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3 years ago

5.6 as a decimal in simplest form

Mathematics

1 answer:

tamaranim1 [39]3 years ago

5 0

0.056 is the decimal in the simplest form.

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Which equation in standard form has a graph that passes through the point (5, −2) and has a slope of 72 ?

aalyn [17]

Given:

Consider the graph passes through the point (5, −2) and has a slope of $\dfrac{7}{2}$ .

To find:

The equation of graph in standard form.

Solution:

Point slope form of a line is

$y-y_1=m(x-x_1)$

where, m is slope and $(x_1,y_1)$ is the point lies on the line.

Slope is $\dfrac{7}{2}$ and graph passes through (5,-2), so the equation of line is

$y-(-2)=\dfrac{7}{2}(x-5)$

$y+2=\dfrac{7}{2}(x-5)$

Multiply both sides by 2.

$2(y+2)=7(x-5)$

$2y+4=7x-35$

$4+35=7x-2y$

$39=7x-2y$

The required equation is $7x-2y=39$ .

Therefore, the correct option is B.

4 0

3 years ago

Find the mean of 8, 7, 9 graphically. What is the sum of the numbers?

Mrrafil [7]

Answer:

Sum=24

Mean=8

Step-by-step explanation:

To find the sum of all the numbers you simply would do 8+7+9.

That is 24.

When finding mean you add up all your numbers then divide by the total number of number (I know that sounds a bit confusing. In this case there are only 3 numbers, so that is the one you will use for dividing)

Your equation would look like this

Once you divide you should get 8 as your mean or average in other terms.

5 0

4 years ago

150 adult complete a survey 80 are women write the ratio men: women in it's simplest form.

Mashcka [7]

70 were men hope it helps

4 0

3 years ago

Read 2 more answers

Find dx/dt when y=2 and dy/dt=1, given that x^4=8y^5-240 dx/dt=

katrin2010 [14]

Answer:

The value of $\frac{dx}{dt}$ is $\frac{160}{x^3}$ .

Step-by-step explanation:

The given equation is

$x^4=8y^5-240$

We need to find the value of $\frac{dx}{dt}$ .

Differentiate with respect to t.

$4x^3\frac{dx}{dt}=8(5y^4)\frac{dy}{dt}-0$ $[\because \frac{d}{dx}x^n=nx^{n-1},\frac{d}{dx}C=0]$

$4x^3\frac{dx}{dt}=40y^4\frac{dy}{dt}$

It is given that y=2 and dy/dt=1, substitute these values in the above equation.

$4x^3\frac{dx}{dt}=40(2)^4(1)$

$4x^3\frac{dx}{dt}=40(16)(1)$

$4x^3\frac{dx}{dt}=640$

Divide both sides by 4x³.

$\frac{dx}{dt}=\frac{640}{4x^3}$

$\frac{dx}{dt}=\frac{160}{x^3}$

Therefore the value of $\frac{dx}{dt}$ is $\frac{160}{x^3}$ .

4 0

3 years ago

Helppppp pleaseeee!!! Its 5 Marks

alexgriva [62]

Answer:

As the picture shows, we have

V = (x+5)(x-2)x =30x

<=> x(x^2+3x-10) =30x

<=> x^3 + 3x^2-10x-30x = 0(*)

<=> x^3 +3x^2 -40x = 0

<=> x(x-5)(x+8) =0

Solve for (*), there are 3 solution :

0(invalid),

-8(invalid)

=> 3 dimensions of rectangular box: 5, 3, 10

8 0

4 years ago

Read 2 more answers