I decide to bake some cakes.I need 150g of flour for 12 cakes.How much flour do I need for 30 cakes?

A boat traveled 12 miles north and then

katrin [286]

Answer:

25.5 mi

Step-by-step explanation:

12 22.5 X

Let X be the shortest distance we are trying to find.

a units 2 b units 2 c units 2

12 units 2 + (22.5)=c units 2

144+506.25=c units 2

650.25=c units 2

$\sqrt{x} 650.25$ =c units 2

25.5

8 0

2 years ago

A construction worker has a cable that is 8 yards long. He wants to cut it into 9 equal-length pieces. How long will each piece

Troyanec [42]

They would be 32 inches because I did it again and got 32 inches or 0.8 yards
long and if you need any more help or this wasn't correct just tell me :D

4 0

2 years ago

Read 2 more answers

The following selected transactions were completed during August of the current year:

gizmo_the_mogwai [7]

Answer:

The journal entries as well as the T-accounts for August transactions are found in the attached spreadsheet.

Step-by-step explanation:

Initially, the transactions were posted individually in the journal and ultimately the related ones were combined together in the T-accounts.

For instance, cash related transactions are posted in the cash T-account to ascertain the true cash position as at the end of the quarter.

Download xlsx

8 0

2 years ago

Below. What is m∠1?<br><br> m∠1 =

drek231 [11]

Answer:

m∠1 = 70

Step-by-step explanation:

A line is 180 degrees

The angles of a triangle total to 180 degrees

180 - 101 = 79

79 + 31 = 110

180 - 110 = 70

5 0

2 years ago

Part B

alisha [4.7K]

Question:

Consider ΔABC, whose vertices are A (2, 1), B (3, 3), and C (1, 6); let the line segment AC represent the base of the triangle.

(a) Find the equation of the line passing through B and perpendicular to the line AC

(b) Let the point of intersection of line AC with the line you found in part A be point D. Find the coordinates of point D.

Answer:

$y = \frac{1}{5}x + \frac{12}{5}$

$D = (\frac{43}{26},\frac{71}{26})$

Step-by-step explanation:

Given

$\triangle ABC$

$A = (2,1)$

$B = (3,3)$

$C = (1,6)$

Solving (a): Line that passes through B, perpendicular to AC.

First, calculate the slope of AC

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$A = (2,1)$ --- $(x_1,y_1)$

$C = (1,6)$ --- $(x_2,y_2)$

The slope is:

$m = \frac{6- 1}{1 - 2}$

$m = \frac{5}{-1}$

$m = -5$

The slope of the line that passes through B is calculated as:

$m_2 = -\frac{1}{m}$ --- because it is perpendicular to AC.

So, we have:

$m_2 = -\frac{1}{-5}$

$m_2 = \frac{1}{5}$

The equation of the line is the calculated using:

$m_2 = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$m_2 = \frac{1}{5}$

$B = (3,3)$ --- $(x_1,y_1)$

$(x_2,y_2) = (x,y)$

So, we have:

$\frac{1}{5} = \frac{y - 3}{x - 3}$

Cross multiply

$5(y-3) = 1(x - 3)$

$5y - 15 = x - 3$

$5y = x - 3 + 15$

$5y = x +12$

Make y the subject

$y = \frac{1}{5}x + \frac{12}{5}$

Solving (b): Point of intersection between AC and $y = \frac{1}{5}x + \frac{12}{5}$

First, calculate the equation of AC using:

$y = m(x - x_1) + y_1$

Where:

$A = (2,1)$ --- $(x_1,y_1)$

$m = -5$

So:

$y=-5(x - 2) + 1$

$y=-5x + 10 + 1$

$y=-5x + 11$

So, we have:

$y=-5x + 11$ and $y = \frac{1}{5}x + \frac{12}{5}$

Equate both to solve for x

i.e.

$y = y$

$-5x + 11 = \frac{1}{5}x + \frac{12}{5}$

Collect like terms

$-5x -\frac{1}{5}x = \frac{12}{5} - 11$

Multiply through by 5

$-25x-x = 12 - 55$

Collect like terms

$-26x = -43$

Solve for x

$x = \frac{-43}{-26}$

$x = \frac{43}{26}$

Substitute $x = \frac{43}{26}$ in $y=-5x + 11$

$y = -5 * \frac{43}{26} + 11$

$y = \frac{-5 *43}{26} + 11$

Take LCM

$y = \frac{-5 *43+11 * 26}{26}$

$y = \frac{71}{26}$

Hence, the coordinates of D is:

$D = (\frac{43}{26},\frac{71}{26})$

3 0

2 years ago