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2 years ago

3. Why does it tako 3 copies of; to show the same amount as 1 copy of ?? Explain your answer in words

1 answer:

7 0

Because 3/6 is equal to 1/2.
You multiply the 1 (in 1/2) by 3, and the 2 (in 1/2) by 3 to get 3/6. You need to have a common denominator, and 6 was the common denominator.

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Calculate the distance between the points N=(-5, 9) and P=(-1, 2) in the coordinate plane.

GalinKa [24]

Answer:

Step-by-step explanation:

5 0

3 years ago

Restaurant is making hamburgers. The cooks use

likoan [24]

Answer:

70.69

Step-by-step explanation:

46 2/3 divided by 2/3

46.66 divided by 0.66

The anwser is 70.69

If the cook makes twice as much beef then the number of hamburgers would be two times larger which would be 141.39.

Half as much beef would be 70.69 divided by two which is 35.34.

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3 years ago

In January, Wally's Widget World had sales of $12,500 and expenses of $10,200. The profit for January was _____.

likoan [24]

Easy, $12, 500 - $10, 200 = ?

8 0

3 years ago

How do u factor this

PilotLPTM [1.2K]

You can factor a parabola by finding its roots: if

$p(x)=x^2+bx+c$

has roots $x_1,\ x_2$ , then you have the following factorization:

$p(x) = (x-x_1)(x-x_2)$

In order to find the roots, you can use the usual formula

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

In the first example, this formula leads to

$x_{1,2} = \dfrac{-2\pm\sqrt{4+4}}{2} = \dfrac{-2\pm\sqrt{8}}{2} = \dfrac{-2\pm2\sqrt{2}}{2} = 1 \pm \sqrt{2}$

So, you can factor

$x^2-2x-1 = (x-1-\sqrt{2})(x-1+\sqrt{2})$

The same goes for the second parabola.

As for the third exercise, simply plug the values asking

$f(1.5)=-5.25$

you get

$f(-1.5) = 1.5c-3 = -5.25$

Add 3 to both sides:

$1.5c = -2.25$

Divide both sides by 1.5:

$c = 1.5$

7 0

3 years ago

The given two-parameter family is a solution of the indicated differential equation on the interval

mr_godi [17]

Answer:

I do 1 option for you as an example, you need to check the leftover by yourself.

Step-by-step explanation:

for d) y(0) = 0 and y'(pi) =0

$y(0) = C_1e^0cos(0)+ C_2 e^0 sin(0) = 0 \longrightarrow C_1 = 0$

$y(x) ' = C_1e^x cos(x) - C_1e^x sin (x) + C_2e^x sin(x) + C_2e^x cos(x)$

$y(\pi)'=C_1e^\pi cos(\pi)- C_1e^\pi sin(\pi)+ C_2e^\pi sin(\pi) + C_2e^\pi cos (\pi)$

Replace $C_1 = 0$ we have

$y'(\pi) = -C_2e^\pi = 0$

if and only if $C_2 =0$

Hence the given solution does not work.

then, d is NOT the correct answer.

3 0

3 years ago