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Galina-37 [17]
2 years ago
13

Please help i dont get it

Mathematics
2 answers:
natta225 [31]2 years ago
6 0

slope is rise over run so it will be 20/4 or you can reduce it to 5

horsena [70]2 years ago
4 0
I don’t get this either good luck
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A newborn giraffe weighs about 65 kilograms.How much does it erigh in grams?
-BARSIC- [3]

Answer:


Step-by-step explanation:

To find the answer convert the amount kilograms you have to grams

8 0
4 years ago
*Please I need help quickly* Approximate the mean of the frequency distribution for the ages of the residents of a town.
Cerrena [4.2K]

Answer:

The mean age of the frequency distribution for the ages of the residents of a town is 43 years.

Step-by-step explanation:

We are given with the following frequency distribution below;

 

     Age                      Frequency (f)               X                    X \times f

     0 - 9                            30                         4.5                   135

    10 - 19                           32                        14.5                  464

    20 - 29                         12                         24.5                 294

    30 - 39                         20                        34.5                 690

    40 - 49                         25                        44.5                1112.5

    50 - 59                         53                        54.5                2888.5

    60 - 69                         49                        64.5                3160.5

    70 - 79                          13                         74.5                968.5

    80 - 89                  <u>        8        </u>                  84.5         <u>        676        </u>

    Total                      <u>      242      </u>                                  <u>      10389      </u>

Now, the mean of the frequency distribution is given by the following formula;

                          Mean =  \frac{\sum X \times f}{\sum f}

                                     =  \frac{10389}{242}  = 42.9 ≈ 43 approx.

Hence, the mean age of the frequency distribution for the ages of the residents of a town is 43 years.

5 0
3 years ago
Read 2 more answers
Question in the image below:
horrorfan [7]

Answer:

x = 5/33 or 0.151515151

Step-by-step explanation:

7(-6x-2) = 8(3x-4)

step 1: simplify the equation.

-42x - 14 = 24x - 4

step 2: isolate the variable (using the balance method).

-42x - 14 + 14 = 24x - 4 + 14

-42x = 24x + 10

-42x - 24x = 24x - 24x + 10

-66x = 10

step 3: solve for x.

x = 10 ÷ -66

x = 5/33

4 0
3 years ago
Y= x^2 -5 solve for x
dexar [7]

Y= x^2 -5

We need to solve for x, we need to get x alone

Y= x^2 -5

Lets start by removing -5

Add 5 on both sides

y + 5= x^2 -5 + 5

y + 5= x^2

Now to isolate x , we need to remove the square from x

To remove square , take square root on both sides

+-\sqrt{y+5} = \sqrt{x^2}

square and square root will get cancelled

+-\sqrt{y+5} = x

So x = +\sqrt{y+5} and  x = -\sqrt{y+5}


7 0
3 years ago
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
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