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3 years ago

What is the value of x in the equation x – y = 30, when y = 15? 4 8 80 200

Mathematics

2 answers:

AVprozaik [17]3 years ago

5 0

Answer:

Step-by-step explanation:

We are given an equation x - y = 30 which we are to solve for x when the value of y is 15.

To solve this linear equation for x, we just need to substitute the given value of y into the equation to get the value of x.

x - y = 30

x - 15 = 30

x = 30 + 15

x = 45

Therefore, the value of x is 45 when y = 30.

Unfortunately this value is not in the given options but 45 is the correct answer.

yawa3891 [41]3 years ago

3 0

There are many ways of solving the set of equations x + y = 30 and x - y = 2 for x and y.

From x +y = 30 we get x = 30 - y

Now substitute this in the equation x - y = 2

30 - y - y = 2

-2y = -28

y = 14

As x = 30 - y = 30 - 14 = 16

The required solution of the set of equations is x = 16 and y = 14

Toodles~

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Give the equation of the circle centered at the origin and passing through the point(0,10)

beks73 [17]

Answer:

$x^{2} + y^{2} = 100$

Step-by-step explanation:

The equation of a circle has the following format:

$(x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}$

In which r is the radius(half the diameter) and the centre is the point $(x_{0}, y_{0})$

Centered at the origin

This means that $x_{0} = 0, y_{0} = 0$

Passing through the point(0,10)

The radius is the distance of any point in which the circle passes to the centre.

Using the formula for the distance between two points.

$r = D = \sqrt{(0 - 0)^{2} + (10 - 0)^{2}} = 10$

$x^{2} + y^{2} = 10^{2}$

$x^{2} + y^{2} = 100$

3 0

3 years ago

without axtually calculating,how much greater is the product of 98times 50 than the product 97 times 50

olasank [31]

With 98x50 its like there are 98 fiveties. In 97x 50 there are 97 fiveties. There is one more 50 in 98x50.

7 0

3 years ago

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Solve the problem 9 divided 6.21 = ?

bezimeni [28]

Answer:

1.44927536232

Step-by-step explanation:

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3 years ago

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Write an equation of cosine function with amplitude 2 and a period 4pi

Arisa [49]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\

\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}$

now.. hmmm let's see, keeping in mind the template above

so, amplitude of 2, A=2, thus |A| = ±2, so, you can use either

and period of 4π, well, that simply means that

$\bf \cfrac{2\pi }{B}=4\pi \implies \cfrac{2\pi }{4\pi }=B\implies \cfrac{1}{2}=B\\\\
-----------------------------\\\\

\begin{array}{llll}
f(\theta)=&\pm 2cos&\left(\frac{1}{2}\theta \right)\\
&\ \uparrow &\ \uparrow \\
&A&B
\end{array}$

8 0

4 years ago

(1+5/6)(1-5/6) I don't understand it

ipn [44]

(a+b)(a-b) = a^2 - b^2

(1+5/6) (1-5/6) = 1^2 - (5/6)^2 = 1 - 25/36 = 36/36 - 25/36 = (36-25)/36 = 11/36

11/36 is your answer

5 0

3 years ago