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3 years ago

5b/12=5only genius can answer this

Mathematics

2 answers:

Dimas [21]3 years ago

7 0

$\\ \rm\longmapsto \dfrac{5b}{12}=5$

$\\ \rm\longmapsto 5b=12(5)$

$\\ \rm\longmapsto 5b=60$

$\\ \rm\longmapsto b=\dfrac{60}{5}$

$\\ \rm\longmapsto b=12$

Ludmilka [50]3 years ago

7 0

Step-by-step explanation:

$\frac{5b}{12} = 5 \\ by \: cross \: multiplicatin \\ 5b = 5 \times 12 \\ 5b = 60 \\ b = \frac{6 0 }{5} \\ b = 12$

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Contact [7]

Answers:

Equation is $(x+1)^2 + (y+2)^2 = 25$

Center is (-1, -2)

Radius = 5

===========================================

Work Shown:

$x^2+2x+y^2+4y=20\\\\(x^2+2x)+(y^2+4y)=20\\\\(x^2+2x+1)+(y^2+4y)=20+1\\\\(x^2+2x+1)+(y^2+4y+4)=20+1+4\\\\(x+1)^2 + (y+2)^2 = 25\\\\$

center = (h,k) = (-1,-2)

radius = 5

-----------------------------

Explanation:

I grouped up the x and y terms separately. Then I added 1 to both sides to complete the square for the x terms. I cut the 2 from 2x in half, then squared it to get 1. In the next step, I cut the 4 from 4y in half to get 2, which squares to 4. So that's why I added 4 to both sides to complete the square for the y terms.

Each piece is factored using the perfect squares factoring rule which is a^2+2ab+b^2 = (a+b)^2

The last equation is in the form (x-h)^2 + (y-k)^2 = r^2

We can think of x+1 as x - (-1) to show that h = -1

Similarly, y+2 = y-(-2) = y-k to show that k = -2

The center is (h,k) = (-1,-2)

The radius is r = 5 because r^2 = 5^2 = 25 is on the right hand side in the last equation above.

3 0

3 years ago

Seven times a number equals<br> 70<br> Show work

DerKrebs [107]

Answer:

7 times 10 equals 70

Step-by-step explanation:

70 divided by 7 equal 10 and 70 divided by 10 equal 7

3 0

3 years ago

1 – 7k + 3k 2 + 10k + 7 – 5k 2

Katyanochek1 [597]

1) The variables are "k"

2) There are 4 terms

3) The coefficients are k

4) The constants are -7, 3, 10, -5

5) The like terms is -7k, 3k, 10k, -5k

The answers are pretty direct but, hope this answers your question.

Have a great day/night!

8 0

3 years ago

Read 2 more answers

A random sample of 10 shipments of stick-on labels showed the following order sizes.10,520 56,910 52,454 17,902 25,914 56,607 21

sammy [17]

Answer:

Confidence Interval: (21596,46428)

Step-by-step explanation:

We are given the following data set:

10520, 56910, 52454, 17902, 25914, 56607, 21861, 25039, 25983, 46929

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{340119}{10} = 34011.9$

Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821

$S.D = \sqrt{\frac{2711418821}{9}} = 17357.09$

Confidence interval:

$\mu \pm t_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.05} = \pm 2.2621$

$34011.9 \pm 2.2621(\frac{17357.09}{\sqrt{10}} ) = 34011.9 \pm 12416.20 = (21595.7,46428.1) \approx (21596,46428)$

7 0

3 years ago

A college-entrance exam is designed so that scores are normally distributed with a mean of 500 and a standard deviation of 100.

Crank

Answer: 1.25

Step-by-step explanation:

Given: A college-entrance exam is designed so that scores are normally distributed with a mean $(\mu)$ = 500 and a standard deviation $(\sigma)$ = 100.