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2 years ago

I keep coming up with different answers please help me?

Mathematics

2 answers:

Levart [38]2 years ago

8 0

Answer:

the answer is 1600

allsm [11]2 years ago

3 0

The answer is 1,600.

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Solve the systems of equations using substitution.

grandymaker [24]

Answer:

y= 8x+1 (0,1) y=6x+3 (0,1)

= 8×0+1 =6×0+3

y = 1 =3

Step-by-step explanation:

= 8×1+1 (1,9) =6×1+3 (1,9)

y=9 =9

=8×4+1 (4,14) =6×4+3 (4,14)

y=33 =27

=8×2+1(2,4) =6×2+1 (2,4)

y=17 =13

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2 years ago

A function is shown f(x)=2/3x+3 what is the value of f(12)

exis [7]

Answer:

55/18

Step-by-step explanation:

f(x) = 2/3*12 +3

= 1/18 +3

=55/18

7 0

3 years ago

If a triangle is a right triangle then the other two angles must be acute true or false ?

kolezko [41]

Answer:

The sum of the angles in any triangle is 180 degrees. If one of the three angles is 90 degrees (a right angle) the the sum of the other two angles is 90 degrees. Therefore each of the other two angles must be acute.

The statement is TRUE.

Step-by-step explanation:

8 0

3 years ago

A tree 20 feet tall has a shadow of 15 feet. How tall is Biff if his<br> shadow is 36 inches tall?

sweet-ann [11.9K]

I Think the answer is 31 because 20-5 is 15 so 36-5 is 31

3 0

3 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago