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3 years ago

WhAt is the answer to -12 - 3=-51

1 answer:

7 0

Greetings!

Simplify both sides of the equation.
$-12 - 3=-51$
$-15 \neq -51$

Both sides of the equation are not equal.

Hope this helps.
-Benjamin

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Doheney bake sale 1 1/2 dozen cookies were sold in 1/3 hour what is the unit rate for the dozens of cookies sold per hour

netineya [11]

1/3 hour = 20 minutes

1 1/2 dozen = 12 + 6 = 18 cookies

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5 0

3 years ago

Read 2 more answers

Question 3 of 10

MrMuchimi

Answer:

The point-slope form of the line is:

$y + 4 = 4(x+3)$

Hence, option B is correct.

Step-by-step explanation:

We know the point-slope form of the line equation is

$y-y_1=m\left(x-x_1\right)$

where

m is the slope of the line

(x₁, y₁) is the point

Given

The slope m = 4

From the graph, we can take the point (-3, -4)

So, in our case:

(x₁, y₁) = (-3, -4)

m = 4

now substituting m = 4 and (x₁, y₁) = (-3, -4) in the point-slope form of the line equation

$y-y_1=m\left(x-x_1\right)$

$y - (-4) = 4(x - (-3))$

$y + 4 = 4(x+3)$

Therefore, the point-slope form of the line is:

$y + 4 = 4(x+3)$

Hence, option B is correct.

The graph of the line is also attached.

6 0

2 years ago

Jocelyn owns a food truck that sells tacos and burritos. She only has enough supplies to make 150 tacos or burritos. She sells e

timurjin [86]

Answer:

We know that:

Price of a taco = $4

Price of a burrito = $6.50

x = number of tacos sold

y = number of burritos sold.

We must have that:

x + y ≤ 150 (Because she only has supplies to make 150 tacos or burritos, but she can sell less than that)

And also we know that she must sell at least $780, then:

x*$4 + y*$6.50 ≥ $780

Now we found the system of inequalities:

x + y ≤ 150

x*$4 + y*$6.50 ≥ $780

To solve it graphically, we just need to find each one of the region solutions for each equation, and the intersection of these regions will be the solution for the system.

To graph them may be easier to write them as lines, you can do it as follows:

y ≤ 150 - x

y ≥ ($780 - x*$4)/$6.50

in the first equation, we will shade the area below the line, and in the second equation, we will shade the area above the line.

You can see the image below.

Where the accepted solutions are the ones in the darker part, and we only restrict this to positive values of x (because of how we defined the variable x)

Then looking at the image, we can see that one solution can be the point

x = 20, y = 120