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3 years ago

Divide 32 in ratio of 3:7

1 answer:

5 0

To divide by ratio, add up all the numbers in the ratio (3 + 7 = 10), then divide your original number by this number (32/10 = 3.2). Finally, multiply this number by each number in the ratio (3.2 * 3 = 9.6, 3.2 * 7 = 22.4)

9.6 : 22.4

You might be interested in

What does (2m + 7) - (3 - 4m) equal

yaroslaw [1]

(2m + 7) - (3 - 4m) = 2m + 7 - 3 + 4m = (2m + 4m) + (7 - 3)

= <u>6m + 4</u>

5 0

2 years ago

Can y’all help me on question 21?!

Studentka2010 [4]

Answer:

Step-by-step explanation:

96 divided by 4 = 24

7 0

2 years ago

Read 2 more answers

Given the system of linear equations.

Karo-lina-s [1.5K]

Part A:

x + y = 9

4x + y = 6

To solve this system using substitution, begin by isolating either x or y in the first equation. I'll isolate x.

x + y = 9

x = 9 - y

Substitute this expression (9 - y) into the second equation for x and solve.

4x + y = 6

4(9 - y) + y = 6

36 - 4y + y = 6

36 - 3y = 6

-3y = -30

y = 10

To find x, substitute 10 for y into either of the original equations.

x + y = 9

x + 10 = 9

x = -1

Finally, check all work by substituting the x- and y-values into each original equation.

x + y = 9 --> -1 + 10 = 9 --> 9 = 9 --> True

4x + y = 6 --> 4(-1) + 10 = 6 --> -4 + 10 = 6 --> 6 = 6 --> True

The answer for Part A is x = -1 and y = 10; (-1, 10).

Part B:

For graphing, it's easier to get the equations into slope-intercept form. Slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. To get our equations into slope-intercept form, we must simply isolate y.

x + y = 9

y = 9 - x

y = -x + 9

4x + y = 6

y = 6 - 4x

y = -4x + 6

Now that we have our slope-intercept equations, we can easily graph them, since their slopes and y-intercepts are readily visible.

Let's start with the y-intercepts. They are (0, 9) and (0, 6). You can plot those points on the graph.

Now, the slopes. The slope of the first line is -1, this means it declines. To plot this, start where you plotted the y-intercept, count one unit down, and then one unit to the right, and plot that point. Continue doing that and connect the dots, and you will have plotted the first line. The slope of the second line is -4, so it also declines. For this line, count four units down, and then one to the right and plot that point. Likewise, continue this and connect the coordinates, and you will have your line. (See attachment for graph.)

The lines do indeed intersect at (-1, 10); our answer is verified by graphing.

5 0

3 years ago

Some body can help me with a geometric mean maze

Mars2501 [29]

Answer:

See explanation

Step-by-step explanation:

Theorem 1: The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.

Theorem 2: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

1. Start point: By the 1st theorem,

$x^2=25\cdot (49-25)=25\cdot 24=5^2\cdot 2^2\cdot 6\Rightarrow x=5\cdot 2\cdot \sqrt{6}=10\sqrt{6}.$

2. South-East point from the Start: By the 2nd theorem,

$x^2=40\cdot (40+5)=4\cdot 5\cdot 2\cdot 9\cdot 5\Rightarrow x=2\cdot 5\cdot 3\cdot \sqrt{2}=30\sqrt{2}.$

3. West point from the previous: By the 2nd theorem,

$x^2=(32-20)\cdot 32=4\cdot 3\cdot 16\cdot 2\Rightarrow x=2\cdot 4\cdot \sqrt{6}=8\sqrt{6}.$

4. West point from the previous: By the 1st theorem,

$9^2=x\cdot 15\Rightarrow x=\dfrac{81}{15}=\dfrac{27}{5}=5.4.$

5. West point from the previous: By the 2nd theorem,

$10^2=8\cdot (8+x)\Rightarrow 8+x=12.5,\ x=4.5.$

6. North point from the previous: By the 1st theorem,

$x^2=48\cdot 6=6\cdot 4\cdot 2\cdot 6\Rightarrow x=6\cdot 2\cdot \sqrt{2}=12\sqrt{2}.$

7. East point from the previous: By the 2nd theorem,

$x^2=22.5\cdot 30=225\cdot 3\Rightarrow x=15\sqrt{3}.$

8. North point from the previous: By the 1st theorem,

$x^2=7.5\cdot 36=270\Rightarrow x=3\sqrt{30}.$

8. West point from the previous: By the 2nd theorem,

$x^2=12.5\cdot (12.5+13.5)=12.5\cdot 26=25\cdot 13\Rightarrow x=5\sqrt{13}.$

9. North point from the previous: By the 1st theorem,

$12^2=x\cdot 30\Rightarrow x=\dfrac{144}{30}=4.8.$

101. East point from the previous: By the 1st theorem,

$6^2=1.6\cdot (x-1.6)\Rightarrow x-1.6=22.5,\ x=24.1.$

11. East point from the previous: By the 2nd theorem,

$20^2=32\cdot (32-x)\Rightarrow 32-x=12.5,\ x=19.5.$

12. South-east point from the previous: By the 2nd theorem,

$18^2=x\cdot 21.6\Rightarrow x=15.$

13. North point=The end.

6 0

3 years ago

A homogeneous rectangular lamina has constant area density ρ. Find the moment of inertia of the lamina about one corner

frozen [14]

Answer:

$I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

Step-by-step explanation:

By applying the concept of calculus;

the moment of inertia of the lamina about one corner $I_{corner}$ is:

$I_{corner} = \int\limits \int\limits_R (x^2+y^2) \rho d A \\ \\ I_{corner} = \int\limits^a_0\int\limits^b_0 \rho(x^2+y^2) dy dx$

where :

(a and b are the length and the breath of the rectangle respectively )

$I_{corner} = \rho \int\limits^a_0 {x^2y}+ \frac{y^3}{3} |^ {^ b}_{_0} \, dx$

$I_{corner} = \rho \int\limits^a_0 (bx^2 + \frac{b^3}{3})dx$

$I_{corner} = \rho [\frac{bx^3}{3}+ \frac{b^3x}{3}]^ {^ a} _{_0}$

$I_{corner} = \rho [\frac{a^3b}{3}+ \frac{ab^3}{3}]$

$I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

Thus; the moment of inertia of the lamina about one corner is $I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

7 0

2 years ago