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3 years ago

What single decimal multiplier would you use to increase by 9% followed by 3% decrease

Mathematics

1 answer:

MaRussiya [10]3 years ago

7 0

Answer:

= 94

Step-by-step explanation:

To work out the multiplier, first add or subtract the percentage from 100, then convert to a decimal. Example: we want to add 20% to £110. To work out the multiplier, add 20 to 100, to get 120, and then change it to a decimal (divide by 100) to get 1.2. The answer is £110 x 1.2 = £132.

so to workout an increase of 9% followed by 3% we would do:

9% - 3% = 6%. convert = 0.06 = 100-6 = 94 converted into 0.94 x 100 = 94

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What is the slope of this line

kifflom [539]

Answer:

3/4

Step-by-step explanation:

We can find the slope from the two points given

m = (y2-y1)/(x2-x1)

= (1- -2)/(2 - -2)

= (1+2)/(2+2)

= 3/4

8 0

3 years ago

Find the volume of the cylinder whose base area is (4x)2 and height is (5x)2.

Gre4nikov [31]

Answer:

Step-by-step explanation:

The computation of the volume of the cylinder is shown below:

As we know that

The Volume of the cylinder is

= base × height

= (4x)^2 + (5x)^2

=16x^2 + 25x^2

= 41x^3

Hence, the volume of the cylinder is 41x^3

4 0

2 years ago

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

7 0

2 years ago

Reflections, rotations, and translations are types of transformations. A reflection flips an image, a rotation spins an image, a

Ostrovityanka [42]

Answer:

Hey there!

A "flip" can be considered a reflection.

A "spin" is a rotation.

A "slide" is a translation.

None of these transformations change the way a shape looks, or the size of the shape. They only change how the shape is oriented.

Let me know if this helps :)

4 0

2 years ago

Read 2 more answers

Could someone help with a maths question? The attachment has the questions on it.

andre [41]

The days she has recorded: 11 + 10 + 7 + 6 + 3 + 2 = 39
The mode distance (most often): 5 km
Median distance: 6.64 km ~= 6.5 km ~= 7 km by rounding upwards, so 3rd

Cumulative frequency: 11; 21; 28; 34; 37; 39

7 0

1 year ago