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quester [9]
3 years ago
9

Need to answer in equation

Mathematics
2 answers:
worty [1.4K]3 years ago
5 0

Answer: A. is correct

Step-by-step explanation:

Thepotemich [5.8K]3 years ago
4 0

Answer:

A

Step-by-step explanation:

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PLEASE SHOW FULL SOLUTIONS!!! WILL MARK BRAINLIEST!!! SHOW ALL STEPS!!! THANK YOU!!!
ELEN [110]

9514 1404 393

Answer:

  a) BE = 5; DE = 6; EF = 4

  b) ∠EFC ≅ ∠DA.F ≅ ∠BDE  or  <em>b = e = f = i</em>

Step-by-step explanation:

Each short segment is the same length as the marked one it is parallel to.

  E is the midpoint of BC, so BE = EC = 5.

  ADEF is a parallelogram, so DE = A.F = 6.

  D is the midpoint of AB, so AD = DB = 4. ADEF is a parallelogram, so ...

     EF = AD = 4

__

As we have noted, AB║EF and DE║A.F, so corresponding angles and alternate interior angles are congruent. <em>b = e = f = i</em>

5 0
3 years ago
"how many different rectangles" can you draw that have a perimeter of 12, 14
amm1812
Perimeter of rectangle = length + length + width + width

To find the combinations, think of two numbers that each multiplied by 2 and added up to give 12 or 14

Rectangle with perimeter 12

Say we take length = 2 and width = 3
Multiply the length by 2 = 2 × 2 = 4
Multiply the width by 3 = 2 × 3 = 6
Then add the answers = 4 + 6 = 10
This doesn't give us perimeter of 12 so we can't have the combination of length = 2 and width = 3

Take length = 4 and width = 2
Perimeter = 4+4+2+2 = 12
This is the first combination we can have

Take length = 5 and width = 1
Perimeter = 5+5+1+1 = 12
This is the second combination we can have

The question doesn't specify whether or not we are limited to use only integers, but if it is, we can only have two combinations of length and width that give perimeter of 12

length = 4 and width = 2
length = 5 and width = 1
--------------------------------------------------------------------------------------------------------------

Rectangle with perimeter of 14

Length = 4 and width = 3
Perimeter = 4+4+3+3 = 14

Length = 5 and width = 2
Perimeter = 5+5+2+2 = 14

Length = 6 and width = 1
Perimeter = 6+6+1+1 = 14

We can have 3 different combinations of length and width





3 0
3 years ago
If y varies directly as x and y = 5.5 when x = 11, find x when y = 3.5
PtichkaEL [24]

Answer:

x=7

Step-by-step explanation:

y=mx

5.5=m×11

5.5÷11=m

m=1/2

y=mx

3.5=1/2×x

3.5÷1/2=x

x=7

4 0
3 years ago
Read 2 more answers
There is a 70% chance that your car will get stuck in the snow during the next big snow fall. Given that you are already stuck i
nikdorinn [45]

Answer:

63%

Step-by-step explanation:

This is a problem of conditional probability.

The two events that are given are:

  • Car stuck in the snow - Let it be event S. P(S) = 70% = 0.70
  • Require a tow truck - Let it be event T.

We have to find the probability of being stuck in the snow AND requiring a tow truck which can be given as P(S and T)

We are also given the conditional probability, which is P(T | S) = 90% = 0.90

Using the given formula for our case we can modify the formula as:

P(T|S)=\frac{P(S \cap T)}{P(S)}

0.90=\frac{P(S \cap T)}{0.70}\\\\ P(S \cap T)=0.90 \times 0.70\\\\ P(S \cap T)=0.63

Therefore, there is 63% (0.63) chance that you will get stuck in the snow with your car AND require a tow truck to pull you out

4 0
3 years ago
Consider a melody to be 7 notes from a single piano octave, where 2 of the notes are white key notes and 5 are black key notes.
igor_vitrenko [27]

Answer:

35,829,630 melodies

Step-by-step explanation:

There are 12 half-steps in an octave and therefore 12^7 arrangements of 7 notes if there were no stipulations.

Using complimentary counting, subtract the inadmissible arrangements from 12^7 to get the number of admissible arrangements.

\displaystyle \_\_ \:B_1\_\_ \:B_2\_\_ \:B_3\_\_ \:B_4\_\_ \:B_5\_\_

B_1 can be any note, giving us 12 options. Whatever note we choose, B_2, B_{...} must match it, yielding 12\cdot 1\cdot 1\cdot 1\cdot 1=12. For the remaining two white key notes, W_1 and W_2, we have 11 options for each (they can be anything but the note we chose for the black keys).

There are three possible arrangements of white key groups and black key groups that are inadmissible:

WWBBBBB\\WBBBBBW\\BBBBBWW

White key notes can be different, so a distinct arrangement of them will be considered a distinct melody. With 11 notes to choose from per white key, the number of ways to inadmissibly arrange the white keys is \displaystyle\frac{11\cdot 11}{2!}.

Therefore, the number of admissible arrangements is:

\displaystyle 12^7-3\left(\frac{12\cdot 11\cdot 11}{2!}\right)=\boxed{35,829,630}

6 0
2 years ago
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