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GenaCL600 [577]

2 years ago

15

the difference of two numbers is 9 The larger number is 6 less then twice the smaller number. gind the number

2 answers:

solong [7]2 years ago

7 0

The numbers are 15 and 24

KonstantinChe [14]2 years ago

4 0

Hi there!

To solve this problem, we need to set up two equations.

Let x equal the larger number.
Let y equal the smaller number.

x - y = 9
x = 2y - 6

Those are our two equations ^.

Since we know that x is equal to 2y - 6, we can plug the value of x into the equation x - y = 9 and find y.

x - y = 9
x = 2y - 6

2y - 6 - y = 9
2y - y - 6 = 9
y - 6 = 9
y = 9 + 6
y = 15

Now, we do the same to the value of y - plug it into the equation - and solve for x:

y = 15
x - y = 9
x - 15 = 9
x = 9 + 15
x = 24

So, the numbers are 24 and 15.

Hope this helps!

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Suppose lines EF and GH are reflected over the line Y equals X to form the lines JK and LM which statement would be true about t

katrin [286]

Answer:

A) JK and LM will be parallel to each other.

Step-by-step explanation:

On reflection on $y=x$ line the x co-ordinate changes with y co-ordinate and y co-ordinate changes with x co-ordinate

$(x,y)\rightarrow (y,x)$

Points on line EF

$(0,6) , (-5,-2)$

On reflection of this line on $y=x$ the new points we get for line JK are

$(6,0),(-2,-5)$

Points on line GH

$(-4,9),(-9,1)$

On reflection on y=x line the new points we get for line LM are

$(9,-4),(1,-9)$

Slope of line JK

$m=\frac{y_2-y_1}{x2-x1}\\m=\frac{(-5)-0}{(-2)-6} \\m=\frac{-5}{-8}=\frac{5}{8}$

Slope of line LM

$m=\frac{y_2-y_1}{x2-x1}\\m=\frac{(-9)-(-4)}{1-9} \\m=\frac{-9+4}{-8}=\frac{-5}{-8}\\m=\frac{5}{8}$

For two line to be parallel, their slopes will be same.

$m_{JK} =\frac{5}{8} , m_{LM}=\frac{5}{8}$

Since slopes of lines JK and LM are same therefore we can say that these are parallel to each other.

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3 years ago

If VT=2,what is the length of PT?

marusya05 [52]

Step-by-step explanation:

$\underline{ \underline{ \text{Given : }}}$

Perpendicular ( P ) = VT = 2
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$\theta = \tt{60 \degree}$

$\underline{ \underline{ \text{Solution}}} :$

$\tt{ \tan(60 \degree) = \frac{perpendicualar}{base}}$

⟶ $\tt{ \sqrt{3} = \frac{2}{PT}}$

⟶ $\tt{ \sqrt{3} \: PT= 2}$

⟶ $\tt{PT = \frac{2}{ \sqrt{3}} }$

⟶ $\tt{PT= \boxed{\tt{ \frac{2 \sqrt{3} }{{3} }}}}$

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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