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WINSTONCH [101]
3 years ago
11

What's 15⁄45 written as a fraction in simplest form?

Mathematics
1 answer:
vekshin13 years ago
5 0
A: 15/45 = 3/5 in simplest form.
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Solve the equation 5*e^x=15.76
liq [111]
Divide both sides by 5

e^x=3.152  take the natural log of both sides...

x=ln(3.152)

x≈1.148
3 0
3 years ago
Write an equation in standard form of the line passing through the points (3,3) and (-3,5)
Korvikt [17]

Answer:

The equation of line AB  with points (3,3) and (-3,5) is given as

: x + 3y = 12

Step-by-step explanation:

Here, the given points are A (3, 3) and B (-3,5).

Now, slope of any line is given as :

m = \frac{y_2 - y_1}{x_2 - x_1}

or, m = \frac{5-3}{-3 - 3}   = \frac{2}{-6}  = -\frac{1}{3}

Hence, the slope of the line AB is (-1/3)

Now , A POINT SLOPE FORM of an equation is

(y - y0)  = m (x - x0) ; (x0, y0)  is any arbitrary point on line.

So, for the point (3,3) the equation of the line is

y - 3y-3 = -\frac{1}{3} (x-3)   \implies 3y - 9 = 3 -x

Hence, the equation of line AB  with points (3,3) and (-3,5) is given as:

x + 3y = 12

8 0
3 years ago
I need quick help pls!!<br><br> by reason: AAA or ASA, SAA or Cannot be determined or SAS or SSS
Ray Of Light [21]

Answer:

  ΔDCE by ASA

Step-by-step explanation:

The marks on the diagram show AE ≅ DE. We know vertical angles AEB and DEC are congruent, and we know alternate interior angles BAE and CDE are congruent. The congruent angles we have identified are on either end of the congruent segment, so the ASA theorem applies.

Matching corresponding vertices, we can declare ΔABE ≅ ΔDCE.

5 0
3 years ago
X-3&lt;-12 Choose the correct graph for the inequality?
olga55 [171]
The correct graph is the last one
x -3<-12
x<-12+3
x<-9
7 0
2 years ago
HELP PLEASE!
olasank [31]

Answer        

Find out the Area of a triangle .

To proof  

Formula

Area of Triangle

= \frac{1}{2}\times( {x_1(y_{2}-y_{3}) +x_{2}(y_{3} - y_{1})+x_{3}(y_{1}-y_{2})})

Now vertices are D(3, 3) , E(3, −1) , and F(−2, −5) .

= \frac{1}{2} (3\times(-1 +5) + 3\times(-5-3)-2\times(3+1))

Solving the above

= \frac{1}{2}(3\times4+3\times-8 -2\times4)

=\frac{1}{2} (12-24-8)\\ =\frac{1}{2} (-32+12)\\=\frac{1}{2} (-20)

= -10 units²

(Neglected the negative sign as area cannot be negative.)

= 10 units ²

Area of  a triangle is 10 units ²

Hence proved

6 0
3 years ago
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