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Alexus [3.1K]
2 years ago
9

What fraction of the rectangle is shaded? PLS PLS PLS help ASAP! Ty

Mathematics
1 answer:
erma4kov [3.2K]2 years ago
3 0

9514 1404 393

Answer:

  1/3

Step-by-step explanation:

The area of the rectangle is ...

  A = lw = (21)(12)

The area of the shaded triangle is ...

  A = 1/2bh = 1/2(14)(12) = (7)(12)

Then the fraction of the rectangle that is shaded is ...

  (shaded area)/(rectangle area) = (7)(12)/((21)(12)) = 7/21 = 1/3

1/3 of the rectangle is shaded.

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Through: (-5, -4), slope = 0
madam [21]

Answer:

y + 4 = 0(x + 5)

y + 4 = 0

y = -4

Step-by-step explanation:

6 0
3 years ago
Who do you do the problem 7/9+1/7+3/5? please explain thoroughly.
Evgen [1.6K]
To add fractions you want to make sure that the bottom numbers are the same so we have to make them the same, but we have to make sure that the fractions are unchanged so we get the correct answer

so if we find out what number that 9,7, and 5 can all multiply into. we find that 9 times 7 times 5 or 315 is the number that will work

so to get them to all be over 315, we must multiply each fraction by a form of 1/1 or x/x or (number)/(same numbe as on top)

so 7/9 times (35/35)=245/315
1/7 times (45/45)=45/315
3/5 times (56/56)=168/315

we add them togther
245/315+45/315+168/315=(245+45+168)/315=458/315
this simplifies to 1 and 143/315
4 0
3 years ago
Read 2 more answers
Is pi/pi irrational
dedylja [7]

Answer:

No, \frac{\pi}{\pi} is 1 which can be written as \frac{1}{1}. The definition of rationals is any number that can be written as a ratio of an integer to an integer where the bottom integer is not 0.

Step-by-step explanation:

\frac{\pi}{\pi}=1  since 1(\pi)=\pi.

Rational numbers included anything that can be written as a fraction where the top and bottom are integers (bottom integer is not 0).

1 can be written many different ways. It can be written as \frac{1}{1}.

Both the numerator and the denominator of the rewrite of one that I did are integers.

The integers include the following numbers:

{...,-4,-3,-2,-1,0,1,2,3,4,...}.

The integers are the whole numbers and the opposite of the whole numbers.

The whole numbers are {0,1,2,3,4,...}.

The opposite of the whole whole numbers are {...,-4,-3,-2,-1,0}.

3 0
3 years ago
QUESTION 60 What is the mean of the following set of data? Round your answer to two decimal places. {25, 32, 16, 21, 30, 18, 37}
padilas [110]

Question 60:

The answer is B. 25.57.

Mean is calculated by adding up all the values of a data set, and then dividing that sum by the amount of values.

25, 32, 16, 21, 30, 18, 37=179

179÷7=25.57

Question 61:

The answer is A. 102.

Mode is the number in a data set that appears most often. 102 shows up twice, while the other values only appear once. So 102 is the mode.

3 0
3 years ago
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Can someone help me with these 4 geometry questions? Pls it’s urgent, So ASAP!!!!
blagie [28]

<u>Question 4</u>

1) \overline{BD} bisects \angle ABC, \overline{EF} \perp \overline{AB}, and \overline{EG} \perp \overline{BC} (given)

2) \angle FBE \cong \angle GBE (an angle bisector splits an angle into two congruent parts)

3) \angle BFE and \angle BGE are right angles (perpendicular lines form right angles)

4) \triangle BFE and \triangle BGE are right triangles (a triangle with a right angle is a right triangle)

5) \overline{BE} \cong \overline{BE} (reflexive property)

6) \triangle BFE \cong \triangle BGE (HA)

<u>Question 5</u>

1) \angle AXO and \angle BYO are right angles, \angle A \cong \angle B, O is the midpoint of \overline{AB} (given)

2) \triangle AXO and \triangle BYO are right triangles (a triangle with a right angle is a right triangle)

3) \overline{AO} \cong \overline{OB} (a midpoint splits a segment into two congruent parts)

4) \triangle AXO \cong \triangle BYO (HA)

5) \overline{OX} \cong  \overline{OY} (CPCTC)

<u>Question 6</u>

1) \angle B and \angle D are right angles, \overline{AC} bisects \angle BAD (given)

2) \overline{AC} \cong \overline{AC} (reflexive property)

3) \angle BAC \cong \angle CAD (an angle bisector splits an angle into two congruent parts)

4) \triangle BAC and \triangle CAD are right triangles (a triangle with a right angle is a right triangle)

5) \triangle BAC \cong \triangle DCA (HA)

6) \angle BCA \cong \angle DCA (CPCTC)

7) \overline{CA} bisects \angle ACD (if a segment splits an angle into two congruent parts, it is an angle bisector)

<u>Question 7</u>

1) \angle B and \angle C are right angles, \angle 4 \cong \angle 1 (given)

2) \triangle BAD and \triangle CAD are right triangles (definition of a right triangle)

3) \angle 1 \cong \angle 3 (vertical angles are congruent)

4) \angle 4 \cong \angle 3 (transitive property of congruence)

5) \overline{AD} \cong \overline{AD} (reflexive property)

6) \therefore \triangle BAD \cong \triangle CAD (HA theorem)

7) \angle BDA \cong \angle CDA (CPCTC)

8) \therefore \vec{DA} bisects \angle BDC (definition of bisector of an angle)

8 0
2 years ago
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