Simplify./ Simplificar. 6(-2a + 4 + 3a)*

What’s the angle of cda

Anastaziya [24]

Answer:

∠ CDA = 32.3°

Step-by-step explanation:

See the given figure attached to this answer.

Draw perpendiculars from B and C on AD which is BE and CF.

Now, Δ ABE is a right triangle and AE² = AB² - BE² = 7.5² - 6² = 20.25

⇒ AE = 4.5 cm

Now, DF = AD - EF - AE = 24 - 10 - 4.5 = 9.5 cm {Since BC = EF}

And CF = BE = 6 cm.

Now, Δ CFD is a right triangle and

$\tan \beta =\frac{CF}{FD} = \frac{6}{9.5} = 0.6315$ {Where β = ∠ CDA}

⇒ $\beta = \tan^{-1} (0.6315)=32.27$ Degree.

So, ∠ CDA = 32.3° {Correct to one decimal place} (Answer)

8 0

3 years ago

Kevin Horn is the national sales manager for National Textbooks Inc. He has a sales staff of 40 who visit college professors all

Oduvanchick [21]

Answer:

Following are the solution to the given points:

Step-by-step explanation:

Given values:

$38, 40, 41, 45, 48, 48, 50, 50, 51, 51, 52, 52, 53, \\\\$ $54, 55, 55, 55, 56, 56, 57, 59, 59, 59, 62, 62, 62, \\ 63, 64, 65,66, 66, 67, 67, 69, 69, 71, 77, 78, 79, 79$

Total staff=40

In point a:

To calculate the median number first we arrange the value into ascending order and then collect the even numbers of calls that were also made. Its average of the middle terms is thus the median.

The midterms =55 and 59 so,

Median = $\frac{55+59}{2} = \frac{114}{2}=57$

In point b:

First-quarter $Q_1 = \frac{1}{4} \text{number of 8th call}$

$=\frac{1}{4} \times 30 th \\\\= 7.5 th$

The first quarterlies are $7.5th \ \ that \ is = (7+0.5)th\ \ term$

therefore the multiply of 0.5 by calculating the difference of the 7th and 8th term are:

$=0.5 \times 0= 0 \\\\\to Q_1 = 50+0=50 \\\\\to Q_3=\frac{3}{4} \text{number of 8th call} \\\\=\frac{3}{4} \times 30 th \\\\=22.5 \ th = (22+0.5)th \ \ term$

therefore the it is multiply by the 0.5 for the difference of the 22nd and 23rd term:

$= 0.5 \times 1=0.5 \\\\\to Q_3= 63+0.5=63.5$

In point c:

First decile $D_1 = \frac{1}{18} \text{number of 8th call}$

$= \frac{1}{10} \times 30 th \\\\= 3rd \ \ term\\\\\to D_1= 41 \\\\\to D_9= \frac{9}{10} \times 8\ th\ call\\\\=\frac{9}{10} \times 30 th \\\\=27th \ \ term\\\\\to D9 = 67$

In point d:

quartiles are:

$Q_1= 51.25 \\\\Q_3=66$

The right answers for the decile are:

$D_1=45.3 \\\\ D_9=76.4$

As for $D) P_{33} = 53.53$ will be available.

7 0

2 years ago

Pls help me hurry I need help!!

Citrus2011 [14]

C. 9 square roots of 2

To answer this question, it's super important that you understand the ratio of sides for special triangles. This triangle in particular, a 45-45-90 triangle, has a ratio between the legs and hypotenuse of 1:1:

Since we are given the value of the hypotenuse, we know that the value of the two sides multiplied by $\sqrt{2}$ will be 18. Knowing this, we can write out an equation:

u* $\sqrt{2}$ = 18

u = $\frac{18}{\sqrt{2} }$

<u>Multiply both sides by </u> $\sqrt{2}$ <u> in order to get rid of the root in the denominator:</u>

u = $\frac{18*\sqrt{2}}{\sqrt{2} * \sqrt{2}}$

u = $\frac{18*\sqrt{2} }{2}$

u = 9 $\sqrt{2}$

If you'd like me to explain how I got to the answer any further, just ask!

- breezyツ

7 0

2 years ago

What is 13 times 13 my little brother wants to know I told him let you guys help!!

Oksi-84 [34.3K]

Answer:

13 times 13 is 169

Step-by-step explanation:

I did it by doing it in my head

:)

5 0

3 years ago

Read 2 more answers

Tutorial Exercise A boat leaves a dock at 2:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due

ANTONII [103]

Answer:

Both the boats will closet together at 2:21:36 pm.

Step-by-step explanation:

Given that - At 2 pm boat 1 leaves dock and heads south and boat 2 heads east towards the dock. Assume the dock is at origin (0,0).

Speed of boat 1 is 20 km/h so the position of boat 1 at any time (0,-20t),

Formula : d=v*t

at 2 pm boat 2 was 15 km due west of the dock because it took the boat 1 hour to reach there at 15 km/h, so the position of boat 2 at that time was (-15,0)

the position of boat 2 is changing towards east, so the position of boat 2 at any time (-15+15t,0)

Formula : D= $\sqrt{(x2-x1)^2+(y2-y1)^2}$