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Reptile [31]
3 years ago
14

Figure A is a scale image of Figure B. What is the value of x?

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
8 0
X=30
Put in proportions x/45 = 18/27.
Cross multiply and get 27x= 45 x 18
Or 27X = 810.
Next, divide by 27 on both sides to get X=30.
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What is the answer to this?
katrin [286]

Answer:

\huge\boxed{-28}

Step-by-step explanation:

We can substitute the given values of a, b, and c to find the value of this expression.

-4 | a + b - c|

-4 | 3 + (-2) - (-6)|

We know that <em>two negatives make a positive. </em>This means that -2 - (-6) will be the same this as -2+6

-4 | 3 + -2 + 6|

Adding a negative is the same as subtracting a positive.

\4 | 3-2+6|\\\\3-2+6 = 7

-4 \cdot 7 = -28

Hope this helped!

7 0
3 years ago
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Are the following figures similar?
Tcecarenko [31]
No the corresponding angles are not congruent, because the angle measures on the smaller figure are 90, 90, 137, and 43, while the larger figure has angle measures of 90, 90, 136, 44. that is why the following figures are not congruent.
3 0
3 years ago
Need help with this math
xeze [42]
What do you need hepl on becasue i dont know
8 0
3 years ago
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7. A bag of M&amp;Ms contains the following distribution of colors:
lys-0071 [83]

Answer:

a) 21/1240 = 0.017

b) 9/1240 = 7.26*10^-3

c) 3/2480 = 1.21*10^-3

Step-by-step explanation:

There are total 32 M&Ms in the bag

a) Probability that all three are blue M&Ms (without replacement)

         (9/32)*(8*31)*(7/30) = 21/1240 = 0.017

b)     Probability that the first one selected is blue, the second one selected is orange, and the third one selected is red

         (9/32)*(6/31)*(4/30) = 9/1240 = 7.26*10^-3

c)     Probability that the first two selected are red, and the third one selected is yellow

        (4/32)*(3/31)*(3/30) = 3/2480 = 1.21*10^-3

3 0
3 years ago
In figure AB and CD bisect each other at O. State the 3 pairs of equal parts in ∆AOC and ∆BOD. Is ∆AOC ≅ ∆BOD? Give reasons
GuDViN [60]

Answer:

SEE EXPLANATION

Step-by-step explanation:

In\:\triangle AOC \:\&\:\triangle BOD

AO \cong OB.... (given)

\angle AOC \cong\angle BOD

(vertical \: \angle s)

CO \cong OD.... (given)

\therefore \triangle AOC \:\cong\:\triangle BOD

(SAS \: postulate)

So, the 3 pairs of equal parts in ∆AOC and ∆BOD are:

AC = BD

m\angle OAC = m\angle OBD

m\angle OCA = m\angle ODB

7 0
3 years ago
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