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Juli2301 [7.4K]

3 years ago

6

Will is selling T-shirts for $15 and sweatshirts for $20 at the basketball game. Wills club team needs to raise over $200 for th

e upcoming tournament. Sketch graph that will represent both number of T-shirts and sweatshirts he must sell to reach his goal?

1 answer:

Paladinen [302]3 years ago

8 0

Answer:

X = 0 4 8 12

Y = 10 7 4 1

Step-by-step explanation:

Given:

Cost of one T shirt = $15

Cost of one sweater = $ 20

To find :

The graph that will represent both number of T-shirts and sweatshirts = ?

Solution:

Let the number of t-shirts be x

The number of sweaters be y

They want to rise a found of $200

Then the equation to represent this situation is

15x + 20y = 200

By trial and error method

15(12) + 20(1) = 200

15(8) + 20(4) = 200

15(0) + 20(10) = 200

15(4) + 20(7) = 200

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Write a linear equation in point-slope form to model the situation below:

Lilit [14]

Answer:

y - 14.95 = 18.075(x-1)

Step-by-step explanation:

If an internet company charges $14.95 per month and a setup fee, then we can say

1 month = $14.95

If the total cost for 9 months is $159.55

Then we can write both cases as a coordinate (1, 14.95) and (9,159.55)

The equation of a line in point slope form is expresses as

y-y0 = m(x-x0)

Find the slope m

m = 159.55-14.95/9-1

m = 144.6/8

m = 18.075

Substitute m = 18.075 and any of the coordinate in the equation to have;

Using the coordinate (1, 14.95)

y - 14.95 = 18.075(x-1)

Hence a linear equation in point-slope form that model the situation is y - 14.95 = 18.075(x-1)

3 0

2 years ago

I need help fast please!!

kaheart [24]

Solution set is: ( $7,\frac{13}{3}$ )

Option B is correct

Step-by-step explanation:

We need to find the solution to system of equations

$y=\frac{1}{3}x+2\\y=\frac{4}{3}x-5$

Let:

$y=\frac{1}{3}x+2\,\,\,eq(1)\\y=\frac{4}{3}x-5\,\,\,eq(2)$

Putting value of y from eq(2) into eq(1)

$\frac{4}{3}x-5=\frac{1}{3}x+2\\Combining\,\,like\,\,terms:\\\frac{4}{3}x-\frac{1}{3}x=2+5\\Taking\,\,LCM\\\frac{4x-1x}{3}=7\\\frac{3x}{3}=7\\x=7$

So, value of x=7

Putting value of x into equation 1 to find the value of y:

$y=\frac{1}{3}x+2\\y=\frac{1}{3}(7)+2\\y=\frac{7}{3}+2\\Taking\,\,LCM\\y=\frac{7+2*3}{3}\\y=\frac{7+6}{3}\\y=\frac{13}{3}$

So, value of y = $\frac{13}{3}$

Solution set is: ( $7,\frac{13}{3}$ )

Option B is correct

Keywords: System of equations

Learn more about system of equations at:

#learnwithBrainly

3 0

3 years ago

PLEASE HELPPPPPPLPPPPPPPP

solong [7]

<h3>Answer: 36%</h3>

============================================

Explanation:

There are 20 blue cards out of 10+20+25 = 55 cards total.

Divide the 20 over 55 to get

20/55 = 0.363636...

where the "36"s go on forever

So 20/55 = 0.3636 approximately

Move the decimal point over 2 spots to go from 0.3637 to 36.37%

Then round to the nearest percent: 36.37% ---> 36%

The probability of getting a blue card is approximately 36%

3 0

2 years ago

How is mathematics done?

Ad libitum [116K]

Answer:

Mathematics can be completed through the use of logical thinking and the use of equations to figure out various problems.

8 0

3 years ago

Read 2 more answers

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago