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2 years ago

Help im stuck and my pullcrap ahh teacher not her to help and she got me do ixls

Mathematics

1 answer:

marysya [2.9K]2 years ago

3 0

Answer:

c=4

Step-by-step explanation:

c=23

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Find the first four terms of the sequence an = 4n-6

natulia [17]

Answer:

The first four terms of the sequence are-6, -2, 2 and 6

Step-by-step explanation:

The given sequence is $a_n=4n-6$

In order to find the first four term of the sequence we put n =0, 1, 2 and 3.

For n =0

$a_0=4(0)-6=-6$

For n =1

$a_1=4(1)-6=-2$

For n =2

$a_2=4(2)-6=2$

For n =3

$a_3=4(3)-6=6$

Therefore, the first four terms of the sequence are

-6, -2, 2 and 6

5 0

3 years ago

2. A supermarket display consists of boxes of cereal. The bottom row has 27 boxes. Each row has four fewer boxes than the row be

tresset_1 [31]

Answer:

Step-by-step explanation:

) Nth term = F + (N - 1) x D, where F=First term, N=Number of terms, D=Common difference

6th row = 23 + (6 - 1) x -3

= 23 + (5) x -3

= 23 + (-15)

= 8 - number of boxes in the top row.

b) Sum = N/2[2F + (N - 1) x D]

= 6/2[2*23 + (6 - 1) x -3]

= 3 [46 + (5) x -3 ]

= 3 [46 + -15 ]

= 3 [ 31 ]

= 93 - total number of boxes in the entire display.

4 0

3 years ago

4.)star liked to practice her number sense when possible. She

Wittaler [7]

Answer:

56 + 17

Step-by-step explanation:

The numbers could be

56 + 17 because Star was separating how she added the decimal places

She added the tens units first and then the ones units last.

so 50 + 10 first

and then 6 + 7

5 0

2 years ago

For a snack, Sameer ate a plum with 32 calories and several strawberries with 4 calories each. If the total of calories in Samee

irinina [24]

Answer: 6

Step-by-step explanation:

So to find the answer, we would use the equation 32+4x=56 where as x represents the amount of strawberries Sameer ate. Now to solve for x

56-32=4x

24=4x

now divide 4 from bboth sides of the equation getting:

x=6

so Sameer ate 6 strawberries

6 0

3 years ago

10 points, please help me and explain how to do this with answers!

8_murik_8 [283]

$\bf f(x)=log\left( \cfrac{x}{8} \right)\\\\
-----------------------------\\\\
\textit{x-intercept, setting f(x)=0}
\\\\
0=log\left( \cfrac{x}{8} \right)\implies 0=log(x)-log(8)\implies log(8)=log(x)
\\\\
8=x\\\\
-----------------------------$

$\bf \textit{y-intercept, is setting x=0}\\
\textit{wait just a second!, a logarithm never gives 0}
\\\\
log_{{ a}}{{ b}}=y \iff {{ a}}^y={{ b}}\qquad\qquad 
% exponential notation 2nd form
{{ a}}^y={{ b}}\iff log_{{ a}}{{ b}}=y 
\\\\
\textit{now, what exponent for "a" can give you a zero? none}\\
\textit{so, there's no y-intercept, because "x" is never 0 in }\frac{x}{8}\\
\textit{that will make the fraction to 0, and a}\\
\textit{logarithm will never give that, 0 or a negative}\\\\
$

$\bf -----------------------------\\\\
domain
\\\\
\textit{since whatever value "x" is, cannot make the fraction}\\
\textit{negative or become 0, , then the domain is }x\ \textgreater \ 0\\\\
-----------------------------\\\\
range
\\\\
\textit{those values for "x", will spit out, pretty much}\\
\textit{any "y", including negative exponents, thus}\\
\textit{range is }(-\infty,+\infty)$
p, li { white-space: pre-wrap; }

----------------------------------------------------------------------------------------------

now on 2)

$\bf f(x)=\cfrac{3}{x^4}$ if the denominator has a higher degree than the numerator, the horizontal asymptote is y = 0, or the x-axis,

in this case, the numerator has a degree of 0, the denominator has 4, thus y = 0

vertical asymptotes occur when the denominator is 0, that is, when the fraction becomes undefined, and for this one, that occurs at $x^4=0\implies x=0$ or the y-axis

----------------------------------------------------------------------------------------------

now on 3)

$\bf f(x)=\cfrac{1}{x}$

now, let's see some transformations templates

$\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}}
\end{array}$

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

now, let's take a peek at g(x)

$\bf \begin{array}{lcllll}
g(x)=&-&\cfrac{1}{x}&+3\\
&\uparrow &&\uparrow \\
&\textit{upside down}&&
\begin{array}{llll}
\textit{vertical shift up}\\
\textit{by 3 units}
\end{array}
\end{array}$

3 0

3 years ago