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3 years ago

PLEASE ANSWER MY RECENT! ITS FOR 47 POINTS AND SUPER EASY! I JUST NEED ADVICE AND NO ONE HAS ANSWERD YET!

Mathematics

2 answers:

Anettt [7]3 years ago

5 0

How do i look at your last question

Ann [662]3 years ago

5 0

Answer:

Step-by-step explanation:

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The population of a city has increased by

Aleksandr-060686 [28]

54954 is your answer

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3 years ago

11. Set up the rational work application problem. Shelby and Mary started a beading

Arada [10]

The first one might be b or c

3 0

3 years ago

Tyler's height is 57 inches. What could be his height in centimeters? Explain your resoning. Nost: 1 inch=2.54 centimeters

AfilCa [17]

Answer:

Tyler's height in centimeters is, 144.78 centimeters.

Step-by-step explanation:

Given the statement: Tyler's height is 57 inches.

To find his height in centimeters.

Using the conversion:

$1 {\tex}inch = 2.54 {\tex}centimeters$

Proportion states that the two ratios or fraction are equal.

Using proportion method:

$\frac{1}{57} = \frac{2.54}{x}$

By cross multiply, we get

$x = 57 \times 2.54 = 144.78$ centimeters.

therefore, his height in centimeter is, 144.78 inches.

4 0

3 years ago

HELP! You might have to zoom in

xeze [42]

Remember slope is $\dfrac{rise}{run}$ . In the graph we can just pick two points on the line to find the rise and run between. Let's say we choose (0,3) and (1,7). Between those points there is 7-3=4 rise and 1-0=1 run. So the slope is $\dfrac{4}{1}=\boxed{4}$

5 0

4 years ago

Can someone help me solve this differentiation/tangent problem?

nirvana33 [79]

A)

$\bf g'(x)=\stackrel{product~rule}{2x\cdot f(x)+x^2\cdot f'(x)}\quad 
\begin{cases}
x=5\\
f(5)=5\\
f'(5)=5
\end{cases}
\\\\\\
g'(5)=2(5)\cdot f(5)+(5)^2\cdot f'(5)\implies g'(5)=50+500\\\\\\ g'(5)=550\\\\
-------------------------------\\\\
g(5)=(5)^2f(5)\implies g(5)=125
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}
\begin{cases}
x=5\\
y=125\\
\stackrel{m}{g'(5)}=550
\end{cases}\implies y-125=550(x-5)
\\\\\\
y-125=550x-2750\implies y=550x+400$

b)

$\bf h'(x)=\stackrel{quotient~rule}{\cfrac{f'(x)(x-6)~~-~~f(x)\cdot 1}{(x-6)^2}}\quad 
\begin{cases}
x=5\\
f(5)=5\\
f'(5)=5
\end{cases}
\\\\\\
h'(5)=\cfrac{f'(5)(5-6)~~-~~f(5)\cdot 1}{(5-6)^2}
\\\\\\
h'(5)=\cfrac{5(-1)-5}{(-1)^2}\implies h'(5)=-10\\\\
-------------------------------\\\\$

$\bf h(5)=\cfrac{f(5)}{5-6}\implies h(5)=\cfrac{5}{-1}\implies h(5)=-5
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\quad 
\begin{cases}
x=5\\
y=-5\\
\stackrel{m}{-10}
\end{cases}\implies y-(-5)=-10(x-5)
\\\\\\
y+5=-10x+50\implies y=-10x+45$

3 0

4 years ago