How many hundreds in 148305

Will give brainliest and thanks

Maru [420]

Answer:

y = -3x - 3

Step-by-step explanation:

First, determine the slope of the red line. As we move from (0, 3) to (1, 0),

x increases by 1 and y decreases by 3. Hence, the slope, m, is

m = rise / run = -3 / 1, or m = -3.

The desired new line, parallel to the red line, has the same slope as the red line: m = -3.

The equation for this new line is y - k = m (x - h), where (h, k): (-1, 0):

We get y - 0 = -3 (x + 1), or y = -3x - 3.

Check: Does this line go thru the given point (-1, 0)?

Is 0 = -3(-1) - 3 true? Yes, it is true.

6 0

3 years ago

(1,-4) x-2y=8 4x-y=8

deff fn [24]

Answer:

False solution; [1⅐, -3 3⁄7]

Step-by-step explanation:

{x - 2y = 8

{4x - y = 8

-¼[4x - y = 8]

{x - 2y = 8

{-x + ¼y = -2 >> New Equation

____________

-1¾y = 6

y = -3 3⁄7 [Plug this back into both equations to get the x-coordinate of 1⅐]; 1⅐ = x

I am joyous to assist you anytime.

3 0

3 years ago

Anyone know the answer? A school held a charity concert over two days, collecting $2,400 the first day and $1,400 the second day

jolli1 [7]

Im not sure try Socratic

6 0

3 years ago

The points A BC :(2, 2,1), :(1,1,3), :(2,0,5) − are the vertices of a right triangle. The radius of the sphere with center at th

ElenaW [278]

Answer:

r=1

Step-by-step explanation:

First we need to know the length of each side of the triangle, so we use the formula of the vector modulus:

$|AB|= \sqrt{(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}$

By doing so, we find:

$|AB|=\sqrt {6}\\|BC|=\sqrt {6}\\|AC|=2\sqrt {5}$

With this we know that the triangle is not right, but, we assume the longest side as the hypotenuse of the problem.

As we have two equal sides, we know that the line between point |AB| and the center of the hypotenuse is perpendicular, therefore, we can calculate it using Pythagoras theorem:

$|BC|^{2}=r^{2}+(\frac{|AC|}{2})^{2}\\\\r^{2}=|BC|^{2}-(\frac{|AC|}{2})^{2}\\\\r^{2}=(\sqrt{6})^{2}- (\frac{2\sqrt{5}}{2})^{2}\\\\r^{2}=6-5=1\\r=1$

4 0

3 years ago

What are the possible numbers of positive, negative, and complex zeros of

o-na [289]

Answer:

Look at changes of signs to find this has 1 positive zero, 1 or 3 negative zeros and 0 or 2 non-Real Complex zeros.

Then do some sums...

Explanation:

f(x)=−3x4−5x3−x2−8x+4

Since there is one change of sign, f(x) has one positive zero.

f(−x)=−3x4+5x3−x2+8x+4

Since there are three changes of sign f(x) has between 1 and 3 negative zeros.

Since f(x) has Real coefficients, any non-Real Complex zeros will occur in conjugate pairs, so f(x) has exactly 1 or 3 negative zeros counting multiplicity, and 0 or 2 non-Real Complex zeros.

f'(x)=−12x3−15x2−2x−8

Newton's method can be used to find approximate solutions.

Pick an initial approximation a0 .

Iterate using the formula:

ai+1=ai−f(ai)f'(ai)

Putting this into a spreadsheet and starting with a0=1 and a0=−2 , we find the following approximations within a few steps:

x≈0.41998457522194 x≈−2.19460208831628

We can then divide f(x) by (x−0.42) and (x+2.195) to get an approximate quadratic −3x2+0.325x−4.343 as follows:

Notice the remainder 0.013 of the second division. This indicates that the approximation is not too bad, but it is definitely an approximation.

Check the discriminant of the approximate quotient polynomial:

−3x2+0.325x−4.343 Δ=b2−4ac=0.3252−(4⋅−3⋅−4.343)=0.105625−52.116=−52.010375

Since this is negative, this quadratic has no Real zeros and we can be confident that our original quartic has exactly 2 non-Real Complex zeros, 1 positive zero and 1 negative one.

3 0

4 years ago