Define intersecting lines

Find the midpoint of A and B where A has coordinates (2,9)

timofeeve [1]

Answer:

$(\dfrac{9}{2}, \dfrac{13}{2})$

Step-by-step explanation:

The coordinates of A = (2,9)

The coordinates of B = (7,4)

Let M (x,y) be the mid point of A and B.

$M(x,y)=(\dfrac{x_1+x_2}{2}),(\dfrac{y_1+y_2}{2})\\\\=\dfrac{2+7}{2}, \dfrac{9+4}{2}\\\\=(\dfrac{9}{2}, \dfrac{13}{2})$

So, the coordinates of mid point of A and B is $(\dfrac{9}{2}, \dfrac{13}{2})$ .

7 0

3 years ago

The critical value for a two-tailed test of h0: β1 = 0 at α = .05 in a simple regression with 22 observations is:

Tresset [83]

At 1.05 skdndbdbxbxbx

4 0

3 years ago

IR = V (solve for R) *

VashaNatasha [74]

Answer:

R = $\frac{V}{I}$

Step-by-step explanation:

IR = V ( isolate R by dividing both sides by I )

R = $\frac{V}{I}$

6 0

2 years ago

Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on

boyakko [2]

Answer:

Mathew invested $600 and $2400 in each account.

Solution:

From question, the total amount invested by Mathew is $3000. Let p = $3000.

Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’

So, the amount invested in second account = 3000 – P

Step 1:

Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest $(I_1)$ earned in first account for one year,

$\text {simple interest}=\frac{\text {pnr}}{100}$

Where

p = amount invested in first account

n = number of years

r = rate of interest

hence, by using above equation we get $(I_1)$ as,

$I_{1}=\frac{P \times 1 \times 3}{100}$ ----- eqn 1

Step 2:

Mathew has paid 8% interest in second account. Let us calculate the simple interest $(I_2)$ earned in second account,

$I_{2} = \frac{(3000-P) \times 1 \times 8}{100} \text { ------ eqn } 2$

Step 3:

Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)

$I = \frac{3000 \times 1 \times 4}{100} ----- eqn 3$

Step 4:

Total simple interest = simple interest on first account + simple interest on second account.

Hence we get,

$I = I_1+ I_2 ---- eqn 4$

By substituting eqn 1 , 2, 3 in eqn 4

$\frac{3000 \times 1 \times 4}{100} = \frac{P \times 1 \times 3}{100} + \frac{(3000-P) \times 1 \times 8}{100}$

$\frac{12000}{100} = \frac{3 P}{100} + \frac{(24000-8 P)}{100}$

12000=3P + 24000 - 8P

5P = 12000

P = 2400

Thus, the value of the variable ‘P’ is 2400

Hence, the amount invested in first account = p = 2400

The amount invested in second account = 3000 – p = 3000 – 2400 = 600

Hence, Mathew invested $600 and $2400 in each account.

3 0

3 years ago

Read 2 more answers

The scale of a map is 1cm:12 km. find the actual distance for 12cm

anzhelika [568]

This is simply a question about ratios. This ratio means that for every 1 cm on the map, there are 12 km in actuality. You can set this problem up in fractions:

1 cm 12 cm
-------- = ----------
12 km x

To solve for x, we cross-multiply, for the moment ignoring units. Remember, your answer will be in km.

1*x=12*12

x=144

If x=144, then for 12 cm, there are 144 km.

6 0

4 years ago