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2 years ago

Simplest form for 4/15X18

Mathematics

2 answers:

pshichka [43]2 years ago

7 0

If you’re asking me to multiply 4/15•18 then that will be 4 and 4/5

kirill115 [55]2 years ago

4 0

4.8 if you’re doing a fraction 4 over 15 x 8

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a college loan of 15000 has a 3% annual simple interest rate what is the interest the student will pay after 5 years

inessss [21]

Answer:

Interest = $ 90000

Step-by-step explanation:

Given, Principal= $15000

Rate = 3%

Time = 5 years

Interest = P × R × T/100

Therefore, Interest = 15000 × 3 × 5/100

(dividing 100 by 5 to get 20)

= 15000 × 3 × 20

= 90000

if it helps don't forget to like and mark me down

7 0

2 years ago

Functions please helpp

rodikova [14]

Answer:

<em>Functions:</em>

3. The chart with arrows

4. The graph

<em>Not Functions:</em>

1. (1,1), (2,2), (3,3), (1,4)

2. (1,17), (0,16), (0,15), (-2,17)

Step-by-step explanation:

Functions do not have repeating domains (x-coordinates).

Hope it helps!

8 0

2 years ago

Right triangle ABC is similar to right triangle DEF. If the side lengths for triangle ABC are 15, 20, 25, respectively, which va

solniwko [45]

Answer:

$A = 15$ $B = 20$ $C = 25$

$D = 30$ $E = 40$ $F = 50$

$D = 3$ $E = 4$ $F = 5$

Step-by-step explanation:

Given

Similar Triangles: ABC and DEF

$A = 15$

$B = 20$

$C = 25$

Required

Determine the sides of DEF

No options were given, so I will solve on a general terms.

Since both triangles are similar, then the following relationship exists.

DEF = ABC * n

i.e.

$D = A * n$ $E = B * n$ $F = C * n$

Where

$n = Scale\ Factor$

Assume n = 2.

So, we have:

$D = A * n$

$D =15 *2$

$D = 30$

$E = B * n$

$E = 20 * 2$

$E = 40$

$F = C * n$

$F = 25 * 2$

$F = 50$

Assume $n = \frac{1}{5}.$

So, we have:

$D = A * n$

$D =15 *\frac{1}{5}$

$D = 3$

$E = B * n$

$E = 20 * \frac{1}{5}$

$E = 4$

$F = C * n$

$F = 25 * \frac{1}{5}$

$F = 5$

So, the possible sides are:

$A = 15$ $B = 20$ $C = 25$

$D = 30$ $E = 40$ $F = 50$

$D = 3$ $E = 4$ $F = 5$

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2 years ago

Graph the points a (-5,0) b (-4,3) and c (0,-4) on the same coordinate plane

Delicious77 [7]

The marks are by 1 so just count

5 0

2 years ago

Let S be the surface defined by x 2 + 2y 3 + 3z 4 = 6. Let T be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−

aleksandrvk [35]

The tangent to $C$ through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces $S$ and $T$ at that point.

Let $f(x,y,z)=x^2+2y^3+3z^4$ . Then $S$ is the level curve $f(x,y,z)=6$ . Recall that the gradient vector is perpendicular to level curves; we have

$\nabla f(x,y,z)=(2x,6y,12z^2)$

so that the gradient of $f$ at (1, 1, 1) is

$\nabla f(1,1,1)=(2,6,12)$

For the surface $T$ , we have

$\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0$

so that $\vec r(1,0)=(1,1,1)$ . We can obtain a vector normal to $T$ by taking the cross product of the partial derivatives of $\vec r(u,v)$ , and evaluating that product for $u=1,v=0$ :

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left(u-2ve^v,-1,\dfrac{2e^v}u\right)$

$\left(\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right)(1,0)=(1,-1,2)$

Now take the cross product of the two normal vectors to $S$ and $T$ :

$(2,6,12)\times(1,-1,2)=(24,8,-8)$

The direction of vector (24, 8, -8) is the direction of the tangent line to $C$ at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by $t\in\Bbb R$ . Then adding (1, 1, 1) shifts this line to the point of tangency on $C$ . So the tangent line has equation

$\vec\ell(t)=(1,1,1)+t(24,8,-8)=(1+24t,1+8t,1-8t)$

7 0

2 years ago