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3 years ago

Please help me with this question

Mathematics

1 answer:

vichka [17]3 years ago

5 0

Write the ratio out.

Number of hands Betty won : Number of hands Adam won

5:20

Both can be simplified by the common factor of 5.

1:4 is the solution.

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Answer:

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Step-by-step explanation:

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3 years ago

3 brother are having dinner

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Read 2 more answers

Which statement best describes the "cult of domesticity" that emerged in the late 19th century?

12345 [234]

Answer:

a movement one single movement cultivating one group

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2 years ago

$9500 is invested, part of it at 11% and part of it at 8%. For a certain year, the total yield is $937.00. How much was invested

netineya [11]

Answer:

5900 at 11%

3600 at 8%

Step-by-step explanation:

x= invested at 11%

y= invested at 8%

x+y=9500

.11x+.08y=937

Mulitply the first equation by .11

.11x+.11y= 1045

Subtract this and the second equation

(.11x+.11y)-(.11x+.08y)=1045-937

.03y=108

y=3600

SOlve for x

x+3600=9500

x=5900

8 0

3 years ago

1) find the equation of the line parallel to

In-s [12.5K]

Answer:

1) The equation of the line parallel to x-5y=6 and through (4,-2) is 5y = x -14

2) The equation of the line perpendicular to y= -2/5x + 3 and through (2,-1) is 2y = 5x -12

Solution:

1) find the equation of the line parallel to x-5y=6 and through (4,-2).

Given, line equation is x – 5y = 6

We have to find the line equation that is parallel to given line and passing through the point (4, -2)

Now, let us find slope of the given line.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-5}=\frac{1}{5}$

Now, we know that, slope of parallel lines are equal.

So, slope of required line is 1/5 and it passes through (4, -2)

Now, using point slope form

$y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on line. }$

$y-(-2)=\frac{1}{5}(x-4) \rightarrow 5(y+2)=x-4 \rightarrow 5 y+10=x-4 \rightarrow x-5 y=14$

Hence, the line equation is 5y = x -14

2) find the equation of the line perpendicular to y= -2/5x + 3 and through (2,-1)

$\text { Given, line equation is } y=-\frac{2}{5} x+3 \rightarrow 5 y=-2 x+15 \rightarrow 2 x+5 y=15$

We have to find the line equation that is perpendicular to given line and passing through the point (2, -1)

Now, let us find slope of the given line.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-2}{5}=-\frac{2}{5}$

Now, we know that, product of slopes of perpendicular lines equals to -1.

So, slope of required line $\times$ slope of given line = -1

slope of required line = $-1 \times \frac{5}{-2}=\frac{5}{2}$

And it passes through (2, -1)

Now, using point slope form

$\begin{array}{l}{\text { Line equation is } y-(-1)=\frac{5}{2}(x-2) \rightarrow 2(y+1)=5(x-2)} \\\\ {\rightarrow 2 y+2=5 x-10 \rightarrow 5 x-2 y=12}\end{array}$

Hence, the line equation is 2y = 5x -12

4 0

3 years ago