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3 years ago

Five notebooks cost $20 at this rate how much would 4 notebooks coat?

Mathematics

2 answers:

Andrews [41]3 years ago

7 0

Answer:

Step-by-step explanation:

$20/5notebooks=$4 each

4$*4notebooks=16$

vagabundo [1.1K]3 years ago

3 0

Answer:

$16

Step-by-step explanation:

if 5=20 then 20÷5=4 and 4x4=16

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BELP ME WITH THIS PLZ!!!!!!

kati45 [8]

Answer:

The entire area of the sailboat is 60cm²

Step-by-step explanation:

You can find the area of this shape by breaking it down into simpler shapes and adding up their individual areas.

In this case, the areas we'll use are the rectangle at the bottom, and the pair of triangles at the top.

Because the two triangles can be put together to form a single triangle, we don't need to measure them independently. We can simply take the total length of their bases, multiply it by their height, and divide by two. This follows the rule that the area of a triangle is equal to the area of the square that contains it divided by two.

(2cm + 3cm) × 6cm

= 5cm × 6cm

= 30cm²

The rectangle's area is of course equal to its width times its height, so we can say:

2.5cm × 12cm

= 30cm²

The total area of the shapes then is 30cm² + 30 cm², giving us a total area of 60cm²

3 0

3 years ago

What is the answer to #3?

SSSSS [86.1K]

Any it's your choice. You can do -3x times -9x then -9x times -3x or -2y times 3y then 3y times -2y. Remember to multiply the number after = too. Nice handwriting btw

5 0

3 years ago

FIRST 2 ANSWERS GET A FREE 20 POINTS!! BE FAST! :P

Romashka-Z-Leto [24]

Answer:

thank uuuuuu

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

For the differential equation 3x^2y''+2xy'+x^2y=0 show that the point x = 0 is a regular singular point (either by using the lim

Svetlanka [38]

Given an ODE of the form

$y''(x)+p(x)y'(x)+q(x)y(x)=f(x)$

a regular singular point $x=c$ is one such that $p(x)$ or $q(x)$ diverge as $x\to c$ , but the limits of $(x-c)p(x)$ and $(x-c)^2q(x)$ as $x\to c$ exist.

We have for $x\neq0$ ,

$3x^2y''+2xy'+x^2y=0\implies y''+\dfrac2{3x}y'+\dfrac13y=0$

and as $x\to0$ , we have $x\cdot\dfrac2{3x}\to\dfrac23$ and $x^2\cdot\dfrac13\to0$ , so indeed $x=0$ is a regular singular point.

We then look for a series solution about the regular singular point $x=0$ of the form

$y=\displaystyle\sum_{n\ge0}a_nx^{n+k}$

Substituting into the ODE gives

$\displaystyle3x^2\sum_{n\ge0}a_n(n+k)(n+k-1)x^{n+k-2}+2x\sum_{n\ge0}a_n(n+k)x^{n+k-1}+x^2\sum_{n\ge0}a_nx^{n+k}=0$

$\displaystyle3\sum_{n\ge2}a_n(n+k)(n+k-1)x^{n+k}+3a_1k(k+1)x^{k+1}+3a_0k(k-1)x^k$
$\displaystyle+2\sum_{n\ge2}a_n(n+k)x^{n+k}+2a_1(k+1)x^{k+1}+2a_0kx^k$
$\displaystyle+\sum_{n\ge2}a_{n-2}x^{n+k}=0$

From this we find the indicial equation to be

$(3(k-1)+2)ka_0=0\implies k=0,\,k=\dfrac13$

Taking $k=\dfrac13$ , and in the $x^{k+1}$ term above we find $a_1=0$ . So we have

$\begin{cases}a_0=1\\a_1=0\\\\a_n=-\dfrac{a_{n-2}}{n(3n+1)}&\text{for }n\ge2\end{cases}$

Since $a_1=0$ , all coefficients with an odd index will also vanish.

So the first three terms of the series expansion of this solution are

$\displaystyle\sum_{n\ge0}a_nx^{n+1/3}=a_0x^{1/3}+a_2x^{7/3}+a_4x^{13/3}$

with $a_0=1$ , $a_2=-\dfrac1{14}$ , and $a_4=\dfrac1{728}$ .

6 0

4 years ago

Assume that Q, R, and S are the angles of a triangle, with opposite sides q, r, and s respectively. Select all of the following

ikadub [295]

One could be

s^2 = q^2 + r^2 - 2*q*r cos S

4 0

4 years ago

Read 2 more answers