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iVinArrow [24]
3 years ago
9

What is the sum of -1/2 + 2/3?

Mathematics
2 answers:
alekssr [168]3 years ago
6 0
0.16666666666 is the sum of your question






satela [25.4K]3 years ago
6 0

the answer is 1/6 because you have to convert them to have a common denominator, so you would have -3/6+ 4/6

You might be interested in
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
Explain why you can subtract exponents when you are dividing two powers with the same base.
masya89 [10]

When you divide numbers, you make the value smaller.  So subtract the exponents will make the value with that base smaller.


3 0
3 years ago
Read 2 more answers
"Gina wrote the following paragraph to prove that the segment joining the midpoints of two sides of a triangle is parallel to th
Aleonysh [2.5K]
What is the flaw in Gina’s proof?
A) Points D and E must be constructed, not simply labeled, as midpoints.

B) Segments DE and AC are parallel by construction.  THIS IS THE FLAW. THE SEGMENTS WERE NOT DRAWN NOR PROPERLY IDENTIFIED.

C) The slope of segments DE and AC is not 0.

D) The coordinates of D and E were found using the Distance between Two Points Postulate 
6 0
3 years ago
Read 2 more answers
1.What is the equation of the line perpendicular to  that passes through ? Write your answer in slope-intercept form. Show your
Juliette [100K]

Answer:

1. Use a compass to make arc marks which intersect above and below then connect.

2. y=\frac{1}{3}x + 2

Step-by-step explanation:

1. To construct a perpendicular line, use a compass to draw arc marks from one end of the segment through point P. Then repeat this again at the other end. This means at point P there will be two intersecting arc marks. Repeat the process down below with the same radius as used above. Then connect the two intersections.

2. The point slope form of a line is (y-y_1)=m(x-x_1) where x_1=-3\\y_1=1. We write  

(y-1)=m(x--3)\\(y-1)=m(x+3)  

Since the line is to be perpendicular to the line shown it will have the negative reciprocal to the slope of the function 3x+y =-8. To find m, rearrange the function to be y=-8-3x. The slope is -3 and the negative reciprocal will be 1/3.

(y-1)=\frac{1}{3}(x+3)  

Simplify for slope intercept form.

(y-1)=\frac{1}{3}(x+3)\\(y-1)=\frac{1}{3}x+1\\y=\frac{1}{3}x + 2


3 0
3 years ago
Can someone help me out with this ??
nalin [4]

Answer:

1. b > -2

2. x <= 3

Step-by-step explanation:

Question 1:

-2(b + 5) < -6

Divide both sides by -2. Remember to change the inequality sign.

b + 5 > 3

Subtract 5 from both sides.

b > -2

Question 2:

-(x - 10) >= 7

Divide both sides by -1. Remember to change the inequality sign.

x - 10 <= -7

Add 10 to both sides.

x <= 3

6 0
3 years ago
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