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3 years ago

Ray MN and ML are two sides of an angle. What is a name of this angle?

Mathematics

1 answer:

Airida [17]3 years ago

6 0

LMN, NML or Angle M. Depends on context for all of these. They all work though.

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Your cell phone bill in August was $61.43, which was $21.75 more than your bill in July. Your cell phone bill in July was $13.62

irina [24]

Answer:

$53.30

Step-by-step explanation:

61.43-21.75=39.68 is the cell phone bill in July.

39.68+13.62= 53.30 is the phone bill in June.

3 0

3 years ago

Read 2 more answers

If QT is 9 and TP is 8 Find the length of the following: RP RT RS

anyanavicka [17]

Answer:

i. RP = 17

ii. RT = 15

iii. RS = 30

Step-by-step explanation:

Given that; QT = 9 and TP = 8.

From the diagram, join R to P. Thus RP = QP (radius of the circle)

RP = QT + TP

= 9 + 8

RP = 17

Applying Pythagoras theorem to triangle TRP;

RT = $\sqrt{(RP)^{2} - (TP)^{2} }$

= $\sqrt{17^{2} - 8^{2} }$

= $\sqrt{225}$

= 15

∴ RT = 15

But, RT = TS = 15.

So that;

RS = 15 + 15

= 30

RS = 30

Therefore; RP = 17, RT = 15 and RS = 30.

6 0

3 years ago

3) Solve the problem using your graphing calculator. Round all para meter values to the nearest hundredth.

Mama L [17]

Answer:

mmm, well, not much we can do per se, you'd need to use a calculator.

I'd like to point out you'd need a calculator that has regression features, namely something like a TI83 or TI83+ or higher.

That said, you can find online calculators with "quadratic regression" features, which is what this, all you do is enter the value pairs in it, to get the equation.

Step-by-step explanation:

8 0

2 years ago

-2x + 15y = -24 2x + 9y = 24

pashok25 [27]

Answer:

dont know

Step-by-step explanation:

sorry

3 0

3 years ago

What are the endpoint coordinates for the midsegment of △BCD that is parallel to BC?

natulia [17]

Answer:

The endpoint coordinates for the mid-segment are: $(-2,-1)$ and $(1,0)$

Step-by-step explanation:

According to the given diagram, the coordinates of the vertices of $\triangle BCD$ are: $B(-3,1), C(3,3)$ and $D(-1,-3)$

Now, the endpoints of the mid-segment of $\triangle BCD$ which is parallel to $BC$ will be the mid-points of sides $BD$ and $CD$ .

Formula for the coordinate of mid-point : $(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})$ , where $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are two endpoints.

So, the mid-point of $BD$ will be: $(\frac{-3-1}{2},\frac{1-3}{2})=(\frac{-4}{2},\frac{-2}{2})=(-2,-1)$

and the mid-point of $CD$ will be: $(\frac{3-1}{2},\frac{3-3}{2})=(\frac{2}{2},\frac{0}{2})=(1,0)$

Thus, the endpoint coordinates for the mid-segment of $\triangle BCD$ that is parallel to $BC$ are $(-2,-1)$ and $(1,0)$

4 0

3 years ago

Read 2 more answers