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3 years ago

2. Your bank account has -15 in it. You deposit $7 per day.

Mathematics

1 answer:

kobusy [5.1K]3 years ago

5 0

2.a it should have 20 dollars in it
2.b it should take 5 days
i don't know the number 3

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Amari claims that a rectangle is sometimes a parallelogram but a parallelogram is always a

Maslowich

Answer:

The statement is false.

Step-by-step explanation:

A parallelogram is a figure of four sides, such that opposite sides are parallel

A rectangle is a four-sided figure such that all internal angles are 90°

Here, the statement is:

"A rectangle is sometimes a parallelogram but a parallelogram is always a

rectangle."

Here if we found a parallelogram that is not a rectangle, then that is enough to prove that the statement is false.

The counterexample is a rhombus, which is a parallelogram that has two internal angles smaller than 90° and two internal angles larger than 90°, then this parallelogram is not a rectangle, then the statement is false.

The correct statement would be:

"A parallelogram is sometimes a rectangle, but a rectangle is always a parallelogram"

6 0

3 years ago

How do I answer this question?

ankoles [38]

It's 78.38
You have to square root

8 0

3 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.