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2 years ago

Sam spent the day at the mall. First, he bought three mules for $30

Mathematics

1 answer:

Dmitriy789 [7]2 years ago

4 0

Answer:

180

Step-by-step explanation:

30x3=90-30+50

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What is the measure of m?<br> 5<br> 15<br> E<br> n<br> m = [?]

AlekseyPX

<h3>Answer: 10</h3>

====================================================

Explanation:

The two smaller triangles are proportional, which lets us set up this equation

5/n = n/15

Cross multiplying leads to

5*15 = n*n

n^2 = 75

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Apply the pythagorean theorem on the smaller triangle on top, or on the right.

a^2+b^2 = c^2

5^2+n^2 = m^2

25+75 = m^2

100 = m^2

m^2 = 100

m = sqrt(100)

m = 10

7 0

3 years ago

Michele correctly solved a quadratic equation using the quadratic formula as shown below.

Fofino [41]

Answer:

The original equation must have been:

$7x^{2} -5x-2=0$

Step-by-step explanation:

A quadratic equation is in the form of :

$ax^{2} +bx+c=0$

And it is solved in the form of :

$x=\frac{-b+\sqrt{b^{-4ac} } }{2a}$ and $x=\frac{-b-\sqrt{b^{-4ac} } }{2a}$

Now the given equation is :

$x=\frac{-(-5)+-\sqrt{(-5)^{2} -4(7)(-2)} }{2(7)}$

We can see that here;

a = 7

b = -5

c = -2

So, the original equation must have been:

$7x^{2} -5x-2=0$

5 0

3 years ago

Which function represents the graph of h(x)=2|x+3|−1 after it is translated 2 units right?

harina [27]

$\bf ~\hspace{10em}\textit{function transformations} \\\\\\ \begin{array}{llll} f(x)= A( Bx+ C)^2+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \end{array}\qquad \qquad \begin{array}{llll} f(x)=\cfrac{1}{A(Bx+C)}+D \\\\\\ f(x)= A sin\left( B x+ C \right)+ D \end{array} \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}$

$\bf ~~~~~~\textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if } B\textit{ is negative}\\ ~~~~~~\textit{reflection over the y-axis} \\\\ \bullet \textit{ horizontal shift by }\frac{ C}{ B}\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the right}$

$\bf ~~~~~~if\ \frac{ C}{ B}\textit{ is positive, to the left}\\\\ \bullet \textit{ vertical shift by } D\\ ~~~~~~if\ D\textit{ is negative, downwards}\\\\ ~~~~~~if\ D\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{ B}$

now with that template in mind

$\bf h(x)=\stackrel{A}{2}|\stackrel{B}{1}x+\stackrel{C}{3}|\stackrel{D}{-1}\qquad \qquad \stackrel{\textit{C=C-2 a translation to the right}}{h(x)=2|x\boxed{+3-2}|-1} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill h(x)=2|x+1|-1~\hfill$