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3 years ago

Helpppp pleaseee X•1 + x/1=

Mathematics

1 answer:

dem82 [27]3 years ago

6 0

x * 1 + x/1 =

x + x =

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You want to enlarge a 4 x 5 inch photo to fit into a 1 inch wide frame

skelet666 [1.2K]

The 4•x by 5•x inner dimensions of the 53 inches outer perimeter picture frame, the 1 inch wide frame and the 4 × 5 inch picture to be enlarged, gives;

a. The outer perimeter equation is presented as follows;

18•x + 8 = 53 inches

b. The scale factor required for the enlargement of the picture is 2.5

<h3>Which method can be used to write the equation and find the scale factor?</h3>

The dimensions of the photo = 4 × 5 inch

Width of the frame = 1 inch

Outer perimeter of the frame = 53 inches

The width of the inside of the picture frame = 4•x

The height of the inside of the picture frame = 5•x

a. The equation for the outer perimeter of the picture frame is therefore;

2 × (4•x + 1 + 1) + 2 × (5•x + 1 + 1) = 53

Which gives;

2 × (4•x + 2) + 2 × (5•x + 2) = 53

8•x + 4 + 10•x + 4 = 53

A simplified equation for the outer perimeter of the picture frame is therefore;

18•x + 8 = 53

b. Solving the above equation, we have;

18•x + 8 = 53

18•x = 53 - 8 = 45

18•x = 45

Therefore;

x = 45 ÷ 18 = 2.5

x = 2.5

The width and height of the picture frame are therefore;

Width = 4•x = 4 × 2.5 = 10

Height = 5•x = 5 × 2.5 = 12.5

Which gives;

Width = 10 inches

Height = 12.5 inches

The enlargement factor, sf, is given by the ratio of a side of the picture frame to a side of the picture as follows;

sf = 12.5/5 = 2.5

The factor by which to enlarge the picture frame is 2.5

Learn more about linear scale factors here:

brainly.com/question/19381630

#SPJ1

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2 years ago

Luke makes fruit cakes for a stall at a village fete. It costs Luke £1.80 for

zalisa [80]

Answer:

2.43

Step-by-step explanation:

1.80 x 0.35 + 1.80

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3 years ago

NEED HELP ASAP (ILL GIVE BRAINLIST)

strojnjashka [21]

Answer:

the answer is 33 9/11

Step-by-step explanation:

multiply the fractions

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2 years ago

Read 2 more answers

You have two summer jobs. At the Burger Palace you earn $8 per hour. At the community center you teach soccer clinics for $10 pe

SIZIF [17.4K]

The complete question in the attached figure
we have that
x-------------> number of hours works at Burger Palace -----> <span>$8
</span>y-------------> number of hours works at <span>community center</span> -----> $10
<span>
8x+10y>=200

using a graph tool
see the attached figure

the answer is the option B
</span><span>
</span>

8 0

3 years ago

What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po

emmasim [6.3K]

$\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA$
$\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA$

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

$\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k$
$\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j$

The cross product is

$\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j$

So, the flux is given by

$\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA$
$\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du$
$\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv$
$\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv$

where $t=-\sin v$ in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

$\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi$

5 0

3 years ago