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Igoryamba
2 years ago
10

Write the equation in slope-intercept form. 2/3(6y+9)=3/5(15x-20)

Mathematics
1 answer:
12345 [234]2 years ago
5 0
\frac{2}{3}(6y+9)=\frac{3}{5}(15x-20)\ \ \ | multiply\ by\ 15\\\\
10(6y+9)=9(15x-20)\\\\
60y+90=135x-180\ \ \ | subtract\ 90\\\\
60y=135x-270\ \ \ | divide\ by\ 60\\\\
Slope\ intercept\ form:\ y=2.25x-4.5
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A bakery sells pastries in boxes of 7 each. If the bakery has 27 pastries, how many boxes can be filled?​
skad [1K]

Answer:

3 with a few left over so 4

Step-by-step explanation:

4 0
2 years ago
Which set of measurements could represent the three sides of a triangle?
yawa3891 [41]

Answer:

The side lengths of a right triangle is 11cm, 60cm and 61cm, that could be selected from the given measurements.

Step-by-step explanation:

The measurements are,

                  7cm, 11cm, 54cm, 60cm, 61cm, 65cm

Step:1

                 To check the right angle triangle, Pythagorean theorem can be used.

                For a Pythagorean theorem,

                                     ..........................(1)

               The side values are lower than the hypotenuse,

                                                        ...................................(2)

               Where,

                         a,b - side values

                            c - Hypotenuse

               For right angle triangle,  c > a, b

               Alternative : 1

               Take, a = 7cm, b = 11cm

               From eqn (2),

                                                   =  = 13.04

              The above value is not equal to the any one of the values of ( 54cm. 60cm, 61cm, 65cm ), So its not an sides of right triangle.

               Alternative : 2

               Take, a = 7cm, b = 54cm

               From eqn (2),

                                                   =  = 54.45

              The above value is not equal to the any one of the values of (60cm, 61cm, 65cm ), So its not an sides of right triangle.

               Alternative : 3

               Take, a = 7cm, b = 60cm

               From eqn (2),

                                                   =  = 60.406

              The above value is not equal to the any one of the values of (61cm, 65cm ), So its not an sides of right triangle.

               Alternative : 4

               Take, a = 7cm, b = 61cm

               From eqn (2),

                                                   =  = 61.40

              The above value is not equal to the values of (65cm ), So its not an sides of right triangle.

                 Alternative : 5

               Take, a = 11cm, b = 54cm

               From eqn (2),

                                                   =  = 55.1089

              The above value is not equal to the any one of the values of (60cm, 61cm, 65cm ), So its not an sides of right triangle.      

                Alternative : 6

               Take, a = 11cm, b = 60cm

               From eqn (2),

                                                  =  = 61

              The above value is equal to the values of (61cm ), So its an sides of right triangle. The three sides are 11, 60 and 61.

Step:2

            Check for solution,

                                     

                                           

Result:

            The side lengths of a right triangle is 11cm, 60cm and 61cm, that could be selected from the given measurements.                

Step-by-step explanation: The side lengths of a right triangle is 11cm, 60cm and 61cm.

4 0
3 years ago
Calculate the area of the following composite figure.
Alexeev081 [22]
\bold{ANSWER:}
100 units^2

\bold{EXPLANATIONS:}

4 0
2 years ago
Read 2 more answers
Wally takes $40 out of his bank account every week.If he had a balance of $100 to begin with,what is his balance after 4 weeks?
vfiekz [6]

Answer:-$60

Step-by-step explanation: the answer is -$60 because multiply 40 by 4 you get $160 and you only have $100 so then you would 100-160=-60

6 0
3 years ago
Read 2 more answers
1. Consider the following hypotheses:
Andrej [43]

Answer:

See deductions below

Step-by-step explanation:

1)

a) p(y)∧q(y) for some y (Existencial instantiation to H1)

b) q(y) for some y (Simplification of a))

c) q(y) → r(y) for all y (Universal instatiation to H2)

d) r(y) for some y (Modus Ponens using b and c)

e) p(y) for some y (Simplification of a)

f) p(y)∧r(y) for some y (Conjunction of d) and e))

g) ∃x (p(x) ∧ r(x)) (Existencial generalization of f)

2)

a) ¬C(x) → ¬A(x) for all x (Universal instatiation of H1)

b) A(x) for some x (Existencial instatiation of H3)

c) ¬(¬C(x)) for some x (Modus Tollens using a and b)

d) C(x) for some x (Double negation of c)

e) A(x) → ∀y B(y) for all x (Universal instantiation of H2)

f)  ∀y B(y) (Modus ponens using b and e)

g) B(y) for all y (Universal instantiation of f)

h) B(x)∧C(x) for some x (Conjunction of g and d, selecting y=x on g)

i) ∃x (B(x) ∧ C(x)) (Existencial generalization of h)

3) We will prove that this formula leads to a contradiction.

a) ∀y (P (x, y) ↔ ¬P (y, y)) for some x (Existencial instatiation of hypothesis)

b) P (x, y) ↔ ¬P (y, y) for some x, and for all y (Universal instantiation of a)

c) P (x, x) ↔ ¬P (x, x) (Take y=x in b)

But c) is a contradiction (for example, using truth tables). Hence the formula is not satisfiable.

7 0
3 years ago
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