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3 years ago

13

Mrs. Stoddard has 264 crayons for her art class of 22 students. How many crayons will each student get if the crayons are divide

d equally? Use a model to help you solve the problem.

1 answer:

DedPeter [7]3 years ago

8 0

Answer:

12

Step-by-step explanation:

264/22

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Determine the slope of the line that has the following coordinates: (5, 9)(11, - 3)

AlladinOne [14]

Answer:

$x_1 = 5, x_2 =11, y_1 =9, y_2 = -3$

The slope can be founded with this formula:

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

And replacing we got:

$m =\frac{-3 -9}{11-5}= -2$

And the best answer for this case would be :

$m=-2$

Step-by-step explanation:

For this case we have the following two points given:

$x_1 = 5, x_2 =11, y_1 =9, y_2 = -3$

The slope can be founded with this formula:

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

And replacing we got:

$m =\frac{-3 -9}{11-5}= -2$

And the best answer for this case would be :

$m=-2$

7 0

3 years ago

HELP ME PLEASE!!!!!!!!!!!!!!!!

Pepsi [2]

The ordered pairs of the solution to the system of equation are [(3, 6), (-3, 18)]

<h3>System of equations</h3>

Given the following system of equations expressed as:

f(x) = x² - 2x + 3 and f(x) = -2x + 12

Since they are both function of x, hence;

x² - 2x + 3 = -2x + 12

x² - 2x + 3 + 2x - 12 = 0

x² - 9 = 0

x² = 9

x = ±√9

x = ±3

If x = 3

f(x) = -2(3) + 12

f(x) = 6

If x = -3

f(x) = -2(-3) + 12

f(x) =18

Hence the ordered pairs of the solution to the system of equation are [(3, 6), (-3, 18)]

Learn more on system of equation here: brainly.com/question/14323743

#SPJ1

6 0

2 years ago

WILL GIVE BRAINLIEST find the equation of the axis of symmetry for the parabola y= -x^2-5x

timama [110]

Answer:

The axis of symmetry is $-\frac{5}{2}$ .

Check the degree of your polynomial
Plug your numbers into the axis of symmetry formula
Write down the equation of the axis of symmetry

Please Mark Brainliest If This Helped!

6 0

2 years ago

Let <img src="https://tex.z-dn.net/?f=i" id="TexFormula1" title="i" alt="i" align="absmiddle" class="latex-formula"> be the imag

VLD [36.1K]

$Hey~freckledspots!\\----------------------$

$We~will~solve~for~i^{425}!$

$Rule~of~exponent: a^{b + c} = a^ba^c\\Apply:~i^{425}~=~i^{424}i\\ \\Rule~of~exponent: a^{bc} = (a^{b})^c\\Apply: i^{424} = i(i^2)^{212} \\\\Rule~of~imaginary~number: i^2 = -1\\Apply: i(i^2)^{212} = -1^{212}i\\\\Rule~of~exponent~if~n~is~even: -a^n = a^n\\Apply: -1^{212}i = 1^{212}i\\\\Simplify: 1^{212}i = 1i\\Multiply: 1i * 1 = i\\----------------------\\$

$Now~let's~solve~1^{14}!\\\\Rule~of~exponent: a^{b + c} = a^ba^c\\Apply: i^{14} = (i^2)^7\\\\Rule~of~imaginary~number: i^2 = -1\\Apply: (i^2)^7 = -1^7\\\\Rule~of~exponent~if~n~is~odd: (-a)^n = -a^n\\Apply: -1^7 = -1^7\\\\Simplify: -1^7 = -1\\----------------------\\Now,~we~have: i-1+i^{-14}+i^{44}\\----------------------$

$Now~lets~solve~i^{-14}\\\\Rule~of~exponent: a^{-b} = \frac{1}{a^b} \\Apply: i^{-14} = \frac{1}{i^{14}} \\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: \frac{1}{i^{14}} = \frac{1}{(i^2)^7}\\ \\Rule~of~imagianry~number: i^2 = -1\\Apply: \frac{1}{(i^2)^7} = \frac{1}{-1^7} \\\\Simplify: \frac{1}{-1^7} = \frac{1}{-1} \\\\Rule~of~fractions: \frac{a}{-b} = -\frac{a}{b} \\Apply: \frac{1}{-1} = -\frac{1}{1} = -1\\----------------------\\Now,~we~have: i-1-1+i^44\\----------------------$

$Now~let's~solve~i^{44}!\\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: i^{44} = (i^2)^{22}\\\\Rule~of~imaginary~numbers: i^2 = -1\\Apply: (i^2)^{22} = -1^{22}\\\\Rule~of~exponent~if~n~is~even: (-a)^n = a^n\\Apply: -1^{22} = 1^{22}\\\\Simplify: 1^{22} = 1\\----------------------\\Now,~we~have~i-1-1+1\\----------------------$

$Now~let's~simplify~the~expression!\\\\= i-1-1+1 \\= 1 + i -2\\= -1+i\\----------------------$

$Answer:\\\large\boxed{-1+i}\\----------------------$

$Hope~This~Helped!~Good~Luck!$

8 0

4 years ago

Read 2 more answers

What picture shows that angle A and angle B are supplementary angles?<br><br>

Sindrei [870]

That picture doesn’t show that they are supplementary angles it’s shows that they are complementary angles.

7 0

3 years ago