27. The scatter plot shows the number of votes cast in U.S. presidential elections since 1932.

Help!

Vinvika [58]

A<b+c so if b=15 and c=40 then
a<15+40
a<45

so a cant possibly be d. 50ft

6 0

2 years ago

Express the product in simplest form.

Dmitry [639]

The product in simplest form is (x - 4)

Solution:

Given expression is:

$\frac{8}{2x+8} \times \frac{x^2-16}{4}$

We have to find the product in simplest form

In the given expression,

2x + 8 = 2(x+ 4)

We know that,

$a^2-b^2=(a+b)(a-b)$

Therefore,

$x^2-16 = x^2-4^2 =(x+4)(x-4)$

Substitute these in given expression

$\frac{8}{2(x+4)} \times \frac{(x+4)(x-4)}{4}$

Cancel the common factors,

$\frac{8}{2(x+4)} \times \frac{(x+4)(x-4)}{4} = x - 4$

Thus the product in simplest form is (x - 4)

8 0

2 years ago

Y = x – 6 x = –4 what is the solution to the system of equations? (–8, –4) (–4, –8) (–4, 4) (4, –4

fredd [130]

Answer:

(- 4, - 10 )

Step-by-step explanation:

Given the 2 equations

y = x - 6 → (1)

x = - 4 → (2)

Substitute x = - 4 into (1) for corresponding value of y

y = - 4 - 6 = - 10

Solution is (- 4, - 10 )

4 0

2 years ago

What is the 5th term in the sequence below? 255, 270, 285, 300,

lesya [120]

Answer:

315

Step-by-step explanation:

each one adds 15.

300 + 15 = 315

8 0

2 years ago

For each initial value problem, determine whether Picard's Theorem can be used to show the existence of a unique solution in an

Alinara [238K]

Answer:

Part a: $f , \, f_y$ is continuous at the initial value (0,0) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.

Part b: $f_y$ is not continuous at the initial value (0,0) so due to Picardi theorem there does not exist an interval such that the IVP has a unique solution.

part c: $f , \, f_y$ is continuous at the initial value (0,1) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.

Step-by-step explanation:

Part a

as $y^{' }=ty^{4/3}$

Let

$f(t,y)=ty^{4/3}$

Now derivative wrt y is given as

$f_y=\frac{4}{3}ty^{1/3}$

Finding continuity via the initial value

$f$ is continuous on $R^2$ also $f_y$ is also continuous on $R^2$

Also

$f , \, f_y$ is continuous at the initial value (0,0) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.

Part b

as $y^{' }=ty^{1/3}$

Let

$f(t,y)=ty^{1/3}$

Now derivative wrt y is given as

$f_y=\frac{1}{3}ty^{-2/3}$

Finding continuity via the initial value

$f$ is continuous on $R^2$ also $f_y$ is also continuous on $R^2$

Also

$f_y$ is not continuous at the initial value (0,0) so due to Picardi theorem there does not exist an interval such that the IVP has a unique solution.

Part c

as $y^{' }=ty^{1/3}$

Let

$f(t,y)=ty^{1/3}$

Now derivative wrt y is given as

$f_y=\frac{1}{3}ty^{-2/3}$

Finding continuity via the initial value

$f$ is continuous on $R^2$ also $f_y$ is also continuous on $R^2$ when $y\neq 0$

Also

$f , \, f_y$ is continuous at the initial value (0,1) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.

6 0

2 years ago