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frosja888 [35]
2 years ago
5

Please help ASAP please

Mathematics
1 answer:
valkas [14]2 years ago
6 0

Orange is a reflection of green because they have the same points. The difference is one has all positive and the other has one negative.

Not actual points just an example:

Orange (5,5), (5,5), (5,5)

Green (-5,5), (-5,5), (-5,5)

They are the same shape but Orange is a reflection of green

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Can somebody please help me with this math question please
kondaur [170]

Answer:

Step-by-step explanation:

isosceles triangles are ones that have two equal sides and angles. this would mean that ab is equal to bc. then you substitute the values in for AB=BC, 3x-4=5x-10. you bring the variables to one side and you flip everything which is then 2x=6 and then divide by 2. this would be x=3. then you have to figure out the AB. 3(3)-4=5. then you can always check for BC, 5(3)-10=5. and they are equal! hope i helped!

4 0
2 years ago
Hii please help i’ll give brainliest
brilliants [131]

Answer:

0.5 or 1/2

Step-by-step explanation

I dont know if they want the answer in fraction form.

6 0
2 years ago
Two sides of a triangle
True [87]

step 1: open brackets n rmb to flip the signs

step 2 (optional): rearrange to make the equation clearer to add/minus

length of 3rd side:

(8x + 3) - (2x + 1) -(5x + 2)

= 8x + 3 - 2x -1 - 5x -2

= 8x -2x -5x +3 -1 -2

= x

Therefore, length of 3rd side is X cm.

---

If perimeter = 75cm,

8x +3 = 75

8x = 75-3

8x = 72

x = 9 cm

Therefore,

1st side = 2(9) +1

= 19 cm

2nd side = 5(9) +2

= 47 cm

3rd side = 9 cm

4 0
3 years ago
In right ABC, AN is the altitude to the hypotenuse. FindBN, AN, and AC,if AB =2 5 in, and NC= 1 in.
Rama09 [41]

From the statement of the problem, we have:

• a right triangle △ABC,

,

• the altitude to the hypotenuse is denoted AN,

,

• AB = 2√5 in,

,

• NC = 1 in.

Using the data above, we draw the following diagram:

We must compute BN, AN and AC.

To solve this problem, we will use Pitagoras Theorem, which states that:

h^2=a^2+b^2\text{.}

Where h is the hypotenuse, a and b the sides of a right triangle.

(I) From the picture, we see that we have two sub right triangles:

1) △ANC with sides:

• h = AC,

,

• a = ,NC = 1,,

,

• b = NA.

2) △ANB with sides:

• h = ,AB = 2√5,,

,

• a = BN,

,

• b = NA,

Replacing the data of the triangles in Pitagoras, Theorem, we get the following equations:

\begin{cases}AC^2=1^2+NA^2, \\ (2\sqrt[]{5})^2=BN^2+NA^2\text{.}\end{cases}\Rightarrow\begin{cases}NA^2=AC^2-1, \\ NA^2=20-BN^2\text{.}\end{cases}

Equalling the last two equations, we have:

\begin{gathered} AC^2-1=20-BN^2.^{} \\ AC^2=21-BN^2\text{.} \end{gathered}

(II) To find the values of AC and BN we need another equation. We find that equation applying the Pigatoras Theorem to the sides of the bigger right triangle:

3) △ABC has sides:

• h = BC = ,BN + 1,,

,

• a = AC,

,

• b = ,AB = 2√5,,

Replacing these data in Pitagoras Theorem, we have:

\begin{gathered} \mleft(BN+1\mright)^2=(2\sqrt[]{5})^2+AC^2 \\ (BN+1)^2=20+AC^2, \\ AC^2=(BN+1)^2-20. \end{gathered}

Equalling the last equation to the one from (I), we have:

\begin{gathered} 21-BN^2=(BN+1)^2-20, \\ 21-BN^2=BN^2+2BN+1-20 \\ 2BN^2+2BN-40=0, \\ BN^2+BN-20=0. \end{gathered}

(III) Solving for BN the last quadratic equation, we get two values:

\begin{gathered} BN=4, \\ BN=-5. \end{gathered}

Because BN is a length, we must discard the negative value. So we have:

BN=4.

Replacing this value in the equation for AC, we get:

\begin{gathered} AC^2=21-4^2, \\ AC^2=5, \\ AC=\sqrt[]{5}. \end{gathered}

Finally, replacing the value of AC in the equation of NA, we get:

\begin{gathered} NA^2=(\sqrt[]{5})^2-1, \\ NA^2=5-1, \\ NA=\sqrt[]{4}, \\ AN=NA=2. \end{gathered}

Answers

The lengths of the sides are:

• BN = 4 in,

,

• AN = 2 in,

,

• AC = √5 in.

7 0
1 year ago
What additional information is needed to prove the triangles are congruent by the SAS Postulate?
Kitty [74]
You need to know the length of two sides and the degree of one angle.
7 0
3 years ago
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