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3 years ago

Y=-2(x-3)^2 +10 Y=4x

Mathematics

1 answer:

shepuryov [24]3 years ago

7 0

Answer:

i might be wrong but 8 if not (2,8) if not sorry

Step-by-step explanation:

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The concession stand sells hot dogs and soda during Beck high school football games John bought six hotdogs and four Sotos and p

siniylev [52]

Question is not proper, Proper Question is given below;

The concession stand sells hot dogs and drinks during their high school foot ball game. John bought 6 hot dogs and 4 soda and paid $6.70. Jessica bought 4 hot dogs and 3 Soda and paid $4.65. What is the price of a hot dog and a Soda?

Answer:

Hot dog cost for $0.75 and Soda cost for $0.55.

Step-by-step explanation:

Let the Price of each Hot dog be x.

Also Let the price of each Soda be y.

Now Given:

John bought 6 hot dogs and 4 Soda and paid $6.70.

Now Total Money paid by John is equal to Sum of Number of hot dogs he bought multiplied by Cost of each hot dogs and Number of Soda he bought multiplied by cost of each soda.

Framing in equation form we get;

$6x+4y=6.70$

Dividing by 2 on both sides we get;

$\frac{6x}{2}+\frac{4y}{2}=\frac{6.70}{2}\\\\3x+2y=3.35 \ \ \ \ equation\ 1$

Also Given:

Jessica bought 4 hot dogs and 3 Soda and paid $4.65.

Now Total Money paid by John is equal to Sum of Number of hot dogs he bought multiplied by Cost of each hot dogs and Number of Soda he bought multiplied by cost of each soda.

Framing in equation form we get;

$4x+3y=4.65 \ \ \ \ equation \ 2$

Multiplying equation 1 by 3 we get;

$3(3x+2y)=3.35\times3\\9x+6y=10.05 \ \ \ \ equation \ 3$

Also Multiplying equation 2 by 3 we get;

$2(4x+3y)=4.65\times 2\\8x+6y = 9.3 \ \ \ \ equation\ 4$

Now Subtracting equation 4 from equation 3 we get;

$(9x+6y)-(8x+6y)=10.05-9.3\\9x+6y-8x-6y= 0.75\\x=\$0.75$

Now Substituting the value of x in equation 1 we get;

$3x+2y=3.35\\\\3\times 0.75+2y = 3.35\\\\2.25+2y=3.35\\\\2y = 3.35-2.25\\\\2y = 1.1\\\\y=\frac{1.1}{2}= \$0.55$

Hence Hot dog cost for $0.75 and Soda cost for $0.55.

4 0

3 years ago

8 out of 40 men wear glasses. write this as a fraction in its lowest terms

Aleksandr-060686 [28]

Answer:

8/40 ..

Step-by-step explanation:

5 0

3 years ago

Please help me with this please please please

ExtremeBDS [4]

Answer: I think it would be 40000/160/500

8 0

3 years ago

Read 2 more answers

6 math questions, answer all please for all points

Sergeeva-Olga [200]

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 1

Recall that the projection of a vector $u$ onto $v$ is $\displaystyle proj_vu=\biggr(\frac{u\cdot v}{||v||^2}\biggr)v$ .

Identify the vectors:

$u=\langle-10,-7\rangle$

$v=\langle-8,4\rangle$

Compute the dot product:

$u\cdot v=(-10*-8)+(-7*4)=80+(-28)=52$

Find the square of the magnitude of vector v:

$||v||^2=\sqrt{(-8)^2+(4)^2}^2=64+16=80$

Find the projection of vector u onto v:

$\displaystyle proj_vu=\biggr(\frac{u\cdot v}{||v||^2}\biggr)v\\\\proj_vu=\biggr(\frac{52}{80}\biggr)\langle-8,4\rangle\\\\proj_vu=\biggr\langle\frac{-416}{80} ,\frac{208}{80}\biggr\rangle\\\\proj_vu=\biggr\langle\frac{-26}{5} ,\frac{13}{5}\biggr\rangle\\\\proj_vu=\langle-5.2,2.6\rangle$

Thus, B is the correct answer

Problem 2

Treat the football and wind as vectors:

Football: $u=\langle42\cos172^\circ,42\sin172^\circ\rangle$

Wind: $v=\langle13\cos345^\circ,13\sin345^\circ\rangle$

Add the vectors: $u+v=\langle42\cos172^\circ+13\cos345^\circ,42\sin172^\circ+13\sin345^\circ\rangle\approx\langle-29.034,2.481\rangle$

Find the magnitude of the resultant vector:

$||u+v||=\sqrt{(-29.034)^2+(2.481)^2}\approx29.14$

Find the direction of the resultant vector:

$\displaystyle \theta=tan^{-1}\biggr(\frac{2.841}{-29.034}\biggr)\approx -5^\circ$

Because our resultant vector is in Quadrant II, the true direction angle is 6° clockwise from the negative axis. This means that our true direction angle is $180^\circ-5^\circ=175^\circ$

Thus, C is the correct answer

Problem 3

We identify the initial point to be $R(-2,12)$ and the terminal point to be $S(-7,6)$ . The vector in component form can be found by subtracting the initial point from the terminal point:

$v=\langle-7-(-2),6-12\rangle=\langle-7+2,-6\rangle=\langle-5,-6\rangle$

Next, we find the magnitude of the vector:

$||v||=\sqrt{(-5)^2+(-6)^2}=\sqrt{25+36}=\sqrt{61}\approx7.81$

And finally, we find the direction of the vector:

$\displaystyle \theta=tan^{-1}\biggr(\frac{6}{5}\biggr)\approx50.194^\circ$

Keep in mind that since our vector is in Quadrant III, our direction angle also needs to be in Quadrant III, so the true direction angle is $180^\circ+50.194^\circ=230.194^\circ$ .

Thus, A is the correct answer

Problem 4

Add the vectors:

$v_1+v_2=\langle-60,3\rangle+\langle4,14\rangle=\langle-60+4,3+14\rangle=\langle-56,17\rangle$

Determine the magnitude of the vector:

$||v_1+v_2||=\sqrt{(-56)^2+(17)^2}=\sqrt{3136+289}=\sqrt{3425}\approx58.524$

Find the direction of the vector:

$\displaystyle\theta=tan^{-1}\biggr(\frac{17}{-56} \biggr)\approx-17^\circ$

Because our vector is in Quadrant II, then the direction angle we found is a reference angle, telling us the true direction angle is 17° clockwise from the negative x-axis, so the true direction angle is $180^\circ-17^\circ=163^\circ$

Thus, A is the correct answer

Problem 5

A vector in trigonometric form is represented as $w=||w||(\cos\theta i+\sin\theta i)$ where $||w||$ is the magnitude of vector $w$ and $\theta$ is the direction of vector $w$ .

Magnitude: $||w||=\sqrt{(-16)^2+(-63)^2}=\sqrt{256+3969}=\sqrt{4225}=65$

Direction: $\displaystyle \theta=tan^{-1}\biggr(\frac{-63}{-16}\biggr)\approx75.75^\circ$

As our vector is in Quadrant III, our true direction angle will be 75.75° counterclockwise from the negative x-axis, so our true direction angle will be $180^\circ+75.75^\circ=255.75^\circ$ .

This means that our vector in trigonometric form is $w=65(\cos255.75^\circ i+\sin255.75^\circ j)$

Thus, C is the correct answer

Problem 6

Write the vectors in trigonometric form:

$u=\langle40\cos30^\circ,40\sin30^\circ\rangle\\v=\langle50\cos140^\circ,50\sin140^\circ\rangle$

Add the vectors:

$u+v=\langle40\cos30^\circ+50\cos140^\circ,40\sin30^\circ+50\sin140^\circ\rangle\approx\langle-3.661,52.139\rangle$

Find the magnitude of the resultant vector:

$||u+v||=\sqrt{3.661^2+52.139^2}\approx52.268$

Find the direction of the resultant vector:

$\displaystyle\theta=tan^{-1}\biggr(\frac{52.139}{-3.661} \biggr)\approx-86^\circ$

Because our resultant vector is in Quadrant II, then our true direction angle will be 86° clockwise from the negative x-axis. So, our true direction angle is $180^\circ-86^\circ=94^\circ$ .

Thus, B is the correct answer

5 0

2 years ago

01:49:1

Keith_Richards [23]

Answer:

The correct option is C:

C) The representative sample contained more girls than boys.

Step-by-step explanation:

It is given that a random sample is chosen from a total students of 160 students. The sample can be of 10,15 or any small numbers of students as compared to 160. However, a sample cannot be of 160 students as it is defined as a population in this case.

A random sample is always unbiased. Which means that the sample chosen should have around the same proportion of girls to boys as it is in the population of 160.

We know that:

Total boys in 160 = 65

Total girls in 160 = 95

Proportion of girls to boys = 95/65 = 1.462

Which means that for every 1 boy, there are 1.462 of girls.

The same ratio is held in a random sample, hence the total number of girls will be greater than boys

6 0

3 years ago