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3 years ago

It's a comon core math question the teacher is nice but.... She dosent like helping students out

Mathematics

1 answer:

Luba_88 [7]3 years ago

7 0

Answer: Well i mean thats a personal question but tell your teacher that she not being much help.

Step-by-step explanation:

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PLEASE ANSWER!!!!! I WILL MARK YOU AS BRAINLIEST!!!!! IT IS HIGHLY APPRECIATED!!! :)

kkurt [141]

Answer:

A number less than 4 was rolled 18 times.

The number cube was rolled 50 times. The relative frequency of rolling a number less than 4 is 36%.

3 0

3 years ago

Match Term Definition Diameter A) a segment between two points on a circle that passes through its center Circumference B) a pie

Viefleur [7K]

The correct answer to this question is:

Diameter - B) a segment between two points on a circle that passes through its center radius

Circumference - F) the distance around a circle

Radius - C) Circle A and a line segment connecting points B and C which are on the circle.

secant - E) Circle A and a line segment connecting points B and C which are on the circle.

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arc - A) a piece of the circumference of a circle circumference.

5 0

3 years ago

Read 2 more answers

The cost per student of a ski trip varies inversely as the number of students who attend. It will cost each student $250 if 24 s

mars1129 [50]

ANSWER

30 students

EXPLANATION

The cost per student varies inversely as the number of students.

Inverse proportion is written as:

$\begin{gathered} y\propto\frac{1}{x} \\ y=\frac{k}{x} \\ \text{where k = constant of proportionality} \end{gathered}$

Let the cost per student be y.

Let the number of students be x.

It will cost each student $250 if 24 students attend. This means that:

$\begin{gathered} 250=\frac{k}{24} \\ \Rightarrow k=250\cdot24 \\ k=6000 \end{gathered}$

If the cost is down to $200, it means that y is now $200.

That is:

$\begin{gathered} 200=\frac{6000}{x} \\ \Rightarrow x=\frac{6000}{200} \\ x=30\text{ students} \end{gathered}$

Therefore, 30 students could attend.

5 0

2 years ago

Describe the solutions in words to the inequality 4 < c + 2

MAVERICK [17]

Answer:

All real numbers are greater than four

Step-by-step explanation:

4+2=6

6 0

3 years ago

Find the equation of the tangent line to the curve when x has the given value.

Leviafan [203]

Answer:

14) The equation of the tangent line to the curve $f(x)=-2-x^2$ at x = -1 is $y=2x-1$

15) The rate of learning at the end of eight hours of instruction is $w'(8) = 5 \frac{items}{hour}$

Step-by-step explanation:

14) To find the equation of a tangent line to a curve at an indicated point you must:

1. Find the first derivative of f(x)

$f(x)=-2-x^2\\\\\frac{d}{dx} f(x)=\frac{d}{dx}(-2-x^2)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dx} f(x)=\frac{d}{dx}\left(-2\right)-\frac{d}{dx}\left(x^2\right)\\\\f'(x)=-2x$

2. Plug x value of the indicated point, x = -1, into f '(x) to find the slope at x.

$f'(-1)=-2(-1)=2$

3. Plug x value into f(x) to find the y coordinate of the tangent point

$f(-1)=-2-(-1)^2=-3$

4. Combine the slope from step 2 and point from step 3 using the point-slope formula to find the equation for the tangent line

$y-y_1=m(x-x_1)\\y+3=2(x+1)\\y=2x-1$

5. Graph your function and the equation of the tangent line to check the results.

15) To find the rate of learning at the end of eight hours of instruction you must:

1. Find the first derivative of f(x)

$w(t)=15\sqrt[3]{t^2} \\\\\frac{d}{dt}w= \frac{d}{dt}(15\sqrt[3]{t^2})\\\\w'(t)=15\frac{d}{dt}\left(\sqrt[3]{t^2}\right)\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\\\f=\sqrt[3]{u},\:\:u=\left(t^2\right)\\\\w'(t)=15\frac{d}{du}\left(\sqrt[3]{u}\right)\frac{d}{dt}\left(t^2\right)\\\\w'(t)=15\cdot \frac{1}{3u^{\frac{2}{3}}}\cdot \:2t\\\\\mathrm{Substitute\:back}\:u=\left(t^2\right)$

$w'(t)=15\cdot \frac{1}{3(t^2)^{\frac{2}{3}}}\cdot \:2t\\w'(t)=\frac{10t}{\left(t^2\right)^{\frac{2}{3}}}$

2. Evaluate the derivative a t = 8

$w'(t)=\frac{10t}{\left(t^2\right)^{\frac{2}{3}}}\\\\w'(8)=\frac{10\cdot 8}{\left(8^2\right)^{\frac{2}{3}}}\\\\=\frac{80}{\left(8^2\right)^{\frac{2}{3}}}\\\\\left(8^2\right)^{\frac{2}{3}}=16\\\\=\frac{80}{16}\\\\w'(8) = 5 \frac{items}{hour}$

The rate of learning at the end of eight hours of instruction is $w'(8) = 5 \frac{items}{hour}$

7 0

3 years ago

It's a comon core math question the teacher is nice but.... She dosent like helping students out​

It's a comon core math question the teacher is nice but.... She dosent like helping students out