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Alina [70]
3 years ago
12

What is another name for the point located at (0 0)?

Mathematics
1 answer:
GenaCL600 [577]3 years ago
8 0
The horizontal axis in the coordinate plane is called the x-axis. The vertical axis is called the y-axis. The point at which the two axes intersect is called the origin. The origin is at 0 on the x-axis and 0 on the y-axis.



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What is 2.08×10to the second power in standard form
BabaBlast [244]

0.0208

you move it two decimal places to the left.

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4 years ago
Jasper Diaz apostrophe Balance Sheet. Total assets are 15,800 dollars. Total liabilities are 4,400 dollars.
Thepotemich [5.8K]

Answer:

the total value of jaspers assets is:  $15,800

the total value of jaspers liabilities is: $4,400

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
20.Solve the triangle: b = 100, c = 100, y = 70". If it is not possible, say so.This triangle is not solvable.Q = 40", a = 68.4,
saveliy_v [14]

Answer:

\begin{gathered} \alpha=\text{ 40\degree} \\ \beta=\text{ 70\degree} \\ a=68.4 \end{gathered}

Step-by-step explanation:

By the theorem of intern angles of a triangle, we know that the sum of all intern angles of a triangle must be 180°, therefore:

*There are two equal angles because there are two equal sides

\begin{gathered} \alpha=180-70-70 \\ \alpha=\text{ 40\degree} \\ \beta=\text{ 70\degree} \end{gathered}

Now, applying the law of sines, we can find side a. The law of sines is represented by the following equation:

\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}

Hence, if b=100, alpha=40° and betha=70°, solve for a:

\begin{gathered} \frac{a}{\sin (40)}=\frac{100}{\sin (70)} \\ a=\frac{100\cdot\sin (40)}{\sin (70)} \\ a=\text{ 68.4} \end{gathered}

7 0
1 year ago
Point C is graphed on a number line at –7. Point D is 13 units away from point C on the number line.
STatiana [176]
(-7)+13=6
(-7)-13=(-20)

Therefore A and C are the possible coordinates
6 0
3 years ago
Find the points of intersection between x^2+y^2=45 and -3x+y=15​
MatroZZZ [7]

Step-by-step answer:

Given:

Circle C1: x^2+y^2 = 45

Line L1: -3x+y=15

Need to find the points of intersection.

Solution:

basically we need to solve for the roots of equations C1 and L1.

Here, we can use substitution of L1 into C1.

Rewrite L1 as : y=3x+15

substitute into C1:

x^2+(3x+15)^2 = 45

Expand

x^2 + 9x^2+90x+225 = 45

Rearrange terms:

10x^2+90x+180 = 0

Simplify

x^2+9x+18 = 0

Factor

(x+6)(x+3) = 0

so

x=-6 or x=-3

Back-substitute x into L1 to calculate y:

x=-6, y=3*x+15 = 3(-6)+15 = -3  => (-6,-3)

x=-3, y=3*x+15 = 3(-3) + 15 = 6 => (-3, 6)

Therefore the intersection points are (-6,-3) and (-3,6)

Check using equation C1:

(-6)^2+(-3)^2 = 36+9 = 45  ok

(-3)^2+(6)^2 = 9 + 36 = 45 ok

Check using equation L1:

Point (-6,-3) : y = 3x+15 = 3(-6) +15 = -3  ok

Point (-3,6) : y = 3x+15 = 3(-3)+15 = 6 ok.

4 0
4 years ago
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