All categories

2 years ago

Help asp, I can’t find the answers:P

Mathematics

1 answer:

zepelin [54]2 years ago

8 0

Answer:

the answer will be B

Step-by-step explanation:

It does increase

You might be interested in

The number of ducks and pigs in a field totals 38. The total number of legs among them is 94 assuming each duck has exactly two

Reika [66]

Answer:

22 pigs and 3 ducks

Step-by-step explanation:

22 pigs would make 88 legs and 3 ducks will make 6 legs so u add that together and you get 94 legs (smile) ik i'm smart lol

8 0

3 years ago

Two angles of a triangle are 30°and 80°. Find the third angle.

stiv31 [10]

Answer:

The third angle of the triangle is 70°.

Step-by-step explanation:

The two angles of a triangle (say Δ ABC) are given to be 30° and 80°.

We have to find the third angle.

Let us assume that ∠ A = 30° and ∠ B = 80°, then we have to find ∠ C.

Now, we know that, ∠ A + ∠ B + ∠ C = 180° {Property of a triangle}

⇒ 30° + 80° + ∠ C = 180°

⇒ ∠ C = 180° - 30° - 80° = 70°

Therefore, the third angle of the triangle is 70°. (Answer)

7 0

3 years ago

Please help asap!!!

sp2606 [1]

Answer:

$y=\frac{1}{6} (x+5)^2-4.5$

$y=\frac{-1}{20} (x-10)^2+1$

Step-by-step explanation:

focus at (-5, -3), and directrix y = -6

Directrix y=-6 so its a vertical parabola

so equation is

(x-h)^2 = 4p(y-k)

(h,k) is the center

P is the distance between focus and vertex

distance between focus and directrix = 2p

distance between -3 and y=-6 is 3

2p = 3

p = 3/2 or p = 1.5

Focus is (h, k+p)

given focus is (-5, -3) so h= -5 and k+p = -3

k+p=-3, plug in 1.5 for p

k + 1.5 = -3

subtract 1.5 on both sides

k = -4.5

(x-h)^2 = 4p(y-k)

$(x+5)^2= 4(1.5) (y+4.5)$

$(x+5)^2= 6(y+4.5)$

divide by 6 on both sides

then subtract 4.5 on both sides

$y=\frac{1}{6} (x+5)^2-4.5$

focus at (10, -4), and directrix y = 6.

Directrix y=6 so its a vertical parabola

so equation is

(x-h)^2 = 4p(y-k)

distance between focus and directrix = 2p

distance between -4 and y=6 is -4-6=-10

2p = -10

p = -5

Focus is (h, k+p)

given focus is (10, -4) so h= 10 and k+p = -4

k+p=-4, plug in 5 for p

k - 5 = -4

add 5 on both sides

k = 1

(x-h)^2 = 4p(y-k)

$(x-10)^2= 4(-5) (y-1)$

$(x-10)^2= -20(y-1)$

divide by -20 on both sides and add 1 on both sides

$y=\frac{-1}{20} (x-10)^2+1$

8 0

3 years ago

My question is in the attachment.<br>

Nikolay [14]

Answer:

5 and 25

Step-by-step explanation:

let the son's age be x then the father's age is x + 20

In 5 years

son = x + 5 and father = x + 20 + 5 = x + 25

Then

x + 25 = 3(x + 5) ← father is three times as old as son

x + 25 = 3x + 15 ( subtract x from both sides )

25 = 2x + 15 ( subtract 15 from both sides )

10 = 2x ( divide both sides by 2 )

5 = x and x + 20 = 5 + 20 = 25

si=on is 5 and father is 25

8 0

2 years ago

Read 2 more answers

Solve the system of equations by row-reduction. At each step, show clearly the symbol of the linear combinations that allow you

adell [148]

Answer:

1) The solution of the system is

$\left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right$

2) The solution of the system is

$\left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right$

Step-by-step explanation:

1) To solve the system of equations

$\left\begin{array}{ccccccc}&3x_2&-5x_3&=&89\\6x_1&&+x_3&=&17\\x_1&-x_2&+8x_3&=&-107\end{array}\right$

using the row reduction method you must:

Step 1: Write the augmented matrix of the system

$\left[ \begin{array}{ccc|c} 0 & 3 & -5 & 89 \\\\ 6 & 0 & 1 & 17 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 2: Swap rows 1 and 2

$\left[ \begin{array}{ccc|c} 6 & 0 & 1 & 17 \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 3: $\left(R_1=\frac{R_1}{6}\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 4: $\left(R_3=R_3-R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]$

Step 5: $\left(R_2=\frac{R_2}{3}\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]$

Step 6: $\left(R_3=R_3+R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & \frac{37}{6} & - \frac{481}{6} \end{array} \right]$

Step 7: $\left(R_3=\left(\frac{6}{37}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 8: $\left(R_1=R_1-\left(\frac{1}{6}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 9: $\left(R_2=R_2+\left(\frac{5}{3}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 10: Rewrite the system using the row reduced matrix:

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right] \rightarrow \left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right$

2) To solve the system of equations

$\left\begin{array}{ccccccc}4x_1&-x_2&+3x_3&=&12\\2x_1&&+9x_3&=&-5\\x_1&+4x_2&+6x_3&=&-32\end{array}\right$

using the row reduction method you must:

Step 1:

$\left[ \begin{array}{ccc|c} 4 & -1 & 3 & 12 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 2: $\left(R_1=\frac{R_1}{4}\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 3: $\left(R_2=R_2-\left(2\right)R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 4: $\left(R_3=R_3-R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 5: $\left(R_2=\left(2\right)R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 6: $\left(R_1=R_1+\left(\frac{1}{4}\right)R_2\right)$

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 7: $\left(R_3=R_3-\left(\frac{17}{4}\right)R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & - \frac{117}{2} & \frac{117}{2} \end{array} \right]$

Step 8: $\left(R_3=\left(- \frac{2}{117}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 9: $\left(R_1=R_1-\left(\frac{9}{2}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 10: $\left(R_2=R_2-\left(15\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 11:

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]\rightarrow \left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right$

8 0

3 years ago