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3 years ago

HELPPPPPPPPPPPPPPPP 6TH GRADE MATH AP EX HELPPPPPPPPPPPPPPPPPPP

Mathematics

2 answers:

PIT_PIT [208]3 years ago

6 0

The answer is A, 42 cubic units

madreJ [45]3 years ago

3 0

The answer is a ; you would use length x width x height or base x height so 8 * 1 1/2 * 3 1/2 = 42

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A teenager’s heart pumps an average of 7200 L of blood every 24 hours. What is the rate of change of volume of blood?

love history [14]

Answer:

300L per hour

Step-by-step explanation:

The rate of change of volume of blood pumped by the teenager's heart expresses the volume of blood pumped with respect to hour.

Rate of change of volume of blood = $\frac{volume of blood}{time}$

= $\frac{7200L}{24}$

= 300L/hour

The rate of change of volume of blood by the teenager's heart is 300 litres per hour. This implies that his/ her heart pumps 300 litres of blood every hour.

7 0

3 years ago

58 yd<br> 73 yd<br> Use 3.14 Instead Of Pie Symbol

defon

Answer:

$771096.8 \: yd^{3}$

Step-by-step explanation:

If you are trying to find the volume, you would use the formula:

$\pi \: r ^{2} (h)$

You would plug in the radius (r), which is 58, and then plug in the height, which is 73.

Make sure you remember to use 3.14 and not the pi key if you are using a calculator.

The final answer is 771,096.8 yd^3

7 0

3 years ago

Simplify the given expression. Assume that no variable equal 0.

marissa [1.9K]

Answer:

The answer I think is correct is B

Step-by-step explanation:

I hope that's corrected?

5 0

2 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

Solve for x in the equation 2x2-5x+1=3

Butoxors [25]

2x^2 - 5x + 1 = 3

Subtract 3 from both sides
2x^2 - 5x + 1 - 3 = 3 - 3

2x^2 - 5x - 2 = 0

Use quadratic formula with A = 2, B = - 5, C = - 2

x = - b +- sqrt b^2 - 4ac/2a

x = - ( - 5) +- sqrt (- 5)^2 - 4(2)( - 2)/(2)(2)

x = 5 +- sqrt 41/4

x = 5/4 + 1/4sqrt 41 or x = 5/4 + - 1/4 sqrt 41

Answer: x = 5/4 + 1/4 sqrt 41
x = 5/4 + - 1/4 sqrt 41

Hope that helps!!!!! (D)

5 0

3 years ago