What is 3/5 of 12? Please answer!

5. Which expression is equivalent to x2 - 17 x - 60?

Leni [432]

15x-60
this is the answer

5 0

3 years ago

Geometry and perimeter help plz :(

Novay_Z [31]

Answer:

11.)4.79mi

12.)8cm

13.)10.6km

14.)5.93

15.)2yd

16.)5.41in

17)2y

18)2m

19)11.8yd

20)4.51km

Step-by-step explanation:

11.)15.8x2=31.6 31.6/6.6=4.79

12.)64/8=8

13.)26x2=54 54/4.9=10.6

14.)8.6x2=17.2 17.2/5.93

15.)2/1=2

16.)29.2/5.4=5.41

17)4/2=2

18)10/5=2

19)139.2/11.8=11.8

20)20.3/4.5=4.51

3 0

3 years ago

Find the measure Of K

Mariana [72]

Answer:

K = 10

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

sin theta = opp/ hyp

sin K = 11 / 61

Taking the inverse sin of each side

sin ^-1 ( sin K) = sin ^-1 (11/61)

K =10.38885782

K = 10

7 0

3 years ago

A sundae at an ice-cream shop costs $3, and each extra ounce of toppings costs

anyanavicka [17]

Answer:

The number of ounce of topping that can be get is 4 .

Step-by-step explanation:

Given as :

The cost of a sundae = $3

The cost for each extra ounce of toppings = $0.50

Let The number of ounce of topping = n

The total spending cost on sundae with n extra topping = $5

Now, According to question

The total spending cost on sundae with n extra topping = The cost of a sundae + The number of ounce of topping × The cost for each extra ounce of toppings

Or , $3 + n × $0.50 = $5

or, 3 + 0.50 n = 5

or , 0.50 n = 5 - 3

or, 0.50 n = 2

∴ n = $\dfrac{2}{0.50}$

I.e n = 4

So, The number of ounce of topping = n = 4

Hence The number of ounce of topping that can be get is 4 . Answer

7 0

3 years ago

How do i solve that question?

yawa3891 [41]

a) The solution of this ordinary differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ .

b) The integrating factor for the ordinary differential equation is $-\frac{1}{x}$ .

The particular solution of the ordinary differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ .

<h3>How to solve ordinary differential equations</h3>

a) In this case we need to separate each variable ( $y, t$ ) in each side of the identity:

$6\cdot \frac{dy}{dt} = y^{4}\cdot \sin^{4} t$ (1)

$6\int {\frac{dy}{y^{4}} } = \int {\sin^{4}t} \, dt + C$

Where $C$ is the integration constant.

By table of integrals we find the solution for each integral:

$-\frac{2}{y^{3}} = \frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32} + C$

If we know that $x = 0$ and $y = 1$ , then the integration constant is $C = -2$ .

The solution of this ordinary differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ . $\blacksquare$

b) In this case we need to solve a first order ordinary differential equation of the following form:

$\frac{dy}{dx} + p(x) \cdot y = q(x)$ (2)

Where:

$p(x)$ - Integrating factor
$q(x)$ - Particular function

Hence, the ordinary differential equation is equivalent to this form:

$\frac{dy}{dx} -\frac{1}{x}\cdot y = x^{2}+\frac{1}{x}$ (3)

The integrating factor for the ordinary differential equation is $-\frac{1}{x}$ . $\blacksquare$

The solution for (2) is presented below:

$y = e^{-\int {p(x)} \, dx }\cdot \int {e^{\int {p(x)} \, dx }}\cdot q(x) \, dx + C$ (4)

Where $C$ is the integration constant.

If we know that $p(x) = -\frac{1}{x}$ and $q(x) = x^{2} + \frac{1}{x}$ , then the solution of the ordinary differential equation is:

$y = x \int {x^{-1}\cdot \left(x^{2}+\frac{1}{x} \right)} \, dx + C$

$y = x\int {x} \, dx + x\int\, dx + C$

$y = \frac{x^{3}}{2}+x^{2}+C$

If we know that $x = 1$ and $y = -1$ , then the particular solution is:

$y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$

The particular solution of the ordinary differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ . $\blacksquare$

To learn more on ordinary differential equations, we kindly invite to check this verified question: brainly.com/question/25731911

3 0

3 years ago