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k0ka [10]
3 years ago
8

Points for late night math helpers! Thank you guys :)

Mathematics
1 answer:
ale4655 [162]3 years ago
3 0
You are awesome! i hope you have an amazing day/night :)
You might be interested in
Identify the reflection rule on a coordinate plane that verifies that triangle A(-1,7), B(6,5), C(-2,2) and A'(-1,-7), B'(6,-5),
malfutka [58]

Answer:

Option B is correct.

The reflection rule is followed here is,  (x,y) \rightarrow (x, -y).

Explanation:

Since, the triangle is reflected over the x-axis , then only y- coordinates will get changed but the x coordinates will remain the same.

A(-1,7) \rightarrow A'(-1,-7)

B(6,5) \rightarrow B'(6,-5)

C(-2,2) \rightarrow C'(-2,-2)

Therefore, the only reflection rule on the coordinate plane that verifies the triangle are congruent when reflected over x-axis is, (x,y) \rightarrow (x, -y)


3 0
4 years ago
PLEASE HELP PLEASEEEEEEEEEEEEEEEEE​
viva [34]
The answer would be D=2
7 0
3 years ago
Read 2 more answers
Slope from graph, please help I don't get this.
dlinn [17]
The slope would be 4/3 because rise over run


Mark with crown!
3 0
3 years ago
Read 2 more answers
The number of apples in a box decreased from 200 to 150. What is the percent decrease of the number of apples int he box? *
ladessa [460]

Answer:

25%

Step-by-step explanation: 50 is 25% of 200

6 0
3 years ago
Read 2 more answers
Evaluate the integral (3 - 2x)dx for x = -1 to x = 3<br> byinterpreting it in terms of areas.
suter [353]

Answer:

4 unit^2

explanation is given at the end.

Step-by-step explanation:

What this integral represents is the net area between the function f(x) = 3 - 2x

and the x-axis, between the range of x between -1 and 3.

\int^{3}_{-1} {3-2x} \, dx \\\left|3x - 2\dfrac{x^2}{2}\right|^{3}_{-1}\\\left|3x - x^2\right|^{3}_{-1}

we have integrated the equation, and now we're going to put the limits find the area under the function f(x)

\left|3x - x^2\right|^{3}_{-1}\\(3(3)-(3)^2)-(3(-1)-(-1)^2)

-------------

if these seems like a big jump, try to understand it through this:

\left|3x - x^2\right|^{b}_a\\(3(b)-(b)^2)-(3(a)-(a)^2)\\

we've only distributed the limits each time to the same integrated expression.

--------------

coming back to our solution:

(3(3)-(3)^2)-(3(-1)-(-1)^2)

(0)-(-4)

4

so our area is 4

the area above the x-axis is positive, and the area below the x-axis is negative. Since, our answer is +4. <u>We now know that if within this range [-1,3] the area above and below the x-axis exist, there is more area above the x-axis than below the x-axis</u><u>. In other words, the net area is above the x-axis and that is equal to 4 unit^2 </u>

5 0
3 years ago
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