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Blizzard [7]
3 years ago
7

Can you pls help this is 6th grade math

Mathematics
2 answers:
goblinko [34]3 years ago
4 0
The answer is b ..................
wlad13 [49]3 years ago
3 0

Answer: 12 red 9 blue

Step-by-step explanation:

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Find the equation of the line with slope =-7 and passing through (-4,-5) .write your equation in point slope and slope intercept
Leona [35]

Point-slope form:

y - y₁ = m(x - x₁)         "m" is the slope


You know:

m = -7

(-4 , -5) = (x₁ , y₁)

Plug these into the equation:

y - y₁ = m(x - x₁)

y - (-5) = -7(x - (-4))

y + 5 = -7(x + 4)



Slope-intercept form:

y = mx + b       "m" is the slope, "b" is the y-intercept (the y value when x = 0)


You can solve this two ways:


#1:  Since you know m = -7, plug it into the equation

y = mx + b

y = -7x + b

To find "b", plug in the point (-4, -5) into the equation

y = -7x + b

-5 = -7(-4) + b

-5 = 28 + b    Subtract 28 on both sides

-33 = b


y = -7x - 33



#2: Use the point-slope form and change it to slope-intercept form by isolating/getting "y" by itself.

y + 5 = -7(x + 4)     First distribute/multiply -7 into (x + 4)

y + 5 = -7x - 28     Subtract 5 on both sides

y = -7x - 33

7 0
3 years ago
Micah was asked to add the following expressions:
scZoUnD [109]

Given expression: \frac{3x^2-x-9}{x^2+3x+2}+\frac{-2x^2+2x+5}{x^2+3x+2}


We need to add the given rational expressions.


First step: Combined like terms in the numerator and kept the common denominator.


\frac{3x^2-x-9-2x^2+2x+5}{x^2+3x+2} =\frac{x^2+x-4}{x^2+3x+2}


Second step they applied : Canceling the like term x^2 and got \frac{x-4}{3x+2}.

Note: We can't cancel like terms in top and bottom like this.

We can cancel out common factors in top an bottom.

<h3>Therefore, Micah did not add the expressions correctly.</h3>
3 0
4 years ago
Find the domain for the expression <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B%5Csqrt%7Bx%2B3%7D%7D%20%20%20" id="Tex
sashaice [31]

Two rules come into play here:

1) We cannot divide by zero. Thux, x cannot = -3.

2) The domain of the square root function is (0, infinity). This is the dominant limiting factor.

x + 3 must be > 0. Thus, x must be > -3

The domain of this expression is (-3, infinity).

3 0
3 years ago
Which is one of the transformations applied to the graph of f(x) = X^2 to change it into the graph of g(x) = -x^2 +16x - 44
ycow [4]

Answer: First a horizontal shift of 8 units, then a reflection over the x-axis, and then a vertical shift of 20 units.

Step-by-step explanation:

Let's construct g(x) in baby steps.

Ok, we start with f(x) = x^2

The first thing we have is a horizontal translation of A units (where A is not known)

A vertical translation of N units to the right, is written as:

g(x) = f(x - N)

Then we have:

g(x) = (x - A)^2 = x^2 - 2*A*x + A^2

Now, you can see that actually g(x) has a negative leading coefficient, which means that we also have an inversion over the x-axis.

Remember that if we have a point (x, y), a reflection over the x-axis transforms our point into (x, -y)

Then if we apply also a reflection over the x-axis, we have:

g(x) = -f(x - A) = -x^2 + 2*A*x - A^2 = -x^2 + 16*x - 44

Then:

2*A = 16

A*A = 44.

The first equation says that A = 16/2 = 8

But 8^2 is not equal to 44.

Then we need another constant coefficient, which is related to a vertical translation.

If we have a relation y = f(x), a vertical translation of N units up, will be

y = f(x) + N.

Then:

g(x) = -f(x - A) + B

-x^2 + 2*A*x - A^2 + B = x^2 + 16*x - 44

Now we have:

2*A = 16

-A^2 + B = - 44

From the first equation we have A = 8, now we replace it in the second equation and get:

-8^2 + B = -44

B = -44 + 64 = 20

Then we have:

The transformation is:

First an horizontal shift of 8 units, then a reflection over the x-axis, and then a vertical shift of 20 units.

8 0
3 years ago
A box of cereal can be purchased in the large size, which is 15 ounces for $3.64, or the family size, which is 24 ounces for $4.
Darina [25.2K]

Answer:

Family size at $0.20 per ounce

5 0
2 years ago
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