9.3, 9 1/3, 37/4, and 9.34 least to greatest

Why do non STEM majors detest taking algebraic courses at the collegiate level?

yKpoI14uk [10]

Answer:

Because they find it hard to pass in the past.

Step-by-step explanation:

Non-STEM majors detest taking algebraic courses at the collegiate level because based on their past experiences in high school, whereby they probably do not excel in algebra, numbers, or mathematical subjects in general, the thought of going through advanced numbers and equations can be overwhelming.

Nobody wants to have or at least start college with poor grades. Hence, because non-STEM students find algebraic courses hard to pass in the past, they detest taking the course that has to do with it at the college level.

7 0

3 years ago

Taylor leaves her house and walks to school She walks 40 meters north (vertically and 9 meters east hozontally . How far Taylor

OleMash [197]

Answer:

40 m

Step-by-step explanation:

4 0

3 years ago

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Please answer. x-4/3 - 10 The photo is below.

wariber [46]

X-4/3=10
Multiply 3 on both sides
x-4=30
Add 4 to both sides
Final Answer: x=34

5 0

3 years ago

Find the length of the function over the given interval. y = 3x from x = 0 to x = 2

olganol [36]

The length of the function y = 3x over the given interval [0, 2] is 3.2 units

For given question,

We have been given a function y = 3x

We need to find the length of the function on the interval x = 0 to x = 2.

Let f(x) = 3x where f(x) = y

We have f'(x) = 3, so [f'(x)]² = 9.

Then the arc length is given by,

$\int\limits^a_b {\sqrt{1+[f'(x)]^2} } \, dx\\\\= \int\limits^2_0 {\sqrt{1+9} }\, dx\\\\=\sqrt{10}\\\\ =3.2$

This means, the arc length is 3.2 units.

Therefore, the length of the function y = 3x over the given interval [0, 2] is 3.2 units

Learn more about the arc length here:

brainly.com/question/10729208

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3 0

2 years ago

Assume that the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder. Based on this assumption,

kompoz [17]

If the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder, then its volume is

$V_{flask}=V_{sphere}+V_{cylinder}.$

Use following formulas to determine volumes of sphere and cylinder:

$V_{sphere}=\dfrac{4}{3}\pi R^3,\\ \\V_{cylinder}=\pi r^2h,$

wher R is sphere's radius, r - radius of cylinder's base and h - height of cylinder.

Then

$V_{sphere}=\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi \left(\dfrac{4.5}{2}\right)^3=\dfrac{4}{3}\pi \left(\dfrac{9}{4}\right)^3=\dfrac{243\pi}{16}\approx 47.71;$
$V_{cylinder}=\pi r^2h=\pi \cdot \left(\dfrac{1}{2}\right)^2\cdot 3=\dfrac{3\pi}{4}\approx 2.36;$
$V_{flask}=V_{sphere}+V_{cylinder}\approx 47.71+2.36=50.07.$

Answer 1: correct choice is C.

If both the sphere and the cylinder are dilated by a scale factor of 2, then all dimensions of the sphere and the cylinder are dilated by a scale factor of 2. So

R'=2R, r'=2r, h'=2h.

Write the new fask volume:

$V_{\text{new flask}}=V_{\text{new sphere}}+V_{\text{new cylinder}}=\dfrac{4}{3}\pi R'^3+\pi r'^2h'=\dfrac{4}{3}\pi (2R)^3+\pi (2r)^2\cdot 2h=\dfrac{4}{3}\pi 8R^3+\pi \cdot 4r^2\cdot 2h=8\left(\dfrac{4}{3}\pi R^3+\pi r^2h\right)=8V_{flask}.$

Then

$\dfrac{V_{\text{new flask}}}{V_{\text{flask}}} =\dfrac{8}{1}=8.$

Answer 2: correct choice is D.

8 0

4 years ago

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