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2 years ago

How do you graph 2 3/7 on the number line ?

Mathematics

1 answer:

Karo-lina-s [1.5K]2 years ago

4 0

Answer:

You would divide the right side of the 2 into 7 parts and place a dot on the 3rd line closest to the number 2.

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Entre Juan, Pedro, María y Luis tienen 62 cartas. Pedro tiene un tercio de las cartas de Juan. María tiene 3 cartas menos que Ju

fenix001 [56]

Answer:

Luis has 30 cards

Step-by-step explanation:

To solve this problem, we need to setup a system of equations.

If we call the amount of cards Juan has "J", Pedro "P", Maria "M" and Luis "L", we have that:

J + P + M + L = 62

P = J/3

M = J - 3

L = 2J

Substituting P, M and L in the first equation, we have:

J + J/3 + J - 3 + 2J = 62

13J/3 = 62 + 3

13J = 65 * 3

J = 15 cards

The amount of cards Luis has is:

L = 2J = 2 * 15 = 30 cards

8 0

2 years ago

Balloon A is charged, and balloon C is charged. If balloon A approaches balloon C, there’ll be a force of between them.

ipn [44]

Answer:

Step-by-step explanation:im not good at long answers but it should push each other away.

5 0

3 years ago

Please help I need help with this badly asap !

evablogger [386]

Answer:

The third option

Step-by-step explanation:

It perfectly represents -24/3.

7 0

2 years ago

Using a graphing utility, find the exact solutions of the system. Round to the nearest hundredth and choose a solution to the sy

Margaret [11]

Answer:

Part 1) The exact solutions are

$(\frac{-1+\sqrt{21}} {2},4+\sqrt{21})$ and $(\frac{-1-\sqrt{21}} {2},4-\sqrt{21})$

Part 2) (1.79, 8.58)

Step-by-step explanation:

we have

$y=x^{2} +3x$ ----> equation A

$y=2x+5$ ----> equation B

we know that

When solving the system of equations by graphing, the solution of the system is the intersection points both graphs

Find the exact solutions of the system

equate equation A and equation B

$x^{2} +3x=2x+5\\x^{2} +3x-2x-5=0\\x^{2} +x-5=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2} +x-5=0$

$a=1\\b=1\\c=-5$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(1)(-5)}} {2(1)}$

$x=\frac{-1\pm\sqrt{21}} {2}$

The solutions are

$x_1=\frac{-1+\sqrt{21}} {2}$

$x_2=\frac{-1-\sqrt{21}} {2}$

Find the values of y

First solution

For $x_1=\frac{-1+\sqrt{21}} {2}$

$y=2(\frac{-1+\sqrt{21}} {2})+5$

$y=-1+\sqrt{21}+5\\\\y=4+\sqrt{21}$

The first solution is the point $(\frac{-1+\sqrt{21}} {2},4+\sqrt{21})$

Second solution

For $x_2=\frac{-1-\sqrt{21}} {2}$

$y=2(\frac{-1-\sqrt{21}} {2})+5$

$y=-1-\sqrt{21}+5\\\\y=4-\sqrt{21}$

The second solution is the point $(\frac{-1-\sqrt{21}} {2},4-\sqrt{21})$

Round to the nearest hundredth

First solution

$(\frac{-1+\sqrt{21}} {2},4+\sqrt{21})$ -----> $(1.79,8.58)$

$(\frac{-1-\sqrt{21}} {2},4-\sqrt{21})$ -----> $(-2.79,-0.58)$

see the attached figure to better understand the problem

6 0

3 years ago

7/8 + ___ = 1 what is the missing numbers

Bezzdna [24]

the missing number is 1/8 i hope this helps chu

4 0

3 years ago