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2 years ago

Suppose 3 people each called 3 friends on the phone.

Mathematics

1 answer:

Usimov [2.4K]2 years ago

3 0

Answer: 3^3 x 3^3 x 3

Because 3x3x3x3x3=243

243 people were called all together.

You might be interested in

Evaluate the expression 0.03[3-0.5

Dafna11 [192]

0.03(3-0.5)
0.03(2.5)
0.075

8 0

2 years ago

Please help I am on number 1 please explain

jeka57 [31]

A % is out of a hundred, so you don't need the %/100. Then, you wouldn't do cross multiplication, you would just divide 15/27. Remember, it would be 15/27 because it's percent markup, not the percent the price increased by. So, doing that simple calculation on a calculator you get 0.5555. This converted to a percent is just moving the decimal to the right twice, or multiplied by a 100. That would give you 55.55%, and that's your answer. Make sense?

5 0

3 years ago

-1+5x≥-16 I dont know how to solve

olasank [31]

Answer: $x \ge -3$

Explanation:

Follow PEMDAS in reverse to undo what's happening to x.

We first add 1 to both sides, then divide both sides by 5 to fully isolate x.

Refer to the steps below to see what I mean.

$-1 + 5x \ge -16\\\\5x \ge -16+1\\\\5x \ge -15\\\\x \ge -15/5\\\\x \ge -3\\\\$

The inequality sign stays the same the entire time. The only time it flips is when you divide both sides by a negative number.

The solution set for x is anything -3 or larger.

If x was an integer, then we could say the solution set is {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, ...}

7 0

1 year ago

The question is in the screenshot

Sidana [21]

Answer:

b. -36/77

Step-by-step explanation:

As 0 < x < $\pi$ /2 => tan x > 0

As 0< y < $\pi$ /2 => tan y > 0

We have the formula:

$\frac{1}{sin^{2}x } =1 + cot^{2}x$ => $\frac{1}{(8/17)^{2} }$ = $1+cot^{2}x$ => $cot^{2} x=225/64$

As tan x = 1/(cotx) => $tan^{2}x = \frac{1}{cot^{2}x} =\frac{1}{225/64} = \frac{64}{225}$

As tan x > 0 => tan x = 8/15

$\frac{1}{cos^{2}y } = 1+tan^{2}y$ => $\frac{1}{(3/5)^{2} } = 1 + tan^{2} y$ => $tan^{2}y=\frac{16}{9}$

As tan y > 0 => tan y = 4/3

As tan (x-y)= $\frac{tan x - tan y}{1+ tanx * tany}$ $=\frac{(8/15)-(4/3)}{1+(8/15)*(4/3)} = \frac{-36}{77}$

8 0

2 years ago

Write an equation that has a graph with the shape of y = x^2, but shifted left 3 units.

svetoff [14.1K]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf % template detailing
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

$\bf \bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

now, with that template in mind, let's see

$\bf y=x^2\implies 
\begin{array}{llll}
y=(&1x&-0)^2\\
&B&C
\end{array}$

so, just change C to +3, thus C/B is 3/1

6 0

2 years ago