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Ierofanga [76]
3 years ago
8

Please Help, I can't figure out this problem

Mathematics
1 answer:
Otrada [13]3 years ago
3 0

Answer:

<h2>The answer is 5 units</h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

d  =  \sqrt{ ({x1 - x2})^{2} +  ({y1 - y2})^{2}  }  \\

where

(x1 , y1) and (x2 , y2) are the points

From the question

The points are (9,-7) and (5, -4)

The distance between them is

d =  \sqrt{ ({9 - 5})^{2} +  ({ - 7 + 4})^{2}  }  \\  =  \sqrt{ {4}^{2}  +  ({ - 3})^{2} }  \\  =  \sqrt{16 + 9}  \\  =  \sqrt{25}

We have the final answer as

<h3>5 units</h3>

Hope this helps you

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Will mark brainliest
SVETLANKA909090 [29]

Answer:

I think the answer should be Mat A cause it has more space.

3 0
2 years ago
Kelvin finished $\frac{2}{3}$ 2 3​​ of his test in a $\frac{1}{2}$ 1 2​​ hour. Express this rate in hours per test.
svet-max [94.6K]

Complete Question

Kelvin finished 2/3​​ of his test in a 1/2​​ hour. Express this rate in hours per test.

Answer:

3/4 hours per test

Step-by-step explanation:

2/3 of his test = 1/2 hour

This rate in hours per test is calculated as:

x = 1/2 /2/3

x = 1/2 hour ÷ 2/3 test

x = 1/2 × 3/2

x = 3/4 hours per test

This rate in hours per test is given as 3/4 hours per test

4 0
2 years ago
Pluto is looking into joining a gym. He has a budget of $68 per month. The gym he wants to join has fees based on the number of
Artist 52 [7]
7(10 visits) ≤ 68
70 ≤ 68
Not Viable


7(9 visits) ≤ 68
63 ≤ 68
Viable


The first solution is not viable because 10 visits will give a total cost of $70 per month, which is $2 over his monthly budget of $68.

The second solution is viable. He could visit the gym a maximum of 9 times per month for a total of $63 a month and that will be under his $68 monthly budget.

Hope this helps! :)
6 0
3 years ago
Complete the following statements.
Charra [1.4K]

Answer:

1.) ax^2 + bx+ c =0

2.) parabola

3.) X=-b2a

4.) maximum

5.) minimum

Step-by-step explanation: hope this helps

4 0
3 years ago
Really need help with this pls help!!
worty [1.4K]

Let's carry this math sentence over to its natural, "shapey" element. We're going to look at each term not as an ordinary number, but as <em>the area of some shape</em>.

x² (read as "x <em>squared"</em>) can be seen as the area of a square with side lengths of x. 2x can similarly be seen as the area of a <em>rectangle </em>with a length of x and a width of 2. (Picture 1)

What's our question actually asking, though? Something about <em>perfect squares</em>. More specifically, we're looking for something to add on that'll <em>make this thing a perfect square</em>. We're trying to find a missing piece we can slot in to make a square, in other words. Problem is, our shapes don't look much like a square if we put them together right now. We need to do a little cutting and gluing first.

First, we're gonna cut the 2x rectangle lengthwise, getting two rectangles with an area of x, a length of 1, and a width of x. Next, we're going to attach them to the x² square, creating this shape that looks, strangely, like a square with a little bit missing from it (picture 2). What we're trying to do is <em>complete this square, </em>to find the area of that little missing chunk.

As it turns out, we have all the information we need for this. Notice that, using the lengths of the x rectangles, we can find that the square's dimensions are 1 x 1, which means that its area is 1 x 1  = 1.

If we tack this new area on to our original expression, we've "completed the square!" We now have a perfect square with side lengths of (x + 1) and an area of (x + 1)² (picture 3).

So, our final expression is x² + 2x + 1, and the missing constant - the area of the "missing square" we had to find to complete our larger one - is 1.

5 0
3 years ago
Read 2 more answers
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