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Marat540 [252]
4 years ago
11

What is the value of 10+/-3-6/

Mathematics
1 answer:
garri49 [273]4 years ago
7 0

Answer:

18

Step-by-step explanation:

-3-6= -8 but since its absolute value it'll be 8 so 8 plus 10 = 18

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Mathematics, 19.11.2019 04:31, RiverH246
Darya [45]

The volume of one tissue box is 170 cubic inches, so the volume of 25 tissue boxes are 4250 cubic inches.

3 0
4 years ago
Type the correct answer in the box.
lana [24]

Correct question is;

Type the correct answer in the box bellow: If cos x = sin(20 + x)° and 0° < x < 90°, the value of x is ...........

Answer:

x = 35°

Step-by-step explanation:

We are told that;

cos x = sin(20 + x)°

and 0° < x < 90°

From trigonometric identities, we know that;

cos x = sin (90 - x)

Applying to our question gives;

sin (90 - x) = sin (20 + x)°

Comparing both sides, we have;

90 - x = 20 + x

Rearranging;

90 - 20 = x + x

70 = 2x

x = 70/2

x = 35°

7 0
3 years ago
CAN SOMEONE HELP ME IN THIS INTEGRAL QUESTION PLS
finlep [7]

Due to the symmetry of the paraboloid about the <em>z</em>-axis, you can treat this is a surface of revolution. Consider the curve y=x^2, with 1\le x\le2, and revolve it about the <em>y</em>-axis. The area of the resulting surface is then

\displaystyle2\pi\int_1^2x\sqrt{1+(y')^2}\,\mathrm dx=2\pi\int_1^2x\sqrt{1+4x^2}\,\mathrm dx=\frac{(17^{3/2}-5^{3/2})\pi}6

But perhaps you'd like the surface integral treatment. Parameterize the surface by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k

with 1\le u\le2 and 0\le v\le2\pi, where the third component follows from

z=x^2+y^2=(u\cos v)^2+(u\sin v)^2=u^2

Take the normal vector to the surface to be

\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial u}=-2u^2\cos v\,\vec\imath-2u^2\sin v\,\vec\jmath+u\,\vec k

The precise order of the partial derivatives doesn't matter, because we're ultimately interested in the magnitude of the cross product:

\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|=u\sqrt{1+4u^2}

Then the area of the surface is

\displaystyle\int_0^{2\pi}\int_1^2\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\int_0^{2\pi}\int_1^2u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv

which reduces to the integral used in the surface-of-revolution setup.

7 0
3 years ago
Madison created two functions.
Degger [83]
Madison created two functions.For Function A, the value of y is two less than four times the value of x. 

The table below represents Function B.
-3,-9
-1,5
1,-1
3,3

In comparing the rates of change, which statement about Function A and Function B is true?

A.

Function A and Function B have the same rate of change.
B.

Function A has a greater rate of change than Function B has.
C.

Function A and Function B both have negative rates of change.
D.

Function A has a negative rate of change and Function B has a positive rate of change.

C is correct
3 0
3 years ago
Read 2 more answers
Find the difference.<br><br> 2g<br><br> g2 − g − 6 − 1<br><br> g − 3
algol [13]
2 is the answer your looking for

3 0
4 years ago
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