How to make a scatter plot

R2-44r+120= -30 + 11r Solve by factoring

Sunny_sXe [5.5K]

Answer:

r = 152/55

Step-by-step explanation:

122 - 44r = - 30 + 11r

122 - 44r - 11r = - 30

- 44r - 11r = - 30 - 122

- 55r = - 30 - 122

- 55r = - 152

4 0

3 years ago

Please help please help

katrin [286]

Answer:

101, 201, 301, 401, 501

Step-by-step explanation:

These are the first five multiples of 101

:D

5 0

3 years ago

Approximate the value of 11−−√ to the nearest tenths place. Plot the approximation on the number line.

posledela

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one.

Approximate the value of $\sqrt{11}$ to the nearest tenths place. Plot the approximation on the number line.

(Please have a look at the attached photo)

My answer:

The value nearest tenths place of $\sqrt{11}$ is: 3.3

Hope it will find you well.

5 0

3 years ago

Assume y≠60 which expression is equivalent to (7sqrtx2)/(5sqrty3)

Drupady [299]

Answer:

The equivalent will be:

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$

Therefore, option 'a' is true.

Step-by-step explanation:

Given the expression

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}$

Let us solve the expression step by step to get the equivalent

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}$

as

$\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}$ ∵ $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=x^{2\cdot \frac{1}{7}}$

$=x^{\frac{2}{7}}$

also

$\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}$ ∵ $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=y^{3\cdot \frac{1}{5}}$

$=y^{\frac{3}{5}}$

so the expression becomes

$\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

$=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$ ∵ $\:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}$

Thus, the equivalent will be:

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$

Therefore, option 'a' is true.

5 0

3 years ago

Suma a 3 numere naturale este 163. Scazand al treilea numar din suma primelor doua numere, obtinem 67. Scazand primul numar din

Vika [28.1K]

Answer: x=63, y=52, z=48

Step-by-step explanation:

The complete question in english is:

The sum of 3 natural numbers is 163. By subtracting the third number from the sum of the first two numbers, we get 67. By subtracting the first number from the sum of the last 2 numbers, we get 37. Determine the 3 numbers

According to this we have a system with three equations and three unknown values:

$x+y+z=163$ (1)

$x+y-z=67$ (2)

$-x+y+z=37$ (3)

Let's begin by isolating $x$ from (1):

$x=163-y-z$ (4)

And substitute (4) in (2):

$163-y-z+y-z=67$ (5)

Isolating $z$ :

$z=48$ (6) We have found the first number

Substituting (6) in (1):

$x+y+48=163$ (7)

Adding (7) to (3):

$\left \{ {{x+y+48=163} \atop {-x+y+z=37}} \right$

We have the following:

$2y+96=200$

Isolating $y$ :

$y=52$ (8) This is the second number

Substituting (6) and (8) in (1):

$x+52+48=163$ (9)

Isolating $x$ :

$x=63$ (10) This is the third number

8 0

4 years ago