Use log b 2=0.693 and /or log b 7 = 1.946
to find log b 14

Answers

Answer 1

Answer: $We\ know:log_a(b\cdot c)=log_ab+log_ac\n\ntherefore:\n\nlog_b14=log_b(2\cdot7)=log_b2+log_b7=0.693+1.946=2.639$

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A certain solution contains 4.3 grams of salt per liter.How much salt is there in 11 liters of the solution?
Round your answer, if necessary, to the nearest tenth.

Answers

Answer:

47.3 grams

Step-by-step explanation:

Given: A solution contains 4.3 grams of salt per liter.

To find: Amount of salt in 11 liters of the solution

Solution:

Amount of salt in 1 liter of the solution = 4.3 grams

So, amount of salt in 11 liters of the solution = 11 × Amount of salt in 1 liter of the solution = 11 × 4.3 = 47.3 grams

Hence, 11 liters of the solution contains 47.3 grams of salt.

Answer: 47.3

Step-by-step explanation: Khan Academy

)Line CP is a perpendicular bisector to line segment AB. If

Answers

Answer:

Step-by-step explanation:

Line CP is a perpendicular bisector to line segment AB

two congruent lines are formed when AB is intersected ( bisector forms two equal in measure and congruent parts)

and

a right angle is formed between the line CP and AB ( perpendicular lines form right angles.)

Marlin's marble jar had 36 red marbles and 64 blue marbles. What was the ratio of red marbles to blue marbles in the jar?

A.
1 : 2

B.
9 : 16

C.
2 : 3

D.
3 : 4

Answers

$36:64=9:16$

36:64 because 36/4=9 and 64/4 = 16

9:16

7x-4x+12=36-5y please show steps

Answers

7x-4x+12=36-5y
move x and y to one side.
7x-4x+5y=36-12
3x+5y=24
HOPE I HELPED!!!

Check the attachment below.

This is all about matrices, I don't really get them. Any help is great.

Answers

See Explanation

Step-by-step explanation:

$\begin{bmatrix} - 3 & 5 & 6\n-3 & 0 & - 3\end{bmatrix} -\begin{bmatrix} 6 & - 1& 1\n0& - 4& - 5\end{bmatrix} \n\n=\begin{bmatrix} - 3-6 & 5-(-1) & 6-1\n-3-0 & 0-(-4) & - 3-(-5)\end{bmatrix} \n\n=\begin{bmatrix} - 3-6 & 5+1 & 6-1\n-3-0 & 0+4 & - 3+5\end{bmatrix} \n\n=\begin{bmatrix} - 9 & 6 & 5\n-3 & 4 & 2\end{bmatrix} \n\n$

$\begin{bmatrix} 4 & - 6 \n-3 & - 6\n2 & 6 \end{bmatrix} + \begin{bmatrix} 5 & 3 \n-5 & 5\n 4 & - 5\end{bmatrix} \n\n= \begin{bmatrix} 4 +5& - 6+3 \n-3+(-5) & - 6+5\n2+4 & 6+(-5) \end{bmatrix} \n\n= \begin{bmatrix} 4 +5& - 6+3 \n-3-5 & - 6+5\n2+4 & 6-5 \end{bmatrix} \n\n= \begin{bmatrix} 9& - 3 \n-8 & - 1\n6 & 1\end{bmatrix} \n\n$