Hi! So, I've got a question. My Math problem is a 2 part problem. Let me just write it out real quick.

Mathematics

1 answer:

Igoryamba2 years ago

4 0

Okay!

This problem is a systems of equations - graphically, your answer will be where all the lines in the problem/system intersect.

You are given two rates at the beginning of the problem:

Keith's rate: 6 pages a day

Tameka's rate: 5 pages a day, 15 page headstart.

When you are given a rate, that is automatically a line! That's what makes these systems of equations really neat.

Try taking Keith and Tameka's rates individually and come up with an equation for the both of them, and then put the two on a graph.

(hint: substitute <em>x</em> for pages)

Good luck!

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3 years ago

PLEASE HELP !!!!!!!!!! Alexia spent 3 minutes working on each of her math problems and 4 minutes on each of her science problem

Lunna [17]

The correct answer to this problem is the third statement which is "Alexia completed 20 math problems and 10 science problem". The following conditions are stated in the problem:
Alexia can answer 3 minutes on each math problem
Alexia can answer 4 minutes on each science problem
She worked more than 60 minutes
We have the inequality 3x + 4y > 60 where x represents the total number of math problem while y represents the total number of the science problem.
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3 years ago

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Area triangles? it says show work so pls do

coldgirl [10]

$\large{\underline{\underline{\pmb{\sf {\color {blue}{Solution:}}}}}}$

Given,

Area = 225 yd²

Base = 30 yd

Height = [To be calculated]

To find:

The height of the given triangle.

We know that, area of a triangle is:

$\frac{1}{2} \times (b \times h) \\ \\ \leadsto \frac{1}{2} \times (30 \times h) \\ \\ \leadsto \frac{1}{2} \times 30h \\ \\ \leadsto \frac{1}{ \cancel{2}} \times \cancel{30}h \\ \\ \leadsto1 \times 15h \\ \\ \leadsto \: h = 15 \: yd$

Therefore, the require height is 15 yd.

$\green\star$ Proof:

$\frac{1}{2} \times (b \times h) \\ \\ \leadsto \frac{1}{2} \times (30 \times 15) \\ \\ \leadsto \frac{1}{ \cancel{2}} \times \cancel{450} \\ \\ \leadsto1 \times 225 \\ \\ area = 225 {yd}^{2}$