Ask question

Ask question

All categories

3 years ago

9

I will give brainliest<3

2 answers:

Ne4ueva [31]3 years ago

8 0

Answer:

4100 workers

Step-by-step explanation:

lina2011 [118]3 years ago

3 0

Answer:

4100

Step-by-step explanation:

If only the mens are 10%, The total is 100.

so we multiply 410 with 10 which gives 4100.

You might be interested in

Belle and Mabel are both 8-week-old puppies of different breeds. Belle has a mass of 12

balandron [24]

Answer:

3:1

Step-by-step explanation:

Take the fraction and simplify it:

12/4 = 3/1

6 0

3 years ago

Dave spent 21 minutes completing a race. He walked 9 minutes and jogged the rest. What

nasty-shy [4]

Dave jogged for 9 minuets and walked for 12 minuets

6 0

2 years ago

Based on the information given say whether or not △ABC∼△FED. Explain your reasoning.

GalinKa [24]

Answer:

Yes, △ABC ∼ △FED by AA postulate.

Step-by-step explanation:

Given:

Two triangles ABC and FED.

m∠A = m∠B

m∠C = m∠A + 30°

m∠E = m∠F = $x$

m∠D = $2x-20$ °.

Now, let m∠A = m∠B = $y$

So, m∠C = m∠A + 30° = $y+30$

Now, sum of all interior angles of a triangle is 180°. Therefore,

m∠A + m∠B + m∠C = 180

$y+y+y+30=180\\3y=180-30\\3y=150\\y=\frac{150}{3}=50$

Therefore, m∠A = 50°, m∠B = 50° and m∠C = m∠A + 30° = 50 + 30 = 80°.

Now, consider triangle FED,

m∠D+ m∠E + m∠F = 180

$2x-20+x+x=180\\4x=180+20\\4x=200\\x=\frac{200}{4}=50$

Therefore, m∠F = 50°

m∠E = 50° and

m∠D = $2x-20=2(50)-20=100-20=80\°$

So, both the triangles have congruent corresponding angle measures.

m∠A = m∠F = 50°

m∠B = m∠E = 50°

m∠C = m∠D = 80°

Therefore, the two triangles are similar by AA postulate.

5 0

3 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0

3 years ago

Aphrodite took out a 30-year loan from her bank for $170,000 at an APR of 7.2%, compounded monthly. If her bank charges a prepay

lora16 [44]

It's 3246.74 --- APEX

7 0

3 years ago