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3 years ago

13

Help with problem please..

1 answer:

alukav5142 [94]3 years ago

7 0

If 13 students is 52% of the school find the other amount of students, once you do I can help further I believe

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The points (9, 9) and (5, 1) fall on a particular line. What is its equation in slope-intercept form?

Ilia_Sergeevich [38]

Your equation would be y= 2x-9

3 0

3 years ago

12 minutes to drive 30 laps; 48 minutes to drive 120 laps

myrzilka [38]

You can divide 12min.➗30lps. Then you divide 48➗120lps.
And the answer will be 12➗30=0.4 then 48➗120=0.4 and the answer is that the ratios or rates are equvalent by dividing.

7 0

3 years ago

Shannon is selling boxes cookies and candy for a fundraiser. She's sold each box of cookies for $5.25. She also sold one one box

defon

Answer:

The minimum boxes of cookies is 39

Step-by-step explanation:

Let

x ----> the number of boxes of cookies sold

y ----> the number of boxes of candies sold

we know that

The word "at least" means "greater than or equal to"

so

The inequality that represent this problem is

$5.25x+8.75y\geq 200$

The solution is the shaded area above the solid line $5.25x+8.75y=200$

using a graphing tool

The solution is the shaded area -----> see the attached figure

Find out the minimum boxes of cookies needed to sell to reach the goal

assuming only cookies are sold

For y=0

$5.25x+8.75(0)\geq 200$

$5.25x\geq 200$

solve for x

$x\geq 38.1$

Round up

therefore

The minimum boxes of cookies is 39

5 0

3 years ago

Choose the solution(s) of the following system of equations x2 + y2 = 6 x2 – y = 6

ElenaW [278]

<h2>Answer</h2>

$x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1$

Or as ordered pairs: $(\sqrt{6} ,0),(-\sqrt{6} ,0),(\sqrt{5} ,-1),(-\sqrt{5} ,-1)$

<h2>Explanation</h2>

Lets solve our system of equations step by step

$x^2+y^2=6$ equation (1)

$x^2-y=6$ equation (2)

1. Solve for $x^2$ in equation (2)

$x^2-y=6$

$x^2=6+y$ equation (3)

2. Replace equation (3) in equation (1) and solve for $y$

$x^2+y^2=6$

$6+y+y^2=6$

$y^2+y=0$

$y(y+1)=0$

$y=0$ or $y=-1$

3. Replace the values of $y$ in equation (3) and solve for $x$

- For $y=0$

$x^2=6+y$

$x^2=6$

$x=\sqrt{6}$ or $x=-\sqrt{6}$

- For $y=-1$

$x^2=6+y$

$x^2=6-1$

$x^2=5$

$x=\sqrt{5}$ or $x=-\sqrt{5}$

So, the solutions of our system of equation are:

$x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1$

6 0

3 years ago

Read 2 more answers

Factor completely: <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20" id="TexFormula1" title=" x^{2} " alt=" x^{2} " align="ab

Maurinko [17]

24 = 6 * 4
and
6 + 4 = 10

so its the last choice.

6 0

3 years ago

Read 2 more answers