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3 years ago

12

(a) How could you calculate 23 x 4 on your calculator if the 4 button is broken? (b) How could you calculate 23 x 21 if the 2 bu

tton on your calculator is broken?

1 answer:

Crank3 years ago

3 0

Answer:

23 x 4

=> 23 x (2 + 2)

=> 23 x 2 + 23 x 2

23 x 21

=> 23 x (3 x 7)

=> (23 x 3) x (23 x 7)

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ANSWER PLEASEE I WILL 15 TO WHOEVER DOES PLEASE

KIM [24]

Answer:

Its either answer one or two. I believe it's answer one.

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3 years ago

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State the value of the discriminant of -6x^2 +2x + 3 = 0

Dmitry [639]

Answer:

76

Step-by-step explanation:

recall that for a quadratic equation in the form

ax² + bx + c = 0, the discriminant is given by

discriminant, D = b²- 4ac

we are given the following:

-6x² + 2x + 3 = 0,

comparing this with the general equation above, it is clear that,

a = -6, b = 2 and c = 3,

substituting these into our equation for discriminant:

D = b²- 4ac

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3 years ago

In another survey, Jane found that 25% of the total number of teenagers surveyed like sandwiches. If 50 teenagers like sandwiche

34kurt

Set up a proportion: 25/100 = 50/n where n = # of teens surveyed
Simplify fraction: 1/4 = 50/n
Cross multiply: n = 200
So 200 teens were surveyed.

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3 years ago

How does the graph of f(x)=3lx+2l+4 relate to its parent function?

Sergeeva-Olga [200]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
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}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\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{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, with that template above in mind, let's see this one

$\bf parent\implies f(x)=|x|
\\\\\\
\begin{array}{lllcclll}
f(x)=&3|&1x&+2|&+4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

A=3, B=1, shrunk by AB or 3 units, about 1/3
C=2, horizontal shift by C/B or 2/1 or just 2, to the left
D=4, vertical shift upwards of 4 units

check the picture below

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3 years ago

Which sum or difference identity would you use to verify that cos (180° - q) = -cos q?

Phantasy [73]

Answer:

$\cos (a-b)=\cos a \cos b+\sin a \sin b$

Step-by-step explanation:

Given : $\cos (180^{\circ}-q)=-\cos q$

We have to write which identity we will use to prove the given statement.

Consider $\cos (180^{\circ}-q)=-\cos q$

Take left hand side of given expression $\cos (180^{\circ}-q)$

We know

$\cos (a-b)=\cos a \cos b+\sin a \sin b$

Comparing , we get, a= 180° and b = q

Substitute , we get,

$\cos (180^{\circ}-q)=\cos 180^{\circ} \cos (q)+\sin q \sin 180^{\circ}$

Also, we know $\sin 180^{\circ}=0$ and $\cos 180^{\circ}=-1$

Substitute, we get,

$\cos (180^{\circ}-q)=-1\cdot \cos (q)+\sin q \cdot 0$

Simplify , we get,

$\cos (180^{\circ}-q)=-\cos (q)$

Hence, use difference identity to prove the given result.

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3 years ago

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