All categories

3 years ago

Find the value of x that makes the equatlon true: 11x=143

Mathematics

1 answer:

kicyunya [14]3 years ago

8 0

Answer:

11(13)=143

Step-by-step explanation:

i divided 143 with 11 and got 13

your welcome for the help :D

You might be interested in

1 3/4 − 4/5 = 35 twentieths − what fraction???<br><br>Plss help will give brainliest :)

7nadin3 [17]

let's firstly convert the mixed fractions to improper fractions and then to do away with the denominators, let's multiply both sides by the LCD of all denominators.

$\stackrel{mixed}{1\frac{3}{4}}\implies \cfrac{1\cdot 4+3}{4}\implies \stackrel{improper}{\cfrac{7}{4}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{7}{4}-\cfrac{4}{5}=\cfrac{35}{20}-\boxed{?}\implies \stackrel{\textit{multipling both sides by }\stackrel{LCD}{20}}{20\left( \cfrac{7}{4}-\cfrac{4}{5} \right)=20\left( \cfrac{35}{20}-\boxed{?} \right)} \\\\\\ 35-16=35-20\boxed{?}\implies 19=35-20\boxed{?}\implies -16=-20\boxed{?} \\\\\\ \cfrac{-16}{-20}=\boxed{?}\implies \cfrac{4}{5}=\boxed{?}$

8 0

3 years ago

Read 2 more answers

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.

sweet-ann [11.9K]

Answer:

2.794

Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then

$\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy$

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

$\bf G(x,y) = (x, y, 3-x^2-y^2)$

with 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)$

we also have

$\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)$

and so

$\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2$

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy$

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

5 0

3 years ago

If (42)^p = 41^4, what is the value of p?

babymother [125]

First you need to make both bases the same:

Lets remove the ^p and ^4

To make the base of 42 equal to 41, you would have 41^x = 41

X - ln(42) / ln(41) = 1.00648904

Now you have 41^1.00648904(p) = 41^4

Now the bases are equal so we need to set the exponents to equal:

1.00648904(p) = 4

Divide both sides by 1.00648904 to solve for P

P = 4 / 1.00648904

P = 3.97421114

4 0

4 years ago

Distribute – 10 (-10x2 + x).<br> Answer:

statuscvo [17]

Answer:

your moms weight

Step-by-step explanation:

look stop if you dont want to do your homework dont do it amd get in trouble idc you are sepose to know

7 0

3 years ago

Is 3/24 or 1/8 greater?

xenn [34]

Answer:

they are = to each other

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers