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3 years ago

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Dr. Potter provides vaccinations against polio and measles. Each polio vaccination consists of 444 doses, and each measles vacci

nation consists of 222 doses. Last year, Dr. Potter gave a total of 606060 vaccinations that consisted of a total of 184184184 doses. How many polio vaccinations and how many measles vaccinations did Dr. Potter give last year

1 answer:

nataly862011 [7]3 years ago

6 0

Answer:

37 polio vaccinations and 23 measles vaccination.

Step-by-step explanation:

Let the number of polio vaccination given=x

Let the number of measles vaccination given=y

Dr. Potter gave a total of 60 vaccinations

x+y=60

Since each polio vaccination consists of 4 doses, and each measles vaccination consists of 2 doses. A total of 184 doses was given out.

4x+2y=184

To find the values of x and y, we solve the two equations simultaneously.

x+y=60 (1)

4x+2y=184 (2)

From (1), x=60-y

Substitute x=60-y into equation 2.

4x+2y=184

4(60-y)+2y=184

240-4y+2y=184

240-184=4y-2y

56=2y

Divode both sides by 2

y=23

Recall earlier that x=60-y

x=60-23=37

Dr Potter gave 37 polio vaccinations and 23 measles vaccination.

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What is the 6th term of the geometric sequence where a1 = -4096 and a4 = 64?

Akimi4 [234]

$\bf \begin{array}{llccll}
term&value\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
a_1&-4096\\
a_2&-4096r\\
a_3&-4096rr\\
a_4&-4096rrr\\
&-4096r^3\\
&64
\end{array}\implies -4096r^3=64
\\\\\\
r^3=\cfrac{64}{-4096}\implies r^3=-\cfrac{1}{64}\implies r=\sqrt[3]{-\cfrac{1}{64}}
\\\\\\
r=\cfrac{\sqrt[3]{-1}}{\sqrt[3]{64}}\implies \boxed{r=\cfrac{-1}{4}}\\\\
-------------------------------$

$\bf n^{th}\textit{ term of a geometric sequence}\\\\
a_n=a_1\cdot r^{n-1}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
r=-\frac{1}{4}\\
a_1=-4096\\
n=6
\end{cases}
\\\\\\
a_6=-4096\left( -\frac{1}{4} \right)^{6-1}\implies a_6=-4^6\left( -\frac{1}{4} \right)^5$

4 0

3 years ago

Read 2 more answers

With a headwind a plane traveled 128 km in 2h. The return trip with a tail wind took 24 min less. Find the air speed of the plan

sertanlavr [38]

Based on the information given, it can be noted that the air speed that the plane has will be 71.1 km per hour.

The number of hours used for the return trip will be:

= 2 hours - 24 minutes.

= 1 hour 36 minutes.

= 1.6 hours.

Therefore, the speed of the air plane will be:

= (128 + 128) / (2 + 1.6)

= 256/3.6.

= 71.1 km per hour.

Learn more about speed on:

brainly.com/question/25860159

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3 years ago

Slope-intercept form of the equation for the line?

vaieri [72.5K]

Answer:

$\large\boxed{y=-\dfrac{3}{10}x+\dfrac{1}{2}}$

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points from the graph (-5, 2) and (5, -1).

Substitute:

$m=\dfrac{-1-2}{5-(-5)}=\dfrac{-3}{10}=-\dfrac{3}{10}$

We have the equation in form:

$y=-\dfrac{3}{10}x+b$

Put the coordinates of the point (5, -1) to the equation:

$-1=-\dfrac{3}{10}(5)+b$

$-1=-\dfrac{3}{2}+b$

$-\dfrac{2}{2}=-\dfrac{3}{2}+b$ <em>add 3/2 to both sides</em>

$\dfrac{1}{2}=b\to b=\dfrac{1}{2}$

4 0

3 years ago

Q3: Identify the graph of the equation and write and equation of the translated or rotated graph in general form. (Picture Provi

natta225 [31]

Answer:

b. circle; $2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

Step-by-step explanation:

The given conic has equation;

$x^2-5x+y^2=3$

We complete the square to obtain;

$(x-\frac{5}{2})^2+(y-0)^2=\frac{37}{4}$

This is a circle with center;

$(\frac{5}{2},0)$

This implies that;

$x=\frac{5}{2},y=0$

When the circle is rotated through an angle of $\theta=\frac{\pi}{3}$ ,

The new center is obtained using;

$x'=x\cos(\theta)+y\sin(\theta)$ and $y'=-x\sin(\theta)+y\cos(\theta)$

We plug in the given angle with x and y values to get;

$x'=(\frac{5}{2})\cos(\frac{\pi}{3})+(0)\sin(\frac{\pi}{3})$ and $y'=--(\frac{5}{2})\sin(\frac{\pi}{3})+(0)\cos(\frac{\pi}{3})$

This gives us;

$x'=\frac{5}{4} ,y'=\frac{5\sqrt{3} }{4}$

The equation of the rotated circle is;

$(x'-\frac{5}{4})^2+(y'-\frac{5\sqrt{3} }{4})^2=\frac{37}{4}$

Expand;

$(x')^2+(y')^2-\frac{5\sqrt{3} }{2}y'-\frac{5}{2}x'+\frac{25}{4} =\frac{37}{4}$

Multiply through by 4; to get

$4(x')^2+4(y')^2-10\sqrt{3}y'-10x'+25 =37$

Write in general form;

$4(x')^2+4(y')^2-10x'-10\sqrt{3}y'-12 =0$

Divide through by 2.

$2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

8 0

3 years ago

Simplify <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B6%5D%7B5%7D%20%20%5Ctimes%20%20%5Csqrt%5B2%5D%7B5%7D%20%20%3D%20" id

jek_recluse [69]

Step-by-step explanation:

$\sqrt[6]{5} \times \sqrt[2]{5}$

$= {5}^{ \frac{1}{6} } \times {5}^{ \frac{1}{2} }$

$= {5}^{ \frac{1}{6} + \frac{1}{2} }$

$= {5}^{ \frac{1 + 3}{6} }$

$= {5}^{ \frac{4}{6} }$

$= {5}^{ \frac{2}{3} }$

$= \sqrt[3]{ {5}^{2} } (ans)$

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3 years ago