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Lunna [17]
3 years ago
15

What is 1+1 ok that’s it

Mathematics
1 answer:
BartSMP [9]3 years ago
3 0

Answer:

Step-by-step explanation:

2

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Maria realiza una encuesta sobre las edades de los postulantes a una universidad y con los datos obtenidos elabora una tabla de
vazorg [7]

Answer:

Step-by-step explanation:

is 10 1/2

5 0
3 years ago
Please help me with the problem
mixas84 [53]
Let's see what we're working on here 

\frac{1}{2}  - 2(m) =?
            ? = 0
Simplify this → \frac{1}{2}

\frac{1}2y} - 2(m) = ?

\frac{2(m)(2)}{2}

\frac{1 - 4(m)}{2} = ?

\frac{1 - 4(m)}{2} = 2

-4(m) + 1 = 0
 Subtract ( - )the number 1 from each of your sides on this part

4(m) = 1 
Multiply( ×) the number -1 to your sides for this part of the equation 

Therefore, the value of m is \frac{1}{4}
m =  \frac{1}{4}


3 0
4 years ago
Read 2 more answers
The sides of a rhombus with angle of 60° are 6 inches. Find the area of the rhombus.
ehidna [41]
1. Check the drawing of the rhombus ABCD in the picture attached.

2. m(CDA)=60°, and AC and BD be the diagonals and let their intersection point be O.

3. The diagonals:
     i) bisect the angles so m(ODC)=60°/2=30°

     ii) are perpendicular to each other, so m(DOC)=90°

4. In a right triangle, the length of the side opposite to the 30° angle is half of the hypothenuse, so OC=3 in.

5. By the pythagorean theorem, DO= \sqrt{ DC^{2}- OC^{2} }=  \sqrt{ 6^{2}- 3^{2} }=\sqrt{ 36- 9 }=\sqrt{ 27 }= \sqrt{9*3}= 
=\sqrt{9}* \sqrt{3} =3\sqrt{3}  (in)

6. The 4 triangles formed by the diagonal are congruent, so the area of the rhombus ABCD = 4 Area (triangle DOC)=4*\frac{DO*OC}{2}=4 \frac{3 \sqrt{3} *3}{2}==2*9 \sqrt{3}=18 \sqrt{3} (in^{2})

6 0
3 years ago
Segments DF, DH, and GF are tangent to the circle. Andrew was asked to find DF. Explain andrew's error.
ANEK [815]

Answer:

DF = 14

Step-by-step explanation:

Andrew was wrong in the assumption that DE = EF

Segments of tangents to a circle from the same external point are congruent

That is

EF = FG = 8

Then

DF = DE + EF = 6 + 8 = 14

5 0
3 years ago
write an equation for a function at has a graph with the given characteristics. the shape of y=|x| is reflected across the y-axi
Luba_88 [7]
Refrection across the y-axis changes the function to y = |-x|.
Vertical stretch by a factor of 1.5 changes the graph to y = 1.5|-x|
Shifting the graph by 4 units downward changes the graph to y = 1.5|-x| - 4

Required equation is y = 1.5|-x| - 4
6 0
3 years ago
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