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3 years ago

Question in picture solve

Mathematics

2 answers:

Temka [501]3 years ago

8 0

Answer:

1:5

Step-by-step explanation:

30 divided by 6 is 5 and so are the rest

kiruha [24]3 years ago

3 0

Answer:

Step-by-step explanation:

10/2=5

20/4=5

hope this helps :3

if it did pls mark brainliest

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the height of an object after t seconds is given by f(t)=-5t^2+20t (a) find the height of the object after 2 seconds, and includ

Leni [432]

Answer:

t=20 4=5 =6 =8 =9 10

Step-by-step explanation:

And that is iy

8 0

3 years ago

A circle in the xy-plane is centered at (3, 0) and has a radius with endpoint (1, 8/3 ). Which answer choice is an equation of t

gayaneshka [121]

Answer: (x - 3)² + y² = $\frac{100}{9}$

<u>Step-by-step explanation:</u>

Circle Formula: (x - h)² + (y - k)² = r² ; (h, k) is the center & r is the radius

Use the distance formula for (3, 0) and (1, $\frac{8}{3}$ )

$r = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$r = \sqrt{(3-1)^{2}+(0-\frac{8}{3})^{2}}$

$r = \sqrt{(2)^{2}+(-\frac{8}{3})^{2}}$

$r = \sqrt{4+\frac{64}{9}}$

$r = \sqrt{\frac{36}{9}+\frac{64}{9}}$

$r = \sqrt{\frac{100}{9}}$

$r = \frac{10}{3}$

****************************************************

Center (h, k) = (3, 0) Radius (r) = $\frac{10}{3}$

(x - h)² + (y - k)² = r²

(x - 3)² + (y - 0)² = ( $\frac{10}{3}$ )²

(x - 3)² + y² = $\frac{100}{9}$

6 0

4 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

3 years ago

Item 14 Question 1 You have at most $60 to spend on trophies and medals to give as prizes for a contest. A drawing of a trophy a

7nadin3 [17]

Answer:

$12x+3y\leq 60$

Step-by-step explanation:

Let

x ----> the numbers of trophies you can buy

y ----> the numbers of medals you can buy

we know that

The term "at most" means "less than or equal to"

The number of trophies multiplied by $12 plus the number of medals multiplied by $3 must be less than or equal to $60

The inequality that represent this situation is

$12x+3y\leq 60$

8 0

3 years ago

Is this right or not because I’m doubting it

olya-2409 [2.1K]

Answer: yes it is

Step-by-step explanation:

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3 years ago