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3 years ago

¿Cual es el número cuya tercera parte más 12 da 26?OK

Mathematics

1 answer:

iren2701 [21]3 years ago

7 0

Answer:

Step-by-step explanation:

x/3+12 = 26

Multiply everything by 3 to eliminate the fraction

x+36=78

subtract 36 on both sides

x=42

42/3=14

14+12=26

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I need help with my algebra 2. I also want to know how to solve this.

Lisa [10]

Add 1 to both sides:

$\sqrt{x+3} = x+1$

In cases like this, we have to remember that a root is always positive, so we can square both sides only assuming that

$x+1 \geq 0 \iff x \geq -1$

Under this assumption, we square both sides and we have

$x+3 = (x+1)^2 \iff x+3 = x^2+2x+1 \iff x^2+x-2 = 0$

The solutions to this equation are

$x = -2,\ x=1$

But since we can only accept solutions greater than -1, we discard $x=-2$ and accept $x=1$ .

In fact, we have

$x=-2 \implies \sqrt{-2+3}-1=0\neq -2$

and

$x=1 \implies \sqrt{4}-1=1$

which is the only solution.

4 0

3 years ago

Write the sum of 5x2 + 2x - 10 and 2x2 + 6 as a polynomial in standard form.

vaieri [72.5K]

Answer:

7x^2+2x-4

Step-by-step explanation:

a standard form is ax+by=c

5x^2 add 2x^2=7x^2 then -2x but when you have a -10 +6 you get -4 giving you the answer of 7x^2 +2x -4

7 0

3 years ago

Read 2 more answers

Find two numbers whos diffrence is 16 and the product is 720

antiseptic1488 [7]

Answer:

the numbers who's difference is 16 and the product is 720 is

36 &20

Explanation:

to finding numbers you have to express the equation in a quadratic equation then Factor it.

since the two numbers have a difference of 16 you can say that the first number is x and the second number is x -16 then their product is 720.

x(x-16)=720

x²-16x=720

x²-16x-720=0

(x-36)(x+20)

there are two possible value of x but since you are looking for the positive value of x, choose the positive value of x the first number is 20.

To find the second number

x-16=20

x=20+16

x=36

the second number is 36

Now to check

36-20=16

20*36=720

8 0

3 years ago

PLEASE HELP! The table shows the number of championships won by the baseball and softball leagues of three youth baseball divisi

Irina18 [472]

Answer:

Question 1: $P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16}}{ \frac{4}{16}} = \frac{1}{2}$

Question 2:

A. $P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

B. $P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

C. $P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}$

Step-by-step explanation:

Conditional probability is defined by

$P(A|B)= \frac{P(A and B)}{P(B)}$

with P(A and B) beeing the probability of both events occurring simultaneously.

Question 1:

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16} }{ \frac{4}{16} } = \frac{1}{2}$

Question 2.A:

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

Question 2.B:

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( Z and B)= \frac{ 1 }{ 16 }[/tex]

By definition,

$P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

Question 3.B

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

then

$P (Y or Z) = P(Y) + P(Z) = \frac{6}{16}$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

$P((YorZ) and B)= \frac{3}{16}$

By definition,

$P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}$

3 0

3 years ago

which best explains whether a triangle with side lengths 2 in., 5 in., and 4 in. is an acute triangle?

sdas [7]

Answer:

Obtuse triangle.

Step-by-step explanation:

Side a = 5

Side b = 2

Side c = 4

Angle ∠A = 108.21° = 108°12'36" = 1.88862 rad

Angle ∠B = 22.332° = 22°19'54" = 0.38976 rad

Angle ∠C = 49.458° = 49°27'30" = 0.86321 rad

Since one angle is above 90 degrees, the triangle is obtuse triangle.

This cannot be an acute triangle.

6 0

3 years ago

¿Cual es el número cuya tercera parte más 12 da 26?OK​

¿Cual es el número cuya tercera parte más 12 da 26?OK