All categories

3 years ago

A student takes 1.5 hours to complete her midterm. How many seconds did it take her to finish the assessment?

Mathematics

2 answers:

ELEN [110]3 years ago

7 0

Answer:

5400

Step-by-step explanation:

valina [46]3 years ago

5 0

Answer: 5,400 seconds

Explanation: 1.5 hours is equal to 90 minutes and there is 60 seconds in a minute all you have to do is multiply 90 x 60 hope this helped(:

You might be interested in

Suppose that a law enforcement group studying traffic

Elodia [21]

4 0.10 that is answer of this question

5 0

3 years ago

Without graphing or solving either system of equations explain how you know what the solution to the system is.

Sonbull [250]

x = 5

y = -1

Step-by-step explanation:

from second equation,

y = -2 + 1
y = -1

by putting the value of y in first equation

y = -x + 4
-1 = -x + 4
-x = -1 -4
-x = -5
x = 5

x = 5 and y = -1

3 0

4 years ago

Y= |x−3| + |x+2| − |x−5|, if x>5<br> I need to know what y equals.

olganol [36]

Answer:

$y \ge9$

Step-by-step explanation:

The given function is

$y = |x - 3| + |x + 2| - |x - 5|$

By the triangle inequality property,

$|x + y| \leqslant |x| + |y|$

and the alternate triangle inequality property;

$|x - y| \geqslant |x| - |y|$

We apply this property to get:

$y \geqslant |x | - |3| + |x | + |2| -( |x | - |5| )$

Expand parenthesis to get:

$y \geqslant |x | - |3| + |x | + |2| - |x | + |5|$

Simplify to get:

$y \geqslant |x | - 3+ |x | + 2- |x | + 5$

Group similar terms to get:

$y \geqslant |x | + |x | - |x | - 3 + 2+ 5$

$y \geqslant |x | + 4$

When x>5,

$y \geqslant |5 | + 4$

That's;

$y \geqslant 5 + 4 \\ y \geqslant 9$

5 0

3 years ago

16) Four students simplified the expression given below :(6 + 5 × 4 ) ÷ 2 , Anshu's solutions is 25 , Ram's solution is 46, Kish

Juli2301 [7.4K]

Answer:

Step-by-step explanation:

6 0

3 years ago

Use the Divergence Theorem to evaluate the following integral S F · N dS and find the outward flux of F through the surface of t

Xelga [282]

Answer:

Result;

$\int\limits\int\limits_S { \textbf{F}} \, \cdot \textbf{N} d {S} = 32\pi$

Step-by-step explanation:

Where:

F(x, y, z) = 2(x·i +y·j +z·k) and

S: z = 0, z = 4 -x² - y²

For the solid region between the paraboloid

z = 4 - x² - y²

div F

For S: z = 0, z = 4 -x² - y²

We have the equation of a parabola

To verify the result for F(x, y, z) = 2(x·i +y·j +z·k)

We have for the surface S₁ the outward normal is N₁ = -k and the outward normal for surface S₂ is N₂ given by

$N_2 = \frac{2x \textbf{i} +2y \textbf{j} + \textbf{k}}{\sqrt{4x^2+4y^2+1} }$

Solving we have;

$\int\limits\int\limits_S { \textbf{F}} \, \cdot \textbf{N} d {S} = \int\limits\int\limits_{S1} { \textbf{F}} \, \cdot \textbf{N}_1 d {S} + \int\limits\int\limits_{S2} { \textbf{F}} \, \cdot \textbf{N}_2 d {S}$

Plugging the values for N₁ and N₂, we have

$= \int\limits\int\limits_{S1} { \textbf{F}} \, \cdot \textbf{(-k)}d {S} + \int\limits\int\limits_{S2} { \textbf{F}} \, \cdot \frac{2x \textbf{i} +2y \textbf{j} + \textbf{k}}{\sqrt{4x^2+4y^2+1} } d {S}$

Where:

F(x, y, z) = 2(xi +yj +zk) we have

$= -\int\limits\int\limits_{S1} 2z \ dA + \int\limits\int\limits_{S2} 4x^2+4y^2+2z \ dA$

$= -\int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 2z \ dA + \int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 4x^2+4y^2+2z \ dA$

$= \int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 4x^2+4y^2 \ dxdy$

$= \int\limits^2_{-2} \frac{(16y^2 +32)\sqrt{-(y^2-4)} }{3} dy$

= 32π.

6 0

4 years ago