,000 mL of seawater, we will find 35 grams of (mostly) sodium chloride, otherwise known as salt.</span>

7 0

3 years ago

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